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THE ELEMENTS 
EVCLIDIS ELEMENTVM LIBRI XV. 
幾何原本 
(Or rather too much!) 
(Clavii long, but interesting preface has not been included, but can be accessed from the resource page as e-text and image. The beginning of Euclic proper is here). 
(The prefaces and translations of Clavii introductions into Chinese have not been included, but can be accessed from the resource page as e-text and image. The Definitions start here
BOOK I. 
EVCLIDIS ELEMENTVM PRIMVM 
幾何原本第一卷之首界說三十六公論十九求作四
泰西利瑪竇口譯
吳淞徐光啟筆受 
DEFINITIONS. 
DEFINITIONES. 
界說三十六則
凡造論。先當分別解說論中所用名目。故曰界說。
凡歷法、地理、樂律、算章、技藝、工巧諸事。有度、有數者。皆依賴十府中。幾何府屬。凡論幾何。先從一點始。自點引之為線。線展為面。面積為體。是名三度。 
1. A point is that which has no part. 
I. PVNCTVM est, cuius pars nulla est.
TOTVS hic primus liber in eo positus est, vt nobis tradat ortus proprietatesque triangulorum, tum quod ad eorum angulos spectat, tum quod adlatera: quæ quidem inter se comparat interdum, interdum vero vnumquodque per se inspicit, & contemplatur. Nam aliquando ex lateribus trianguli angulos considerat, aliquando vero ex angulis latera, secundũ æqualitatem atque inæqualitatem rimatur. Idemqúe variis rationibus inquirit in duobus quandoque triangulis inter secollatis. Deinde aperit nobis parallelarum proprietates, parallelogrammorumqúe contemplatione aggreditur, tum inter se, tum etiam, vt cum triangulis inter easdem parallelas constitutis conferuntur. Vt autem hæc omnia rectius & commodius exequatur Euclides, docet diuisione anguli rectilinei, & linea rectæ in partes aquales, constitutionem lineæ perpendicularis, quo pacto angulus angulo fiat æqualis, & alia huiusmodi. Itaque vt vno verborem totam complectar, in primo libro traduntur, ex Procli sententia, rectilinearum figurarum maximè primæ, ac præcipuæ, triangula inquam, atque parallelogramma, Ante omnia vero Euclides more Mathematicorumrem propositam exorditur à principiis, initio facto à definitionibus, quarum prima punctum explicat, docensillud dici punctum in quantitate continua, quod nullas habet partes. Quæ quidem definitio planius ac facilius percipietur, si prius intelligamus, quantitatens continuam triplices habere partes, vnas secundum longitudinem, alteras secundum latitudinem, & secundum profunditatem altitudinémue alteras; quanquam non omnis quantitas omnes has partes habet, sedquædam vnicas tantum secundum longitudinem; quædam duplices, ita vt illis adijciat partes etiam latitudinis; quædam denique præter duplices has partes, tertias quoque altitudinis, siue profunditatis continet Quantitas enim omnis continua aut longa solum est, aut longa simul, & lata, aut longa, lata, atque profunda. Neque aliam dimensionem habere potest res vlla quanta, vt rectè demonstrauit Ptolemæus in libello de Analemmate, opera Federici Commandini Vrbitatis nuper in pristinam dignitatem restituto, necnon, vt ait Simplicius in libello de Dimensione, quiquidem, quod sciam, adbuc nondum est excusus. Itaque quodin quantitate continua sine magnitudine existit, intelligiturqúe sine omni parte, ita vt neque longum, neque latum, neque prosundum esse cogitetur (vt nimirum excludamus animam rationalem, Nunc vel Instans temporis, & vnitætem, quæ etiam partes non habent) id appellatur ab Euclide, & à Geometris punctum. Huius exemplum in rebus materialibus reperiri nullum potest, nisi velis extremitætem alicuius acus acuttssimæ, similitudine puncti exprimere; quod quidem omni ex parte verum non est, quoniam ea extremitas diuidi potest, & secari infinitè, punctum vero indiuiduum prorsus debet existimari. Denique in magnitudine id concipi debet esse punctum, quod in numero vnitas, quodque intempore instans. Sunt enim & hæc concipienda indiuidua. 
第一界
點者、無分。
無長短、廣狹、厚薄。如下圖。 凡圖十干為識。 干盡用十二支。 支盡用八卦八音。 
2. A line is breadthless length. 
II. LINEA vero, longitudo latitudinis expers.
DEFINIT bîc lineam, primam speciem magnitudinis, quam dicit esse quantitatem longam duntaxat, non autem latam, intellige ntque profundam. A qua enim quantitate excluditur latitudo, ab eadem etiam necessario profunditas remouetur, non autem contra. Lineam autem banc, siue longitudinem absque latitudine, non absurdè concipere, intelligereqúe poterimus ex termino loci alicuius partim illuminati, & partim obumbrati. Finis enim, seu termin us communis lucidi, & obumbrati, longitudo quædam est, ad longitudinem ipsiusnet luminis, & vmbræ extensa, carens omni latitudine, cum sit limes vtriusque. Mathematici quoquè, vt nobis inculcent veram lineæ intelligentiam, imaginantur punctum iam desecriptum superiore definitione, è loco in locum moueri. Cum enim punctum sit prorsus indiuiduum, relinquetur ex isto motu imaginario vestigium quoddam longum omnis expers latitudinis. Vt si punctum A, fluere intelligatur ex A, in B, vestigium effectũ A B, linea appellabitur, cum vero mteruallum inter duo puncta, A, & B, comprebensum sit longitudo quædam, carens omni latitudine, propterea quod punctum A, omnipriuatum dimensione, eam efficere nulla ratione potuerit. Hine factum est, vt alij dixerint, lineam nil esse aliud, quàm punctum fluxum: Alij vero, magnitudinem vno contentã interuallo. Potest enim linea vntcotantum modo, vtpote secundum longitudinem secari, atque diuidi. 
第二界
線、有長無廣。
試如一平面。光照之。有光無光之間。不容一物。是線也。眞平眞圜相遇。其相遇處止有一點。行則止有一 線。線、有直、有曲。 
3. The extremities of a line are points. 
III. LINEÆ autem termini, sunt puncta.
DOCET, quænam sint extrema lineæ cuiusuis, seu termini, dicens lineam terminari, sine claudi vtrinque punctis; Non quod omnis linea terminos habeat; quomodo enim lineæ infinitæ terminos assignare poterimus? qua etiam ratione in linea circulari extremum aliquod deprehendemus? Sed quod linea quælibet habens extrema, in suis extremitatibus puncta recipiat. Vt superior linea A B, extrema habet puncta A, & B. Idemqúe in omnibus lineis terminatis, ac finitis intelligendum est, ita vt earum extremitates sola esse puncta cogitemus. 
第三界
線之界、是點。凡線有界者。兩界必是點。 
4. A straight line is a line which lies evenly with the points on itself. 
IV. RECTA linea est, quæ ex æquo sua interiacet puncta.
TRIPLEX omnino est linea apud Mathematicos, recta, circularis, quam & curuam dicunt, & mixta, siue composita ex vtraque. Ex his describit hocloco Euclides lineam rectam, quam dicit esse eam, quæ æqualiter inter sua puncta extenditur, hoc est, in quanullum punctum intermedium ab extrem is sursum, aut deorsum, vel buc, atque illuc deflectendo subsultat; in qua denique nibil flexuosum reperitur. Hanc nobis ad viuum exprimit filum aliquod tenue summa vi extentum: In eo enim omnes partes mediæ cum extremis æqualem obtinent situm, neque vlla est alia sublimior, aut bumilior, sed omnes æquabiliter inter extremos fines positæ progrediuntur. Proclus bano definitionem exponens ait, tunc demum lineam aliquam ex æquo suæ interiacere puncta, quando æquale occupat spatium ei, quodinter suæ situm est puncta extrema. Vt linea A C B, dicetur recta, quoniam tantum occupat præcisè spatium, quanta est distantia puncti A, a puncto B. Lineæ vero A D B, A E B, A F B, non dicentur rectæ, cum maiora obtineant spatia, quàm sit distantia extremorum punctorum A, & B. Sic etiam vides omnia puncta lineæ A C B, inter quæ est punctum C, æqualiter inter extrema A, & B, iacere, iuxta Euclidis definitionem; quod non cernitur in alijs lineis, quoniam puncta D, E, F, subsultant ab extremis A, & B. Plato rectā lineam perpulcbrè sic definit: Linearecta est, cuius media obumbrant extrema Vt in linea A C B, sipunctū C, aut quoduis aliud modium, vim haberet occultandi, & A, extremum virtutem illuminandi, impedimento vtique esset C, punctum interiectum, ne B, extremum alterum ab A, illuminaretur: Rursus oculus in A, existens extremo, nõ videret aliud extremū B, ob interiectum punctũ C; quis quidem non contingit in lineis non rectis, vt perspicuum est in lineis A D B, A E B, A F B. Archimedes vero ait, lineam rectam esse minimam earum, quæ terminos habent eosdem; qualis est A C B, comparata cum A D B, A E B, A F B. Si enim A C B, nõ esset minima earũ quæ eosdem terminos A, & B, possident, nõ ex æquo interiaceret sua puncta, sed ea potius linea, quæ minor diceretur, quàm A C B. Campanus denique describens rectam lineam, vocat eam breuissimam ex vno puncto in aliud extensionem. Quemadmodum autem Mathematici per fluxum puncti imaginarium concipiunt describi lineam, ita per qualitatem fluxus puncti qualitate lineæ descriptæ intelligunt. Si namque punctum recta fiuere concipiatur per breuissimum spatium, ita vt neque in hanc partem, neque in illam deflectat, sed æquabilem quendam motũ, atque incessum teneat, dicetur linea illa descripta, Recta: Si vero punctum fluens cogitetur in motu vacillare, atque hinc inde titubare, appellabitur linea descripta, mixta: Si denique punctum fiuens in suo motu non vacillet, sed in orbem feratur vniformi quodam motu, atque distantia à certo aliquo pũcto, circa quod fertur, vocabitur descripta illa linea, circularis. Itaque si duo puncta moueantur similibus prorsus motibus, ita vt semper æqualiter inter se distent; describentur ab ipsis duæ lineæ similes, hoc est, si vna earum fuerit recta, erit & alteræ recta: si vero vna fuerit curua, erit & altera eodem omnino mode curua, &c. Lineas non rectas, quæomnes obliquædici possunt, non definit hoc loco Euclides, sed circularem exponet definitione decimæquinta, mistam prorsus omittens, quod eain hisce elementis Geometricis nullum habeat vsum. Sunt autem plurima gener a linearũ mistarum; quædam enim sunt vniformes, quædam difformes. Vniformium rursus alia suntin plano, aliæ in solido. In plano sunt Hyperbole, Parabole, Ellipsis, de quibus agit copiosissimè Apollonius in conicis elementis; linea Conchoideos, de qua Nicomedes; linea Helica, de qua Archimedes in libro de lineis spiralibus tractationem in stituit, & aliæ huiusmodi. In solido, seu superficie curua sunt alterius generis lineæ belicæ, quàmea ab Archimede descripta, qualis est illa, quæ circa cylindrum aliquem conisolmitur; nec nonea, quæ circa conum existit, vel etiam quæ circa sphæram, cuiusmodi sunt spiræilla, quæs Sol describit abortu in occasum, vt in sphæra docuimus. Difformium autem infinitus est numerus, quas non est opus hîcrecensere. Ex his constat, duas tantum esse lineas simplices, rectam, & circularem, omnes autem alias, quæcunque sunt, mist as appellari, quod ex illis componantur. Vnde ingeniosè concludit Aristoteles in libro de Cœlo, iuxtæ triplicem lineam, tres tantum esse motus, duos quidem simplices, rectum & circularem, tertium vero mistum, siue ex illis duobus compositum.
Sed quoniam lineas rectas regula ducere solemus, doceamus, quæ ratione regulam propositam examinare possimus, num linea per illam descripta recta sit, necns. Sit ergoregula A B, secundum cuius latus C D, recta C D, describatur ex puncto C, in punctum D. Deinde conuertatur regula, vt manente eadem parte superiore, punctum C, statuatur in D, & punctum D, in C: & secundum idem latus regulæ C D, recta ducatur ex eodem puncto C, in punctum D. Nam si posterior hæc linea priori omni ex parte congruet, dubitari non debet, quin regulæ A B, in lineis rectis ducendis fidere possimus: Si vero non congruet omni ex parte, latus illud C D, perfectè rectum non erit, sed corrigendum erit diligentius.
 
第四界
直線止有兩端。 兩端之間。 上下更無一點。
兩點之間。 至徑者直線也。 稍曲則繞而長矣。 直線之中。 點能遮兩界。 凡量遠近、皆用直線。甲乙丙是直線。甲丁丙、甲戊丙、甲己丙皆是曲線。 
5. A surface is that which has length and breadth only. 
V. SVPERFICIES est, quæ longitudinem, latitudinemque tantum habet.
POST lineam, quæ est prima quantitatis continuæ species, vnicamqúe habet dimensionem, definit superficiem, quæ secundam magnitudinis speciem constituit, additqúe primæ dimensioni secundum longitudinem, alteram secundum latitudinem. Nam in superficie reperitur non solum longitudo, vt in linea, verum etiam latitudo, sine tamen omni profunditate. Vt quantitas A B C D, inter lineas A B, B C, C D, D A, comprehensa, considerataque secundum longitudinem A B, vel D C, & secundum latitudinem A D, vel B C, omnis expers profunditatis, appellatur superficies. Hanc nobis refert latitudo extremæ cuiusque corporis, si ab ea omnis soliditas intellectu auferatur. Non incongruè etiam, vt ait Proclus, imaginem quasi expressam superficiei nobis exhibent vmbræ corporum. Hæ enim cum interiorem terræ partem penetrare non possint, longæ tantum erunt, & latæ. Mathematici vero, vt nobis eam ob oculos ponant, monent, vt intelligamus lineam aliquam in transuersum moueri: Vestigium enim relictum ex ipso motuerit quidem longum, propter longitudinem lineæ, latum quoque propter motum, qui in transuersum est factus; nulla vero ratione profundum esse poterit, cum linea ipsum describens omni careat profunditate; quare superficies dicetur. Vt si linea A B, fluat versus D C, efficietur superficies A B C D. Alij describentes superficiem dicunt, eam esse corporis terminum: Alij vero, magnitudinem duo bus constantem interuallis. Potest enim superficies diuidi, & secari duobus modis, vno quidem secundum longitudinem, altero vero secundum latitudinem. 
第五界
面者止有長有廣。
一體所見為面。凡體之影極似於面。 無厚之極。想一線橫行、所留之迹卽成面也。 
6. The extremities of a surface are lines. 
VI. SVPERFICIEI autem extrema sunt lineæ.
NON dissimilis est hæc definitio superiori, qua termini lineæ fuere explicati. Vult enim extremitates superficiei esse lineas, quemadmodum lineæ fines extitere puncta. Vt superioris superficiei A B C D, extrema sunt lineæ A B, B C, C D, D A. Eodemque modo in quacunque altera superficie, quæ extrema habet, lineas cogitare oportet in extremitatibus: Non autem in superficie infinita, vel etiam sphærica, quæ corpus sphæricum circumdat. Potest etiã superficies aliqua claudi, & terminari vnica tantum linea, qualis est circularis superficies, vt dicemus in definitione circuli. 
第六界
面之界是線。 
7. A plane surface is a surface which lies evenly with the straight lines on itself. 
VII. PLANA superficies est, quæ ex æquo suas interiacet lineas.
HÆC quoque definitio similitudinem quandam descriptionis lineæ rectæ gerit. Superficies enim, quæ ex æquo lineas suas interiacet, ita vt mediæ partes ab extremis sursum, deorsúmue subsultando, non recedant, appellabitur plana: qualis est superficies perpoliti alicuius marmoris, in qua partes omnes in rectum sunt collocatæ, ita vt nihil habeat incisum angulis, nibil anfractibus, nihil eminens, nihil lacunosum: In hac enim partes intermediæ cum extremis æqualem adeptæ sunt situm, nec vlla est alia sublimior, humiliórue, sed omnes æquabiliter protenduntur. Alij superficiem planam definiunt, dicentes eam esse, cuius partes mediæ obumbrant extrema: Vel esse minimam, siue breuissimam omnium, quæ eadem habent extrema: Vel cuius omnibus partibus recta linea accommodari potest, vt placet Heroni antiquo Geometræ. Vt superficies A B C D, tum demum plana dici debet, quando linea recta A E, circa punctũ A, immobile circumducta, ita vt nunc eadem sit, quæ A B, nune eadem, quæ A F, nunc eadem, quæ A G, & nune eadem, quæ A H, nihil in superficie offendit depressum, aut sublatum, sed omnia puncta superficiei à linearecta tanguntur, & quodammodo raduntur. Quod si minima superficiei particula alus humilior à linea rectanõ tangeretur, vel ipsa linea recta liberè non posset circumduci, propter aliquem tumorem, seu eminentiam in superficie occurrentem, tam non posset nuncupari plana. Itaque vt sit plana, requiritur vt omnibus modis possit recta linea commensurari, hoc est, vt ei applicari possit recta linea secundum A B, & A F, & denique secundum omnes partes. Hæc autem superficies sola erit ea, quam imaginari, & intelligere possumus describi ex motu lineæ rectæ in transuersum, qui super duas alias lineas rectas conficitur. Vt si linea recta A B, per duas rectas A D, B C, feratur, efficietur superficies perfectè plana, iuxta omnes definitiones. Non enim difficile erit huic superficiei traditas descriptiones accommodare. Solent Mathematici superficiem planam frequenter appellare planum, ita vt quando loquuntur de plano, intelligenda semper sit superficies plana. Cæteræomnes superficies, quibus non omni ex parte accommodari potest linea recta, qualis est superficies interior alicuius fornicis, vel exterior alicuius globi, columnæve rotundæ, vel etiam coni, &c. appellantur curuæ, & non planæ. Quamuis enim superficiei columnæ rotundæ se@@ cylindri secundũ longitudinem adaptari possit linearecta, tamen secũdum latitudinem minimè potest. Idemqúe dicendũ est de aliis. Superficies autem curua duplex est, conuexa videlicet, vt exterior superficies sphæræ, vel eylindri; & concaua, vt interior fornicis, siue aicus alicuius. Quoniam vero omnium harum contemplatio pertinet ad Stereometriam, idcirco Euclides hoc primo libro solum planam nobis explicauit, de qua est disputaturus prioribus sex libris. 
第七界
平面一面平在界之內。
平面中間線能遮兩界。平面者。 諸方皆作直線。試如一方面。 用一直繩施於一角。 繞面運轉。 不礙不空。 是平面也。若曲面者。 則中間線不遮兩界。 
8. A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line. 
VIII. PLANVS vero angulus, est duarum linearum in plano se mutuo tangentium, & non in directum iacentium, alterius ad alteram inclinatio.
DECLARAT, quidnam sit angulus planus, dicens; Quandocunque duæ lineæ in plana aliqua superficie inuicem concurrunt, & non in directum constituuntur, efficietur ex huiusmodi concursu, seu inclinatione vnius ad alteram, angulus, qui diciturplanus, propterea quodin plana constituatur superficie. Verbi gratia, quia duælineæ A B, A C, concurrunt in A, & non iacent iu directum, ideo efficiunt angulum A, planum in eadem existent\-e superficie, in qua duæ illæ lineæ constituuntur. Dicentur autem duæ lineæ non in directum iacere, quando altera earum versus concursum protensa non coincidit cum altera, sed veleam secat, vel certè statim post punctum concursus ab earecedit. Quod dixerim propter angulum contactus, qui fit, quando duo circuli secontingunt, veletiam, quando linea recta circulum tangit. Protracta enim recta linea post punctum contactus, quanquam non secet circulum, tamen statim post illud ab eo seiungitur. Eodem pacto circularis illa linea secundum propriam dispositionem, ac formam extensareoedit à recta tangente, quamuis eam non secet. Vnde verè est angulus constitutus in illo contactu: qua de re plura scribemus in proposiiione 16. tertij liber contra Iacobum Peletarium, qui contendit, eam non esse angulum. Quod si duæ lineæ se mutuo tangant iacentes in directum; ita vt alterutra producta congruat toti alteri, non fiet vllus angulus ex illo concursu, cum nulla sit inclinatio, sed ambæ vnam integram lineam constituent. Vt quia recta A B, producta conuenit cum recta B C, non efficietur angulus in B. Sic etiam non fiet angulus in B, ex lineis curuis A B, B C, quia alterutra secundum suam inflexionem, & obliquum ductum extensa, cum altera coincidit. Quare in directũ dicentur iacere. It aque vt lineæ rectæ efficiant angulum, necesse est, vt post concursum productæ se mutuo secent: Curuæ autem lineæ, vel quarum altera curua, altera vero recta existit, angulum constituere verè possunt, etiamsinon se mutuo intersecent; sufficit enim, quod sese contingant, ita vt statim post contactum altera ab alter a separetur, quemadmodum & ante eundem semotæ cernuntur. Consistit autem anguli cuiusuis quantitas in sola inclinatione, non in longitudine linearum; lineæ etenim longius excurrentes non augent suam inclinationem, igitur neque anguli magnitudinem. Sunt & aliaduo genera angulorum, quorum prius solidos comprehendit, de quibus Euclides disserit in Stereometria, quique in corporibus existunt; Posterius vero sphærales, quiin superficie sphæræ constituuntur ex circulorũ maximorũ circumferentiis, & de quibus copiosè agitur in sphæricis elementis Menelai. Horum autem omniũ explicatio in alium locum à nobis reijcitur, cum hîc de solis planis angulis sit futurus sermo. 
第八界
平角者。 兩直線於平面縱橫相遇交接處。
凡言甲乙丙角。 皆指平角。
如上甲乙丙二線。平行相遇。 不能作角。
如上甲乙,乙丙二線。 雖相遇。 不作平角。 為是曲線。所謂角。 止是兩線相遇。 不以線之大小較論。] 
9. And when the lines containing the angle are straight, the angle is called rectilineal. 
IX. CVM autem, quæ angulum continent lineæ, rectæ fuerint, rectilineus ille angulus appellatur.
ANGVLVS omnis planus conficitur aut ex lineis duabus rectis, qui quidem rectilineus dicitur, & de quo solum hîc agit Euclides: aut ex duabus curuis, quem curuilineum vocare licet; aut ex vna curua & altera recta, qui non ineptè mixtus appellatur. Ex hisce porro lineis possunt curuilinei anguli tribus variari modis, & mixti duobus, pro varia inclinatio ne, seu habitudine linearum curuarum, vtpote secundum conuexum, & concauum, ceu in propositis angulis planè, & apertè perspicitur. Rectilineus vero variari non potest ratione inclinationis, habitudinis ve linearum, nisi matorem, vel minorem inclinationem variam velimus dicere habitudinem, quod est absurdum; cum hoc modo augeatur tantum angulus rectilineus, aut diminuatur, quod & aliis εommune est, non autem ita varietur, vt aliud constituat genus. 
第九界
直線相遇作角。為直線角。
平地兩直線相遇。為直線角。本書中所論止是直線角。但作角有三等。今附蓍於此。
一直線角。二曲線角。三雜線角。 如下六圖。(p. 五) 
10. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. 
X. CVM vero recta linea super rectam consistens lineam eos, qui sunt deinceps, angulos æquales intersecerit, rectus est vterque æqualium angulorum: Et quæ insistit recta linea, perpendicularis vocatur eius, cui insistit.
VSVS frequentissimus reperitur in Geometria angulirecti, & lineæ perpendicularis, nec non anguli obtusi, & acuti, propterea docet hoc loco Euclides, quisnam angulus rectilineus apud Geometras appelletur rectus, & quænam linea perpendicularis: In sequentibus autem duabus definitionibus explicabit angulum obtusum, & acutum. Non enim alius dari potest angnlus rectilineus, præter rectum, obtusum, & acutum, Igitur si recta linea A B, rectæ C D, insistens essiciat duos angulos prope punctum B, (qui quidem ideo dicuntur à Maìbematicis esse deinceps, quod eos eadem linea C D, protracta, propo idem punctum B, efficiat) inter se æquales, quod tum demum fiet, quandorecta A B, non magis in C, quam in D, inclinabit, sed æquabiliter rectæ C D, insistet, vocabitur vterque angulus B, rectus, & recta A B, perpendicularis recta C D, cui insistit, Eadem ratione nominabitur recta C B, perpendicularis recta A B: quamuis enim C B, tantum faciat cum A B, vnum angulum, tamen si A B, extenderetur inrectum & continuum versus punctum B, efficeretur alter angulus æqualis priori. Qua vero arte linea duci debeat efficiens cum alter a duos angulos æquales, decebit Euclides propositione 11. & 12. buius primi libri. Itaque vt in Geometria concludamus angulum aliquem esse rectum, aut lineam, quæipsum essicit, ad aliam esse perpendicularem, requiritur, & sufficit, vt probemus angulum, qui est ei deinceps, æqualem illi esse. Pariratione, si dicatur aliquis angulus rectus, aut linea, quæ ipsum constituit, perpendicularis ad aliam, colligere licebit, angulum illi deinceps æqualem quoque esse. Quando enim anguli, qui sunt deinceps, fuerint inter se æquales, nuncupatur vterque illorum rectus, & linea ipsos efficiens, perpendicularis, iuxta banc 10. definitionem: quando autem non fuerint æquales, non dicitur quisquam illorum rectus, vt constabit ex sequentibus duabus definitionibus, & propterea neque linea eos constituens perpendicularis appellatur. Hæc dixerim, vt videas, quidnam liceat ex hac definitione colligere in rebus Geometricis, & quemnam vsum babeant apud Geometr as descriptiones vocabulorum. Non enim magno laborebæc quæ diximus, ad alias definitiones poterunt transferri. 
第十界
直線垂於橫直線之上。若兩角等。必兩成直角。而直線下垂者。謂之橫線之垂線。
量法。常用兩直角。及垂線。垂線加於橫線之上。必不作說角及鈍角。
若甲乙線至丙丁上。則乙之左右作兩角相等。為直角。而甲乙為垂線。
若甲乙為橫線。則丙丁又為甲乙之垂線。何者。丙乙與甲乙相遇。雖止一直角。然甲線若垂下過乙。則丙線上下定成兩直角。所以丙乙亦為甲乙之垂線。如今用矩尺。一縱一橫。互相為直線。互相為垂線。凡直線上。有兩角相連是相等者。定俱直角。中間線為垂線。
反用之。若是直角。則兩線定俱是垂線。 
11. An obtuse angle is an angle greater than a right angle. 
XI. OBTVSVS angulus est, qui recto maior est.
QVANDO recta A B, rectæ C D, insistens non fecerit angulos ad punctum B, æquales, & obeam causam neutrum rectum, sed vnum quidem recto maiorem, alterum vero minorem, dicitur maior angulus obtusus, qualis est angulus B, ad punctum C, vergens, quicontinetur rectis lineis A B, B C. 
第十一界
凡角大于直角。為鈍角。
如甲乙丙角與甲乙丁角不等。而甲乙丙大於甲乙丁。則甲乙丙為鈍角。(p. 六) 
12. An acute angle is an angle less than a right angle. 
XII. ACVTVS vero, qui minor est recto.
VT in præcedenti figura, minor angulus B, ad punctum D, vergens, qui continetur rectis lineis A B, B D, vocatur acutus, Itaque angulus rectus, vt ex dictis colligitur, nullam patitur varietatem, vt vnus altero maior, minórue detur, cum lineaperpendicularis eum efficiens non debeat mag is in vnam partem inclinare, quàm in alteram: Obtusus vero, & Acutus augeri possunt, & minui infinitis modis, cum ab illa inflexibilitate lineæ perpendicularis infinitis etiam modis recta line a possit recedere, vt per spicuum est. Quoniam vero ad quemuis angulum planum constituendum concurrunt duæ lineæ, & aliguando in vno puncto plures existunt anguli, solent Mathematici, vt tollatur confusio, angulum quemlibet exprimere tribus literis, quarum media ostendit punctum, in quo lineæ conficiunt angulum, extremæ vero significant initia linearum, quæ angulum continent. Exempli gratia, in superiore figura angulum obtusum intelligunt per angulum A B C, acutum vero, per angulum A B D, quod diligenter est notandum, vt facile dignoscamus angulos, quorum mentio fit in demonstr ationibus.
IAM vero proposito nobis angulo aliquorectilineo, si experivi velimus, num rectus sit, an obtusus, acutúsue, efficiemus id boc modo. Contineant duæ rectæ A B, A C, angulum A. Ductarecta B C, vtcunque, quæ angulum subtendat, & diuisa bifariam in D, describatur ex D, vt centro ad interuallum D A, circumferentia circuli; quæ siper puncta B, C, transeat, 1 erit angulus A, rectus, vtpote qui in semicir culo B A C, existat: si vero idem semicirculus rectam B C, secet in E, F, erit angulus B A C, obtusus; propterea quod, du ctis rectis E A, F A, 2 angulus E A F, in semicirculo E A F, rectus est, qui quidem pars est anguli B A C. Si denique idem semicirculus rectam B C, productam secet in E, F, erit angulus B A C, acutus; propterea quod ductis rectis E A, F A, 3 angulus E A F, in semictrculo E A F, rectus est, qui quidem maior est angulo B A C.
ALITER idem assequemur boc modo. Describatur ex puncto D, quodrectam B C, dato angulo A, subtensam secat bifariam, semicirculus ad interuallum D B, vel D C: qui si transeat per punctum A, 4 datus angulus erit rectus, vtpote qui in semicirculo existat. Si vero idem semicirculus transeat supra punctum A, datus angulus erit obtusus, Ducta enim recta D A, secante circumferentiam in E, iungantur rectæ E B, E C; 5 eritqúe angulus B E C, in semicirculo rectus. Cum ergo angulus B A C, 6 datus maior sit angulo B E C, erit angulus datus A, recto maior, boc est, obtusus, Si denique semicirculus idem secet rect as A B, A C, erit datus angulus acutus, Sumpto namque puncto E, inter rect as A B, A C, in circumferentia, iungantur rectæ E B, E C; 7 erit que angulus B E C, rectus in semicirculo: 8 quicum maior sit angulo dato A, erit datus angulus A, recto minor, id est, acutus. Non videatur autem mirum cuipiam, quod ad demonstrationem assumamus propositiones, quæ posterius demonstrantur ab Euclide; quod alienum esse videtur à puritate demonstrationum Geometricarum: Non videatur, inquam, mirum, quia cum id, quod boc loco ostendimus, necessarium non sit ad sequentes demonstrationes, poterit commodè differri, donec propositiones requisitæ sint demonstratæ. Satis est, vt praxis huiusce rei boc loco intelligatur. Idem obseruabimus in nonnullis praxibus problematum. Eas enim propriis in locis, quoad eius sieri poterit, proponemus, vt diuisionem angulirectilinei in quotuis partes æquales eo in loco docebimus, vbi Euclides docet diuisionem eiusdem anguli in duas æquales partes, &c. quanquam ad earum praxium demonstrationes necessariæ sint propositiones posterius demonstratæ. FACILIVS idem cognoscemus beneficio nermæ alicuius accuratè fabricatæ, qualem refert instrumentum A B C, constans duabus regulis A E, A F, ad angulum rectum in A, coniunctis. Nam si latus A B, buius normæ, rectæ A B, applicetur, cadente puncto A, in punctum A; si quidem & normæ latus A C, rectæ A C, congruat, erit angulus A, rectus: si vero citra rectam A C, cadat normæ latus A C, erit angulus A, obtusus; si denique latus normæ A C, vltra rectam A C, cadat, acutus erit angulus, vt perspicuum est.
ITA autem norm am examinabimus, num accuratè sit fabricata, nec ne. Descripto semicirculo B A C, ex centro G, cuiusuis magnitudinis, ductaqúe diametro B C, ponatur angulus A, in aliquo puncto circumferentiæ, vt in A, latusqúe vnum normæ, vt A B, per B, punctum extremum diametri transeat. Nam si alterum tunc latus A C, per alterum punctum extremum C, transeat, ritè fabricata erit norma A B C; 9 quod tunc angulus B A C, in semicirculo B A C, rectus 31. ter tij. sit: si vero latus A C, non per C, transeat, emendanda erit norma; quia eius angulus A, tunc rectus non erit. Eadem ratione interiorem partem normæ examinabimus, si angulum D, circumferentiæ applicemus, & latera D E, D F, punctis extremis B, C, &c.
 
第十二界
凡角小於直角。為銳角。
如前圖甲乙丁是。
通上三界論之。直角一而已。鈍角銳角。其大小不等。乃至無數。
是後凡指言角者。俱用三字為識。其第二字。卽所指角也。 如前圖甲乙丙三字。第二乙字。卽所指鈍角。若言甲乙丁。卽第二乙字。是所指銳角。 
13. A boundary is that which is an extremity of anything. 
XIII. TERMINVS est, quod alicuius extremum est.
TRES sunt termini iuxta banc definitionem. Punctum enim cerminus est, seu extremum lineæ: Line a superficiei: & superficies corporis. Corpus autem terminare amplius nibil potest, quod non reperiatur alia quantit as plures babens dimensiones, quam tres. Omne siquidem terminatum super at terminum suum vna dimensione, vt perspicuum est ex adductis exemplis. 
第十三界
界者。一物之始終。
今所論有三界。點為線之界。線為面之界。面為體之界。體不可為界。 
14. A figure is that which is contained by any boundary or boundaries. 
XIV. FIGVRA est, quæ sub aliquo, vel aliquibus terminis comprehenditur.
NON omnis quantit as terminos possidens Figura dici potest, ne lineam finitam figuram appellare cogamur: Sedeæ solum magnitudines, quæ latitudinem babent, nempe superficies terminatæ, & quæ profunditdtem adeptæ quoque sunt, vt solida finita Figuræ nomine appellabuntur. Hæ enim proprie terminis comprebendi dicuntur. Nam linea finita non proprie dicitur punctis extremis comprebendi, cum puncta lineam non ambiant, sed potius punctis terminari dicitur. Itaque termini debent quantitatem, quæ figura dicitur, ambire, & non tantum terminare. Superficies quoque infinita, vel etiam corpus, cum nullis terminis comprebendatur, Figura vocari nulla ratione potest, Figuræ vnico comprebensæ termino sunt, Circulus, Ellipsis, sphæra, sphæroides, & aliæ huiusmodi: Pluribus vero terminis inclusæ figuræ sunt, Triangulum, Quadratum, Cubus, Pyramis, &c. Superficies terminatæ nuncupantur figuræ planæ: solida autem circumscripta, figuræ solidæ, siue corporeæ. Porro quia formas, seu typos variarum figurarum in spicies quamplurimas in sequentibus, planarum quidem in prioribus 10. libris, solidarum vero in posterioribus quinque, propterea nulla hoc loco figura depingenda esse videtur. 
第十四界
或在一界、或在多界之間。為形。
一界之形。如平圜、立圜等物。多界之形。如平方、立方、及平立、三角、六、八角等物。 圖見後卷。 
15. A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another; 
XV. CIRCVLVS, est figura plana sub vna linea comprehensa, quæ peripheria appellatur, ad quam ab vno puncto eorum, quæ intra figuram sunt posita, cadentes omnes rectæ lineæ inter se sunt æquales.
DEFINIT hic circulum, figuram inter planas perfectissimam, docens figur am illam planam, quæ vnica linea circumscribitur; ad quam lineam omnes rectæ lineæ ductæ ab vno puncto, quod intr a figuram existit, sint æquales, vocari circulum. Vt si superficies, seu spatium concludatur vnica linca A B C, babueritqúe hanc conditionem, vt ab aliquo puncto intus suscepto, vtpote à D, omnes rectæ lineæ cadentes ad terminum A B C, quales sunt D A, D B, D C, inter se sint æquales, appellabitur talis figura plana, circulus, alias non, Qua vero ratione in circulo punctum illud medium reperiri debeat, docebit Euclides propositione 1. tertij liber Adiungit quoque Euclides, lineam extrem am circuli, qualis est A B C, appellari Peripberiam, seu, vt Latini exponunt, circumferentiam. Potest circulus etiam bac ratione describi. Circulus est figura plana, quæ describitur à linea rect a finita circa alterum punctum extremum quiescens circumducta, cum in eundem rursus locum restituta fueru, vnde moueri cœperat. Quæ quidem descriptio persimilis est ei, qua ab Euclide spbæra describitur liber 11. Vt si intelligatur recta A D, circa punctum D, quiescens moueri, donec ad eundem redeat locum, à quo dimoueri cœpit, describet ipsarect a totum spatium circulare; punctum vero alterum extremum A, delineabit peripberiam A B C: Erit quoque punctum quiescens D, illud, à quo omnes linea cadentes in peripberiam sunt inter se æquales, propterea quod recta A D, circumducta, omnes lineas, quæ ex D, possunt educi ad peripheriam, æquè metiatur. Igitur Ellipsis, quamuis figura sit planæ vna linea circumscripta, tamen quia in ea non datur punctum, à quo ad ipsam lineam terminantem omnes rectæ lineæ sint æquales. circulus. dici nequit. 
第十五界
圜者一形於平地居一界之間。自界至中心作直線俱等若甲乙丙為圜。丁為中心。則自甲至丁、與乙至丁、丙至丁其線俱等。(p. 七)外圜線為圜之界。內形為圜。
一說。圜是一形。乃一線屈轉一周。復於元處所作。如上圖甲丁線轉至乙丁。乙丁轉至丙丁。丙丁又至甲丁。復元處其中形卽成圜。 
16. And the point is called the centre of the circle. 
XVI. HOC vero punctum, centrum circuli appellatur.
HOC ET, punctum illud intra circulum, à quo omnes lineæ rectæ ad circumferentiam ducta sunt æquales, appellari centrum circuli; quale est præcedentis figuræ punctum D. Vnde perspicuum est, polum alicuius circuli in sphæra, à quo omnes rectæ ad peripberiam circuli cadentes sunt æquales, vt ait Theodosius in sphæricis elementis, nom dici debere centrum circuli, cum punctum illud, quod polus dicitur, existat in superficie spbæræ, non autem in superficie circuli; quæ tamen est necessario requisita conditio, vt punctum aliquod centrum vocetur. Cæterum, vtpunctum aliquod circuli dicatur centrum, satis est, vt œb eo tres duntaxat lineæ cadentes in peripheriam sint æquales inter se, vt demonstrat Euclides propositione 9, lib. 3. Hac enim ratione fiet, vt omnes aliæ ab eodem puncto emissæ inter se sint æquales. 
第十六界
圜之中處。為圜心。。 
17. A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle. 
XVII. DIAMETER autem circuli, est recta quædam linea per centrum ducta, & ex vtraque parte in circuli peripheriam terminata, quæ circulum bifariam secat.
SI in circulo ducatur recta linea A B, per centrum C, it a vt extrema eius A, & B, terminentur in peripheria, appellabitur ea circuli diameter. Non igitur omnis in circulorecta line a ducta diameter dicetur, sed ea solummodo, quæ per centrum vsque ad peripheriam vtrinque extenditur. Vnde plures assignari poterunt in circulo diametri, vnum vero centrum duntaxat. Quod autem Euclides addit, circulum bifariam secari a diametro, perspicuum ex eo esse potest, quod diameter per medium circulum, vtpote per centrum, ducitur. Hinc enim fit, vt propter directum diametriper centrum transitum, vtrinque æquales circumferentiæ abscindantur. Quod tamen Thaletem Milesium hac ratione demonstrasse testatur Proclus. Concipiamus animo, portionem A D B, accommodari, & coaptari portioni reliquæ A E B, itæ vt diameter A B, communis sit vtrique portioni: Si igitur circumferentia A D B, congruat penitus circumferentiæ A E B, manifestum est, duas illas portiones a diametro factas, esse inter se æquales, quandoquidem neutra alteram excedit: Si vero circumferentia A D B, non omni ex parte cadere dicatur super circumferentiam A E B, sed vel extra eam, velintra, vel partim extra, partim intra; tunc ductarecta à centro C, secante circumferentiam A D B, in D, & circumferentiam A E B, in E, erunt duærectæ C D, C E, ductæ ex centro ad circumferentiam eiusdem circuli æquales, per circuli definitionem, cum tamen vna sit pars alterius, quod est ab surdum. Non ergo cadet vna cir cumferentia extra aliam, vel intra, vel partim extra, partim intra, sed ambæ inter se aptabuntur, ideoqúe æquales erunt. quod demonstrandum proponebatur.
EX hac demonstratione constat, diametrum non solum circumferentiam, verum etiam totam aream circuli seoare bifariam. Cum enim semicir cumferentiæ sibi mutuo congruant, vt ostensum est, congruent etiam superficies ipsæ inter diametrum, & vtramque circumferentiam comprehensæ, cum neutra alteram excedat. Quare æquales inter se erunt.
 
第十七界
自圜之一界作一直線。過中心至他界。為圜徑。徑分圜兩平分。
甲丁乙戊圜。自甲至乙、過丙心、作一直線。為圜徑。 
18. A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle. 
XVIII. SEMICIRCVLVS verò est figura, quæ continetur sub diametro, & sub ea linea, quæ de circuli peripheria aufertur.
EXEMPLI gratia, in superiori circulo figura A D B, contentæ sub diametro A B, & peripheria A D B, dicitur semicirculus, quia, vt in præcedenti definitione ostendimus, ea est dimidiata pars circuli. Eadem ratione erit figura A E B, semicirculus. Idem autem punctum C diametrum secans bifariam, centrum est in circulo, & in semicirculo.
QVOD sirecta linea B D, nontranseat per centrum E, secabitur circulus ab ea non bifariam, sed in duas portiones inæquales B A D, B C D, quarum ea, in qua centrum circuli existit, cuiusmodi est portio B A D, maior est, quàm alia B C D, extra quàm centrum E, reperitur, Esse autem portiones B A D, B C D, inæquales, it a probari potest. Concipiatur per centrum E, ducta diameter ad rectam B D, perpendicularis A G. Si igitur dictæ portiones dicantur esse æquales, & portio B C D, intelligatur moueri circarectam B D, vt super portionem B A D, cadat, congruet illa portio huic, & recta F C, rectæ F A, congruet, ob angulos rectos ad F, qui omnes inter se æquales sunt ex defin. 10. cum sint sibi mutuo deinceps. Recta ergo F C, quæ nune eademest, quæ F A, maior erit, quàm E A, pars ipsius F A. Cumergoipsi E A, sit æqualis E C, quod ambæ ducantur è centroad circumferentiam, erit quoque F C, maior quàm E C, pars quàm totum, quod est absurdum. Non igitur portio B C D, portioni B A D, congruet, sed intra eam cadet, cuiusmodi est portio B G D, vt recta F G, eadem tunc existens, quæ F C, minor possit esse quàm E A, vel E C. Sinamque diceretur cadere extra, vt si circulus esset B C D G, cuius centrum E, & portio B C D, caderet extra B G D, qualis est portio B A D, esset rursus F A, eadem tunc existens, quæ F C, maior quàm E G, hoc est, quàm E C, atque ita pars F C, maior rursum foret toto E C. quod absurdum est. Ex quo patet, portionem B A D, in qua cextrum E, existit, maiorem esse reliqua portione B C D, cum hæc æqualis sit portioni B G D, quæ pars est portionis B A D. Cum enim osten sum sit, portionem B C D, circa rectam B D, circumductam non posse congruere portiont B A D, nequé cadere extra, cadet omnino intra, qualis est B G D.
 
第十八界
徑線與半圜之界所作形。為半圜。 
19. Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines. 
XIX. RECTILINEÆ figuræ sunt, quæ sub rectis lineis continentur.
POST definitionem circuli, traditurus iam Euclides descriptiones variarum figurarum, explicat prius, quænam figuræ dicantur rectilineæ. De his enim potissimum sermo futurus est in hisce libris. Omnes igitur figuræ planæ, quæ vndique rectis clauduntur lineis, rectilineæ nuncupantur. Ex quo perspicuum est, figur as plan as curuis lineis comprehensas, dici curuilineas. Eas vero, quæ partim curuis, partim rectis circumscribuntur, appellari mixtas. V aria autem nun@ genera figurarum rectilinearum ab Euclide describentur.
XX. TRILATERÆ quidem, quæ sub tribus.
AFFIRMANS Euclides, eas rectilineas figuras dici trilateras, quætribus rectis lineis circumscribuntur, apertè nobis innuit, quonam modo Triangulum definiri debeat. Cum enim in rectilineis figuris tot sint anguli, quot latera, seurectæ lineæ, ex quibus constant, dicetur triangulum, figura tribus rectis lineis contenta, cuius omnes species iam iam adducentur.
XXI. QVADRILATERÆ verò, quæ sub quatuor.
EADEM ratione erit Quadrangulum, figur a quatuer rectis lineis contenta, cuius variæ species mox subsequentur.
XXII. MVLTILATERÆ autem, quæ sub pluribus, quàm quatuor, rectis lineis comprehenduntur.
QVONIAM species rectilinearum figur arum sunt innumer abiles, propter infinitum numerorum progressum: Nam tres rectæ lineæ claudentes figuram efficiunt primam speciem, sub qua omnia triangula continentur; quatuor constituunt secundam, quæ omnia quadrangula complectitur; quinque tertiam componunt speciem; sex quartam, atque it a deinceps infinitè: Ideo Euclides, ne infinitatem hanc figur arum cogatur persequi, vocat omnes alias figur as rectilineas, quæ pluribus, quàm quatuor, rectis lineis circumscribuntur, generali vocabulo Multilater as; contentus denominatione trilater arum figurarum & quadrilaterarum, fortassis eam ob causam, quod præcipuè in prioribus his libris de Triangulis, atque Quadrangulis sermo habeatur, & quod facilè ad similitudinem harum duarum specierum cæter a omnes à quolibet definiri possint. Quis enim ex dictis non colligat, figuram quinque lineis rectis contentam appellari quinquilateram, & sex lineis comprehensam sexilateram, atque reliquas eodem modo? Sicetiam dici poterunt huiusmodi figur æ quinquangulæ, sexangulæ, septangulæ, &c. 
第十九界
在直線界中之形。為直線形。
第二十界
在三直線界中之形。為三邊形。
第二十一界(p. 八)
在四直線界中之形為四邊形。
第二十二界
在多直線界中之形為多邊形。五邊以上俱是。 
20. Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal. 
XXIII. TRILATERARVM autem figurarum, Æquilaterum est triangulum, quod tria latera habet æqualia.
DESCENDIT iam ad singulas species triangulorum. Quia verò triangula diuidi possunt vel habita ratione laterum, vel angulorum, declar at prius species prioris diuisionis, quæ tres sunt duntaxat, quod tria latera tribus tantum modis sese possint habere. Aut enim omnia æqualia sunt; aut duo tantum, tertio existente vel matore, vel minore; aut omnia inæqualia. Quando igitur omnia tria latera inter se æqualia sunt, dicitur triangulum Æquilaterum. Porro ex æqualitate omnium trium laterum trianguli æquila teri infertur, omnes tres eius angulos æquales quoque esse, ceu ad quiætam propositionem huius libri demonstrabimus.
XXIV. ISOSCELES autem est, quod duo tantum æqualia habet latera.
EX hac rursum æqualitate duorum laterum trianguli Isoscelis efficitur, duos angulos super reliquum latus etiam esse æqualles, vt demonstrabit Euclides propos 5 huius Iibri. Apposuimus autem duo triangula psoscelia, quorum prius habet tertium latus vtrouis æqualium maius, losterius autem idem minus obtinet: ita vt duæ sint species triangui Isoscelis; alterum, cuius tertium latus sit vtrouis æqualium maius, & alterum, cuius tertium latus vtrouis æqualium minus sit.
XXV. SCALENVM vero est, quod tria inæqualia habet latera.
HIC denique exinæqualitate omnium laterum trianguli Scaleni colligitur omnium angulorum inæqualitas, vt ostendetur propositio 18. huius 1. liber Porro ex his constat, eodem modo potuisse diuidi triangulum in tres speties, si æqualitatis angulorum ratio haberetur Cum enim aut omnes tres anguli sint inter se æquales; aut duo tantum, tertio maiore, vel minore existente; aut omnes tres inæquales; erit omne triangulum vel æquiangulum, habens tres omnes angulos æquales: vel duorum tantum angulorum æqualium: vel omnium angulorum inæqualium; quorum primum quidem Æ quilatero, secundum vero Isosceli, tertium denique Scalenorespondet triangulo. Cæterum quanam arte construenda sint triangula huius partitionis super quauis data recta linea finita, trademus propositio 1. huius liber. 
第二十三界
三邊形三邊線等。為平邊三角形。
第二十四界
三邊形。有兩邊線等。為兩邊等三角形。或銳或鈍。(p. 九)
第二十五界
三邊形。三邊線俱不等。為三不等三角形。 
21. Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acuteangled triangle that which has its three angles acute. 
XXVI. AD hæc etiam, trilaterarum figurarũ, Rectangulum quidem triangulum est, quod rectum angulum habet.
NVNC exponit triangulorum species iuxta posteriorem diuisienem, habitaratione vartetatis angulorum. Quia vero tria tantummodo sunt angulorum rectilineorum genera diuersa; (Omnis enim angulus rectilineus vel est rectus, vel obtusus, vel acutus, vt supra diximus,) fit vt tres quoque species triangulorum sub hac consideratione reperiantur. Nam aut vnus angulus trianguli est rectus, & ob eam rem reliqui acuti, vt ex 17 propositio 1. liber constabit; aut obtusus, & ob eandem causam reliqui acuti, aut denique nullus rectus, nullusqúe obtusus, sed omnes acuti. Quando igitur triangulum aliquod habet angulum vnum rectum, vocatur ab Euclide, & aliis Geometris Rectangulum. Potest autem triangulum huiusmodi esse Isosceles, vel scalenum, vt hæ figuræ indicant, æquilaterum autem nulla ratione. Propter æqualitatem enim laterum essent perea, quæ propositio 5. dicemus, omnes etiam anguli æquales, ideoque, cum vnus concedatur rectus, omnes tres recti, quod pugnat cum propositio 17. & 32. huius libri.
XXVII. AMBLYGONIVM autem, quod obtusum angulum habet.
TRIANGVLVM Amblygonium, siue obtusangulum esse quoque potest vel Isosceles, vel scalenum, vt in his figuris cernitur, non autem æquilaterum, alias eadem ratione essent omnes tres anguli per ea, quæ propositio 5. ostendemus, æquales, ideoque cum vnus ponatur obtusus, omnes tres obtusi, quod mulio magis pugnat cum propositio 17. & 32. huius libri.
XXVIII. OXYGONIVM vero, quod tres habet acutos angulos.
OMNE triangulum Oxygonium, siue acutangulum, potest esse vel æquilaterum, vel Isosceles, vel scalenum, vt cernere licet in triangulis quæ in speciebus prioris diuisionis spectanda exhibuimus, nc eadem hic frustra repetantur. Ex dictis igitur palam fit, triangulum quodcunque æquilaterum, esse necessarie Oxygonium: At omne triangulum tam Isosceles, quàm Scalenum, esse vel Rectangulum, vel Amblygonium, vel Oxygonium; atque Isosceles Oxygonium rursum duplex, Isosceles nimirum Oxygontum habens tertium latus vtrouis æqualium maius, atque Isosceles Oxygonium habens tertium latus vtrouis æqualium minus: Vt vnica sit species trianguli æquilateri, quatuor vero Isoscelis, & tres Scaleni: atque in vniuersum octo triangulorum genera; æquilaterum, quod perpetuo Oxygonium esse diximus, Isosceles rectangulum, Isosceles Amblygonium, Isosceles Oxygonium habens tertium latus vtrouis æqualium maius, Isosceles Oxygonium habens tertium latus vtrouis æqualium minus, Scalenumrectangulum, Scalenum Amblygonium, & Scalenum Oxygonium Quæ etiam hisce licebit nominibus immutatis appellare, Rectangulum Isosceles, Rectangulum Scalenum, Amblygonium Isosceles, Amblygonium Scalenum, Oxygonium æquilaterum, Oxygonium Isosceles habens tertium latus vtrouis æqualium maius, Oxygonium Isosceles habens tertium latus vtrouis æqualium minus, & Oxygonium Scalenum. Quare perspicuum est, quamnam connexionem, siue affinitasem habeant inter setriangula vtriusque partitionis. Posse autem daritriangulum Isosceles Oxygonium, cuius duorum laterum æqualium vtrumuis tertio sit minus, vt rectè animaduertit Franciscus Barocius in sua Cosmographia, ostendemus ad propositionem 15. liber 4. In omni porro triangulo, cuius duo quæcunque later a expressè nominantur, solet reliquum latus tertium à Mathematicis appellari Basis, siue illud in situ insimum occupet locum, siue supremũ, & c Hoc te breuiter monere volui, ne putares aliquid latere mysterij in base triãguli, intelligeresque quodlibet latus, omni discrimine remoto, basis nomine posse nũcupari. 
第二十六界
三邊形。有一直角。為三邊直角形。
第二十七界
三邊形。有一鈍角。為三邊鈍角形。(p. 一〇)
第二十八界
三邊形。有三銳角。為三邊各銳角形。
凡三邊形。恆以在下者為底。在上二邊為腰。
 
22. Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia. 
XXIX. QVADRILATERARVM autem figurarum, Quadratum quidem est, quod & æquilaterum, & rectangulum est.
POST figurarum trilaterarum species, exponit iam singulatim quadrilater as figuras, recensendo quinque tantummodo eorum genera, quorum quatuor priora regularia sunt, postertus autem, & quintum irregulare. Prima figura quadrilatera dicitur Quadratum, cuius quidem omnia quatuor latera inter se æqualia existunt, omnesqúe anguli recti. It aque quadrangulum æquilaterum, & non rectangulum; vel contra, rectangulum, & non æquilaterum, nequaquam Quadratum appellabitur. Docebit autem Euclides propositio 46. huius liber quonam modo construendum sit quadratumsuper recta linea proposita finita.
XXX. ALTERA vero parte longior figura est, quæ rectangula quidem, at æquilatera non est.
SECVNDA figura quadrilatera appellatur Altera parte longior, in qua quidem anguli sunt recti, at latera non sunt inter se æqualia, quamuis bina opposita inter se æqualia existant. Vt in altera parte longiori A B C D, latera A B, D C, inter se, & A D, B C, inter se quoque æqualia sunt, cum A B C D, propter angulorum rectitudinem, parallelogrammum sit, vt in boc liber ad propos. 34. ostendemus.
XXXI. RHOMBVS autem, quæ æquilatera, sed rectangula non est.
HÆC figura tertia inter quadrilateras, quæ Rhombus dicitur, oppositas prorsus habet conditiones. & diuersas a conditionibus figuræ altera parte longioris. Habet enim omnia latera æ qualia, angulos vero non rectos, & inæquales, quamuis binioppositi inter se æquales existant. Vt in Rhombo A B C D, anguli A, & C, inter se, & B, & D, quoque inter se æquales sunt, cum A B C D, propter æqualitatem laterum, parallelogrammum sit, ceu ad eandem propositio 34. huius libri demonstrabitur.
XXXII. RHOMBOIDES verò, quæ aduersa & latera, & angulos habens inter se æquales, neque æquilatera est, neque rectangula.
EST hæc figura, quæ Rhomboides vocatur, quadrato omni ex parte opposita. Nam neque eius latera omnia æqualia sunt, neque vllus angulus rectus, sedtan en latera bina opposita, qualia sunt A B, D C, & A D, B C, in Rhomboide A B C D, æqualia inter se, item anguli bini oppositi, quales sunt A, C, & B, D, inter se existunt æquales Hæ igitur quatuor figure quadrilateræ dici possunt regulares; cæteræ vero omnes, quæcunque sint, irregulares.
XXXIII. PRÆTER has autem, reliquæ quadrilateræ figuræ, trapezia appellentur.
RELIQVAS omnes figuras quadrilateras, quæ à prædictis quatuor differunt, ita vt neque later a omnia æqualia, neque omnes angulos æquales, seu rectos, neque latera bina opposita; neque angulos binos oppositos habeant inter sese æquales, generali vocabulo Trapezia nominat: quæ quidem cum infinitis modis variari queant, rectè irregulares nuncupabuntur. Possunt enim duo anguli esse recti, vel vnus obtusus, & aly acuti, vel duo obtusi, & alij acuti, & c. Eademqúe fieri potest quasi diuisio penes latera: Nam vel aliqua æqualia inter se sunt, vel nullum alters est equale, & c. Determinatas porro trapeziorum species nonnullas afferemus post definitionem linearum parallelarum, seuæquidistantium, & parallelogrammi. 
第二十九界
四邊形。四邊線等而角直。為直角方形。
第三十界
直角形。其角俱是直角。其邊兩兩相等。
如上甲乙丙丁形。甲乙邊與丙丁邊自相等。甲丙與乙丁自相等。
第三十一界
斜方形。四邊等。但非直角。(p. 一一)
第三十二界
長斜方形其邊兩兩相等。但非直角。
第三十三界
已上方形四種。謂之有法四邊形。四種之外。他方形。皆謂之無法四邊形。(p. 一二) 
23. Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. 
XXXIV. PARALLELÆ rectæ lineæ sunt, quæ cum in eodem sint plano, & ex vtraque parte in infinitum producantur, in neutram sibi mutuo incidunt.
VT duæ, vel plures rectæ lineæ dicantur parallelæ siue æquidistantes, non sat is est, vt in quamcunque partem, etiam spatio infinito, product æ nunquam ad vnum punctum coeant; sed necesse quoque est, vt in vna plana superficie existant. Multæ siquidem lineærectæ non existentes in eadem superficie plana productæ ad spatium infinitum, nunquam in vnum conueniunt, & tamen non sunt parallelæ dicendæ; quales sunt, exempli gratia, duæ rectæ lineæ in transuersum positæ in medio aëre, & non se tangentes; Hæ etenim nunquam coire possunt. Dicuntur autem duæ rectæ lineæ in eadem existere planæ superficie, quande superficies aliqua plana vni earum accommodata, ita vt omnia puncta illius tangat, & circa illam immobilem circumuoluta, alteri quoque accommodari potest secundum omnia eius puncta, quamuis re ipsa in duabus superficiebus diuersis reperiantur; Vt propositis duabus rectis lineis A B, C D, si superficies aliqua plana rectæ A B, applicetur, omnia cius tangens puncta, ita vt circa illam circumducta tangat quoque omniæ puncta alterius rectæ C D; dicentur huiusmods rectæ duæ lineæ in eadem superficie plana existere, alias non. Si igitur hæ duærectæ lineæ eædem non coëant, etiamsi infinitè producantur tam ad partes A, C, quàm ad B, D, appellabuntur parallellæ, siue æquidistantes. Cæterum planius, perfectiusque intelliges in x i. liber quo modo duæ rectæ lineæ, vel etiam plures in eadem dicantur superficie existere: Satis sit hoc loco breuiter admonuisse, rectè ab Euclide vtramque conditionem esse positam in definitione linearum parallelarum. Debent enim in eodem existere plano, & productæ in vtramuis partem nunquam in vnum conuenire, quanquam hæc productio continuetur ad spatium infinitum. Quod si duæ rectæ lineæ per immensum aliquod spatium extensæ non cernantur coire, constet tamen, eas tandem ex vna parte longius protractas in vnum punctum conuentur as, quamuis ex altera semper magis ac magis inter sedistent, ac disiungantur, nequaquam appellandæ erunt par allelæ. Quotiescunque ergo duæ line æ rectæ dicuntur à quopiam esse parallelæ, is necesse est concedat, illas in vna, eademqúe superficie iacere, & nunquam posse coire. Similiter, si quis concludere velit, duas rectas lineas esse parallelas, hic demonstret prtus oportet, eas in eodem existere plano, & in neutram partem productas coniungi posse. Qua in re non pauci videntur hallucinari, qui ex eo duntaxat conantur ostendere, aliquas rectas lineas esse parallelas, quod in neutram partem coeant, etiamsi infinitè produæantur, nullæ facta prorsus mentione alterius conditionis, quæ easdem lineas in eodem requirit existere plano.
HIC finem imponit Euclides definitionibus primi libri. Quoniams vero hoc eodem libro mentio fiet figuræ, quæ Parallelogrammum, necnon earum, quæ complementa parallelogrammi dicuntur, necessarium esse duximus, duabus definitionibus adiunctis explicare, quid sit Parallelogrammum, & quæ sint parallelogrammi complementa, vt facilius dem on strationes percipiantur.

XXXV. PARALLELOGRAMMVM est figura quadrilatera, cuius bina opposita latera sunt parallela, seu æquidistantia.
VT figura quadrilatera A B C D, siquidem latus A B, æquidistet lateri D C, & latus A D, lateri B C, nuncupatur Parallelogrammum. Sunt autem quatuor solum parallelogramma; Quadratum, figura alter a parte longior, Rhombus, & Rhomboides, quorũ priora duo rectangula, quod omnes angulos habeant rectos, posteriora vero duo non rectangula vocantur, quod nullus in eis angulus existat rectus. Cæterum, quatuor has figuras esse parallelogramma, ostendenus ad propositionem 34. huius liber Itaque possumus quadrilater as figur as, (vt & antiqui Geometræ) diuidere in Parallelogrammum, & Trapezium. Parallelogrammum rursus in rectangulum, & æquilaterum, quale est Quadratum: in nec rectangulum, nec æquilaterum, quale est Rhomboides; in rectangulum, sed non æquilaterum, qualis est figura altera parte longior: & in æquilaterum, sed non rectangulum, cuiusmodi est Rhombus. Trapeziorum quoque aliud quidem habet duo latera opposita parallela, alia vero minimè aliud autem nulla opposita latera habet parallela. Præterea illud prius vel habet duo illa latera quæ non sunt parallela, inter se æqualia, diciturqúe Trapez ium Isosceles: vel inæqualia, Trapeziumqúe Scalenum appellatur. Itaque ex his omnibus septem genera figurarum quadrilaterarum constitus possunt; Quadratum, figura altera parte longior, Rhombus, Rhomboides, Trapezium Isosceles, Trapezium Scalenum, & Trapezium illud irregulare, in quo nulla latera sunt parallela.
XXXVI. CVM vero in parallelogrammo diameter ducta fuerit, duæque lineæ lateribus parallelæ secantes diametrum in vno eodemque puncto, ita vt parallelogrammum ab hisce parallelis in quatuor distribuatur parallelogramma; appellantur duo illa, per quæ diameter non transit, complementa; duo vero reliqua, per quæ diameter incedit, circa diametrum consistere dicuntur.
SIT parallelogrammum A B C D, in quo diameter A C, & linea E F, secans diametrum in G, & parallela existens la teribus A D, B C. Item linea H I, secans diametrum in eodem puncto G, parallelaque lateribus A B, D C, existens. Quæ cum it a sint, per spicuum est, parallelogrammum totum diuisum esse in quatuor par allelogramma, quorum quidem duo E B I G, G F D H, per quæ diameter A C, non transit, vocantur a Geometris complementa siue supplementa reliquorum duorum A E G H, G I C F. quæ dicuntur circa diametrum consisiere, quippecum perea diameter transeat, vt videre est in præsenti figura. 
第三十四界
兩直線於同面行至無窮。不相離。亦不相遠。而不得相遇。為平行線。
第三十五界
一形。每兩邊有平行線。為平行線方形。(p. 一三)
第三十六界
凡平行線方形。若於兩對角作一直線。其直線為對角線。又於兩邊縱橫各作一平行線。其兩平行線與對角線交羅相遇。卽此形分為四平行線方形。其兩形有對角線者。為角線方形。其兩形無對角線者。為餘方形。
甲乙丁丙方形。於丙乙兩角作一線。為對角線。又依乙丁平行。作戊己線。依甲乙平行作庚辛線。其對角線與戊己、庚辛、兩線。交羅相遇於壬。卽作大小四平行線方形矣。則庚壬己丙、及戊壬辛乙、兩方形。謂之角線方形。而甲庚壬戊、及壬己丁辛、謂之餘方形。 
POSTULATES. 
PETITIONES, SIVE POSTULATA 
求作四則
求作者。不得言不可作。 
1. Let the following be postulated: To draw a straight line from any point to any point. 
I. POSTVLETVR, vt à quouis puncto in quoduis punctum, rectam lineam ducere concedatur.
PRIMVM hoc postulatum planum admodum est, sirectè considerentur ea, quæ paulo ante de linea scripsimus. Nam cum linea sis fluxus quidam puncti imaginarius, atque adeo linea recta fluxus directo omnino itinere progrediens, fit vt sipunctum quodptam ad aliud directo moueri intellexerimus, ducta sane sit à puncto ad punctum recta linea: Id quod prima hac petitione postulat Euclides, quemadmodum hic vides à puncto A, ductamesse rectam lineam ad punctum B; ab eodemque aliam ad punctum C; Item aliam ad punctum D; & sic innumere aliæ ab eodem puncto educi possunt ad alia atque aliæ puncta. 
第一求
自此點至彼點。求作一直線。
此求亦出上篇。蓋自此點直行至彼點。卽是直線。
自甲至乙或至丙、至丁。俱可作直線。 
2. To produce a finite straight line continuously in a straight line. 
II. ET rectam lineam terminatam in continuum recta producere.
QVOD sipunctum illud ferri adhuc cogitauerimus motu directo, & qui omnis inclinationis sit expers, producta erit ipsa recta linea terminata, & nunquam erit finis huius productionis, cum punctum illud intelligere possimus moueri ad infinitam distantiam. Sic lines recta A B, producta est primo in continuum ad punctum C. Deinde ad punctum D, &c. 
第二求
(p. 一四)一有界直線。求從彼界直行引長之。
如甲乙線。從乙引至丙。或引至丁。俱一直行。 
3. To describe a circle with any centre and distance. 
III. ITEM quouis centro, & interuallo circulum describere.
IAM vero, si terminatam rectam lineam cuiuscunque quantitatis mente conceperimus applicatam esse secundum alterum extremũ ad quoduis punctum, ipsamque circa hoc punctum fixum circumduci, donec ad eum reuertatur locum, à quo dimoueri cœpit; descriptus erit circulus, essectumqúe, quod tertia petitio iubet. Exemplum habes in his quinque lineis A B, A C. A D, A E, A F, quæ singulæ citra centrum A, circumulutæ singulos circulos descripserunt iuxta quantitatem, seu interuallum ipsarum.
PRÆTER hæc tria postulata, quibus Euclides contentus fuit, sunt multa alia æquè facilia, è quibus duntaxat in medium proferre decreui illud, quod frequentius repetendum erit in progressu totius Geometriæ. Reliqua enim prudens lector ex se vel facilè intelliget.
 
第三求
不論大小。以點為心求作一圜。 
4. That all right angles are equal to one another. 
IV. ITEM quacunque magnitudine data, sumi posse aliam magnitudinem vel maiorem, vel minorem.
OMNIS enim quantitas continua per additionem augeri, per diuisionem vero diminui potest infinitè: Vnde nunquam dabitur quantitas continua adeo magna, quin ea maior dari possit: neque tam parua, quin minor ea possit exhiberi. Hoc idem in numeris verum est, quod ad additionem pertinet. Nam quilibet numerus per continuam additionem vnitatis augeri potest infinitè: quamuis in eius diminutione ad vnitatem indiuiduam deueniatur. 10  
第四求
(p. 一五)
設一度於此。求作彼度。較此度或大或小。凡言度者。或線或面。或體皆是。或言較小作大可作。較大作小不可作。何者。小之至極。數窮盡故也。此說非是。凡度與數不同。數者。可以長。不可以短。長數無窮。短數有限。如百數減半成五十。減之又減。至一而止。一以下不可損矣。自百以上。增之可至無窮。故曰可長不可短也。度者。可以長。亦可以短。長者增之可至無窮。短者減之亦復無盡。嘗見莊子稱一尺之棰。日取其半。萬世不竭。亦此理也。何者。自有而分。不免為有。若減之可盡。是有化為無也。有化為無。猶可言也。令已分者更復合之。合之又合。仍為尺棰。是始合之初。兩無能幷為一有也。兩無能幷為一有。不可言也 
5. That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. 
 
 
COMMON NOTIONS. 
CUMMVNES NOTIONES SIVE Axiomata, quæ & Pronunciata dici solent, vel Dignitates. 
公論十九則
公論者。不可疑。 
1. Things which are equal to the same thing are also equal to one another. 
I. QVÆ eidem æqualia, & inter se sunt æqualia.
Et quod vno æqualium maius est, aut minus; maius quoque est, aut minus altero æqualium. Et si vnum æqualium maius est, aut minus magnitudine quapiam, alterum quoque æqualium eadem magnitudine maius est, aut minus.
FIERI nulla ratione potest, vt duæ quantitates inæquales, æquales sint alteri quantitati. Si enim minor illarum propositæ quantitati æqualis extiterit, excedet eandem necessario maior illarum. Et si maior æqualis fuerit propositæ quantitati, superabitur minor ab esdem. Quare rectè colligitur, quantitates, quæ eidem quantitati æquales fuerint, inter se æquales quoque esse. Reliquæ quoque partes axiomatis à nobis adiectæ, quod frequentem vsum habeant, clarissimæ sunt.
 
第一論
設有多度。彼此俱與他等。則彼與此自相等。 
2. If equals be added to equals, the wholes are equal. 
II. Et si æqualibus æqualia adiecta sint, tota sunt æqualia.
SI enim quantitates conflatæ, siue compositæ, inæquales forent, proculdubio maiori plus esset adiectum, quàm minori, cum antea æquales extiterint. Quare ex additione æqualium quantitatum ad quantitates æquales, conficientur quantitates quoque æquales. 
第二論
有多度等。若所加之度等。則合幷之度亦等。 
3. If equals be subtracted from equals, the remainders are equal. 
III. ET si ab æqualibus æqualia ablata sint, quæ relinquuntur, sunt æqualia.
NAM si reliquæ quantitates forent inæquales, à minore plus fuisset detractum, quàm à maiore. 
第三論
有多度等。若所減之度等。則所存之度亦等。(p. 一六) 
() 
IV. ET si inæqualibus æqualia adiecta sint, tota sunt inæqualia. Et, si inæqualibus inæqualia adiecta sint, maiori maius, & minori minus, tota sunt inæqualia, illud nimirum maius, & hoc minus.
QVIN &, si æqualibus inæqualia adiecta sint, tota erunt inæqualia: quoniam maior quantitas addita vr. iæqualium, maiorem constituit quantitatem, quàm minor alteri æqualium adiecta: quemadmodum & si inæqualibus æqualia adijciantur, composita quantitas ex maiore, maior est, quàm composita ex minore. Alteram partem huius axiomatis nos adiecimus, propter frequentem eius vsum.
V. ET si ab inæqualibus æqualia ablata sint, reliqua sunt inæqualia. Et si ab inæqualibus inæqualia ablata sint, à maiori minus, & à minori maius, reliqua sunt inæqualia, illud nimirum maius, & hoc minus.
SIC etiam, Si ab æqualibus inæqualia ablata sint, reliqua erunt inæqualia: quia maior quantitas ablata relinquet minorem quantitatem, quàm minor; quemadmodum residuum maioris maius est residuo minoris, si æqualia aufer antur ab inæqualibus. Cæterum Euclides non docet, quidnam simpliciter, & absolutè gignatur ex additione quantitatum inæqualium ad quantitates inæquales, vel quid relinquatur post subtractionem inæqualium quantitatum ab inæqualibus quantitatibus; propterea quod nihil certo colligi inde potest, nisi quando maiori maius additur, & à matori minus detrahitur, vt in secundaparte axiomatis dictum est, quam nos ob insignem eius vtilitatem adiecimus. Possúnt enim compositæ quantitates, vel residuæ, esse & inæquales, & æquales, Si enim ad 7 & 5. addantur 4. & 3. efficientur 11. & 8. quæ sunt inæqualia Sic etiam si ex 7. & 5. detrahantur 2. & 1. relinquentur 5. & 4. quæ sunt inæqualia. At vero, si ad 7. & 5. addantur 4. & 6. conficientur 11. & 11. quæ æqualia sunt. Item si detrahantur 3. & 1. ex 7. & 5. remanebunt 4. & 4. qua æqualia quoque existunt.
PORRO in his omnibus pronunciatis, primo excepto, nomine æqualium quantitatum intelligenda est etiam vna & eadem multis communis. Si enim æqualibus idem commune adijciatur, tota fient æqualia: Et si ab æqualibus idem commune detrahatur, residua æqualia erunt. Et si inæqualibus idem commune adijciatur; veleidem communi addantur inæqualia, tota sient inæqualia: & si ab inæqualibus idem commune detrahatur, vel ab eodem commune inæqualia auferantur, residua existent inæqualia.
 
第四論
有多度不等。若所加之度等。則合幷之度不等。
第五論
有多度不等。若所減之度等。則所存之度不等。 
() 
VI. ET quæ eiusdem duplicia sunt, inter se sunt æqualia.
Et quod vnius æqualium duplum est, duplum est & alterius æqualium. SIMILITER, quæ eiusdem sunt triplicia, vel quadruplicia, vel quintuplicia, &c. inter se sunt æqualia. Si enim inæqualia forent, & maius eorum esset duplex, vel triplex, &c. alicuius quantitatis, deficeret vtique minus à duplict vel triplici, &c. Quod si contra, minus esset duplex, vel triplex, &c. quantitatis cuiuspiam, excederet sanè maius duplex ipsum; vel triplex, &c. Hoc autem & ex secundo axiomate comprobari potest ad hunc modum. Si enim duæ quantitates æquales fuerint alieui tertiæ, & vtrique tertia illa addatur, erunt compositæ duplices illius tertiæ,11 sed & inter se æquales, ob idem additamentum. Quod sirursum compositis eadem tertia adijciatur, erunt conflatæ triplices eiusdem tertiæ. Cum igitur 12 & æquales inter se, propter idem additamentum existant; eademqúe sit ratio in cæteris multiplicibus, perspicuum erit axioma propositum. Secundam porro partem huius axiomatis nos apposuimus, quod non raro eius vsus in rebus Geometriois requiratur. 
11. 2. pron;  12. 2. pron 
第六論
有多度俱倍於此度。則彼多度俱等。 
() 
VII. ET quæ eiusdem sunt dimidia, inter se æqualia sunt.
Et contra, Quæ æqualia sunt, eiusdem sunt dimidia. PARI ratione, quæ eiusdem sunt partes tertiæ, vel quartæ, vel quintæ, &c. inter se æqualia sunt. IN his duobus pronunciatis per eandem quantitatem, intelligi debent quantitates etiam æquales. Nam quæ æqualium duplicia sunt, vel triplicia, &c. inter se æqualia quoque sunt: Item, quæ æqualium sunt dimidia, vel tertia, vel quarta, &c. & inter se æqualia necessario existunt. Partem quoque secundam huiusce axiomatis nos adiunximus, propterea quod non minus frequenter, quàm prima, à Geometris vsurpatur. 
第七論
有多度俱半於此度。則彼多度亦等。 
4. [7] Things which coincide with one another are equal to one another. 
VIII. ET quæ sibi mutuo congruunt, ea inter se sunt æqualia.
HOC est, duæ quantitates, quarum vna superposita alteri, neutra alteram excedit, sed ambæ inter se congruunt, æquales erunt. Vt duæ lineæ rectæ dicentur esse æquales, quando vna alteri superposita, eaquæ superponitur, alteri tota congruit, ita vt eam nec excedat, nec ab ea excedatur. Sic etiam duo anguli rectilinei æquales erunt, quãdo vno alteri superposito, is qui superponitur, alterum nec excedit, nec ab eo exceditur, sed lineæ illius cum lineis huius prorsus coincidunt: Ita enim erunt inclinationes linearum æquales, quamuis lineæ interdum inter se inæquales existant.
E CONTRARIO, Quæ inter se sunt æqualia, sibi mutuo congruent, si alterum alteri superponatur. Intelligendum est autem, quantitates sibi mutuo congruentes, esse æquales, secundum id duntaxat, in quo sibi congruunt. Congruit autem longitudo longitudini tantum, superficies superficiei, solidum solido, linearum inclinatio inclinationi linearum, &c.
 
第八論
有二度自相合。則二度必等。以一度加一度之上。 
5. [8] The whole is greater than the part. 
IX. ET totum sua parte maius est.
CVM pars à toto ablata relinquat adhuc aliquid, ne totum ipsum auferatur; perspicuum est, omne totum sua esse parte maius.
IN sequentibus porro pronunciatis interrumpitur ordo Euclidis, propterea quod duo alia axiomata hoc loco inserenda esse censuimus valde necessaria, cum ex ijs axioma 12. quod Euclidi est decimum, demonstrari possit, vt ibi dicemus. In margine tamen numeros apposuimus ordini Euclidis respondentes.
 
第九論
全大於其分。如一尺大於一寸。寸者、全尺中十分中之一分也。 
 
X. DVÆ lineæ rectæ non habent vnum & idem segmentum commune.
NON est difficile istud axioma, si perfectè intelligatur natura rectæ lineæ. Cum enim linea recta directo semper itinere, nullam in partem deflectendo, producatur, fieri nulla ratione potest, vt duæ lineæ rectæ habeant vnam partem, quamuis minimam, communem, præter vnicum punctum, in quo se mutuo intersecant. Quod tamen breuiter Proclus it a demonstrat. Habeant, si fieri potest, duæ rectæ A B, A C, partem communem A D. Ex centro autem D, & interuallo D A, 13 describatur circulus secans duas rectas propositas in punctis B, & C; 14 Erunt igitur duæ circumferentiæ A B, A B C, inter se æquales, (Sunt enim circumferentiæ semicirculorum æqualium, cum A D B, A D C, ponantur esse diametri) pars & totum, quod est absurdum. Non ergo duæ rectæ habent vnum & idem segmentum commune. Quod est propositum.
POSSVNT tamen duæ lineæ rectæ commune habere segmentum, quando vnam & eandem rectam lineam constituunt. Vt in hac figura, rectæ A D, B C, commune habent segmentum C D, quia ambæ vnam rectam constituunt lineam A B. At vero quando duæ rectæ sunt diuersæ, quales fuêre A B, A C, in superiori exemplo, non possunt possidere segmentum aliquod commune, vtrectè à Proclo fuit demonstratum.

XI. DVÆ rectæ in vno puncto concurrentes, si producantur ambæ, necessario se mutuo in eo puncto intersecabunt.
HOC etiam axioma ex natura lineæ rectæ pendet. Quodtamen ita demonstrabimus Coeant duæ rectæ A B, C B, in B. Dicoillas productas se mutuo secare in B, nempe C B, productam cadere in E, supra rectam A B, productam Nam si C B, producta non cadit supra A B, productam, congruet cum A B, producta, ita vt transeat per D, atque ita duæ rectæ A B D, C B D, habebunt idem segmentum commune B D, quod in antecedenti axiomate ostensum est fieri non posse: vel certè infra A B, productam cadet, ita vt C B, producta cadat in F, sitque vna recta linea C B F. Centro igitur B, describatur ad quoduis interuallum circulus A C F D, secans rectas A B, C B, productas in D, F. Quia ergo vtraque recta A B D, C B F, per centrum B, ducitur, erit tam A C D, quàm C F, semicirculus, per defin. 18. ac proinde æquales erunt circumferentiæ A C D, C F. vt ad defin. 17. demonstrauimus, totum & pars. Quod est absurdum. DVO proxima axiomata ab Euclide non ponuntur, quia tamen necessaria sunt ad aliorum axiomatum probationes, ea bîcinseruimus. Tria autem sequentia Euclidis sunt.
XII. ITEM, omnes anguli recti sunt inter se æquales.
Hoc axioma apertissimum esse cuilsbet potest ex 10. definitione, quâ angulus rectus describitur; propterea quod inclinatio linearum angulum rectum constituentium augeri, minuíue nequit, sed prorsus est immutabilis. Efficitur enim rectus angulus à linea perpendiculari, quæ quidem alteri lineæ rectæ ita superstat, vt faciat vtrobique angulos æquales, neque magis in vnam partem, quàm in alteram inclinet. Ex quofit, omnes angulos rectos æquales inter se esse, cum semper sit eadem inclinatio, quamuis lineæ sint inæquales interdum. Conatur tamen Proclus ex 10. definitioneid demonstrare hacratione. Sint duo angulirecti A B C, D E F, quos dico esse inter se æquales. Si enim fieri potest, sintinæquales, sitqúe A B C, maior. Si igitur mente concipiamus punctum E, applicari puncto B, & rectam D E, rectæ A B, cadetrecta E F, interrectas A B, B C, qualis est B G, propterea quod angulus D E F, minor ponitur angulo A B C. 15 Producatur C B, in rectum & con 2. petit tinuum vsque ad H. Cum igitur angulus A B C, sit rectus, 16 erit angulus A B H, illi deinceps æqualis, & rectus quoque: quare maior fin. etiam angulo A B C. 17 Producta autem G B, in rectum & continuum vsque ad 1, cadet portio producta B I, infra C B, productam, vt in præcedenti axiomate est demonstratum. Quare cum angulus A B G, ponatur rectus, 18 fiet angulus A B I, illi deinceps æqualis. Quapropter angulus A B H, maior quoqueerit angulo A B I, pars tofin. to, quod est absurdum. Non ergo inæquales sunt duo anguli recti propositi, sed æquales. Quod est propo situm: eademqúe est ratio in cæteris.
RECTE autem boc loco monet Pappus, axioma istudnon posse conuerti; non enim omnis angulus recto angulo æqualis rectus est, cum & curuilineus recto æqualis esse queat, vt in 5. lib. demon strabimus, quitamen non dicitur rectus, cum non sitrectilineus. Solus igitur angulus rectilineus æqualis angulo recto, rectus nuncupabitur: Et omnes angult recti inter se æquales erunt, sine vlla exceptione.

XIII. ET si in duas rectas lineas altera recta incidens, internos ad easdem´que partes angulos duobus rectis minores faciat, duæ illæ rectæ lineæ in infinitum productæ sibi mutuo incident ad eas partes, vbi sunt anguli duobus rectis minores.
VT si in duas lineas rectas A B, C D, incsdens alia recta E F, faciat duos angulos internos, & ex eadem parte B E F, D F E, minores duobus rectis, vult Euclides, illas tandem conuentur as esse ad aliquod punctum vnum, versus eam partem, in qua duo anguli minores existunt duobus rectis, vt appositum exemplum commonstrat. Ratio huius perspicua est, quoniam quando duo anguli internt, & ex eadem parte æquales sunt duobus rectis, duæ rectæ lineæ in neutram partem coire possunt, sed æquali semper spatio protenduntur, vt propositio 28. huius liber demonstrabitur. Quare si duo anguli interni, & ex eadem parte efficiuntur minores duobus rectis, necesse est ex ea parte dictarum linearum spatium coarctari, ex altera vero magis ac magis dilatari; ideoque eas conuentur as tandem esse aliquando in vnum punctum. Verum quia axioma hoc sub obscurum videri solet tyronibus, imo à numero principiorum reijcitur à Gemino Geometra, Proclo, & alijs, quod non facilè quiuis ei assensum præbeat; præsertim cumreperiantur aliæ quædam lineæ, quarum spatium, licet semper magis ac magis coangustetur (quemadmodum & in duabus rectis A B, C D, accidit, vt ad propositionem 28. huius liber demon strabimus) nunquam tamen in vnum punctum coeunt, etiamsi infinitè producantur, vt constat ex elementis conicis Apollonij Pergæi, & ex linea conchili Nicomedis. Idcirco pleniorem illius explicationem in scholiuns propositio 28. huius liber differimus, vbi illud ex Procli sententia Geometricè demonstrabimus, vt firmè, ac sine vlla dubitatione, tanquam verissimũ, ad propositionis 29. huius liber (vbi primum eius vsus incipit apparere) & ad aliarum propositionum demonstrationes possit assumi. Quod tamen nos aliter quàm Proclus, & quidem magis geometricè demonstrabimus, ita vt nullus dubitatione locus relinquatur. 
第十論
直角俱相等。見界說十。
第十一論
(p. 一七)有二橫直線。或正或偏。任加一縱線。若三線之間。同方兩角。小於兩直角。則此二橫直線。愈長愈相近。必至相遇。
甲乙、丙丁、二橫直線。任意作一戊己縱線。或正或偏。若戊己線旁同方兩角。俱小於直角。或幷之小於兩直角。則甲乙丙丁線。愈長愈相近。必有相遇之處。
欲明此理。宜察平行線不得相遇者。界說卅四加一垂線。卽三線之間。定為直角。便知此論兩角小於直角者。其行不得不相遇矣。
第十二論
兩直線。不能為有界之形。
第十三論
兩直線。止能於一點相遇。
如雲線長界近。相交不止一點。試於丙乙二界。各出直線交於丁。假令其交不止一點。當引至甲則甲丁乙、宜為甲丙乙圜之徑。而甲丁丙、亦如之界說十七夫(p. 一八)甲丁乙。圜之右半也。而甲丁丙。亦右半也界說十七甲丁乙為全。甲丁丙為其分。而俱稱右半。是全與其分等也。本篇九。 
 
XIV. DVÆ rectæ lineæ spatium non comprehendunt.
NVLLAM prorsus habet difficultatem hoc principium. Si enim duæ rectæ lineæ ex vna parte coeant ad efficiendum angulum, necessario ex altera parte semper magis ac magis disiungétur, si producantur, vt in exemplo proposito perspicuum est. Quare vt superficies, spatiúm ve quodpiam rectilineum ex emni parte concludatur, duabus rectis lineis tertia quædam adiungenda est. Ita enim conficietur spatium triangulare, seufigurarum rectilinearum prima. Proclus tamen demon strat hoc principium, hoc modo. Si fieri potest, vt duæ lineæ rectæ claudant superficiem, comprehendãt duæ rectæ A B C, A D C, superficiem A B C D, ita vt duæ illæ rectæ coeant in duobus punctis, A, & C. Facto deinde centro C, 19 describatur circulus interuallo C A, 20 & producantur rectæ A B C, A D C, in rectam, & continuum vsque ad circismferentiam, nempe ad puncta, E, & F Itaque quia rectæ A C E, A C F, transeunt per centrum C, 21 erunt semicirculi A E, A E F, in17. def terse æquales, & idcirco circumferentia quoque A E, circumferentiæ A E F, æqualis erit, parstoti, quod fieri non potest. Non ergo rectæ duæ lineæ spatium comprehendunt. Quod est propositum.
SED quia fortassis aduersarius dicet, rectas A B C, A D C, productas coire iterum in aliquo puncto circumferentiæ, vt in E, vel F, atque adeo non sequi, partem æqualem esse toti, demonstrabimus tune idem axioma hoc modo. Coeant ergo duæ illæ lineæ iterũ, si fieri potest, in E. Sumpto pũcto F, in recta A D C, quocunque, erit A F, minor, quàm F E, cum minor sit, quàm A F C, hoc est, quàm C H E, quæ ipsi A F C, æqualis est, atque adeo multo minor, quam F E. Circulus igitur ex F, ad interuallum F A, descriptus secabitrectam F E, in H, atque adeo C G E, in G. Quoniam igitur A F H, diameter circuliest, erit A I H, semicirculus, vt ad defin. 17. ostendimus: Portio autem A I G, quam aufert recta A B G, & in qua centrum non est, semicirculo minor, vt ad defin. 18. demonstrauimus. Est ergo circumferentia A I G, minor quàm A I H, totum quàm pars, quod est absurdũ. Quod autem minor sit portio A I G, semicirculo, ostendemus, vt suprà. Nam ducta ex centro F, ad rectam A B G, perpendiculari, & circumuoluta portione A I G, circa rectam A B G, cadet circumferentia A I G, intra circumferentiam A K G, ne pars maior sit quàm totum, vt suirà demon strauimiss.
CONSTAT hoc etiam axioma ex definitione lineæ rectæ. Cum enim recta linea sit breuissima extensio ab vno puncto ad aliud, duci poterit vnica tantum linea ab vno pũcto ad aliud. Quare si A B C, recta est, nõ erit A D C, recta. Quod etiam patet ex definitione Platonis. Nam si A B C, est recta, obumbrabũt media illius extremitates eiusde Igitur media pũcta lineæ A D C, nõ obumbrant extrema, cum visus, per rectans A B C, seratur. Non ergo recta est A D C. HIS axis maiis ab Eucl de positis adiungemus nos nonnulla alia ex aliis Geometris decerpta, non minus necessaria ad futur as demonstrationes Problematum atque Theorematum cum Euclidis, tum cæteterum Mathematicorum, quàmea, quæ nobis tradidit Euclides.
 
第十四論
有幾何度等。若所加之度各不等。則合幷之差。與所加之差等。
甲乙、丙丁、線等。於甲乙加乙戊。於丙丁加丁己。則甲戊大於丙己者。庚戊線也。而乙戊大於丁己。亦如之。 
 
XV. SI æqualibus in æqualia adjiciantur, erit totorum excessus, adiunctorum excessui æqualis.
HOC, & sequens pronunciatum desumpsit Proclus ex Pappo. Aequalibus itaque quãtit itibus A B, C D, addantur inæquales B E, D F, sitqúe B E, mator quàm D F. Et ex B E, auferatur B G, æqualis ipsi D F, vt sit G E, excessus, quo quantitas addita B E, superat quantitatem additam D F Quoniam igitur æqualibus A B, C D, addita sunt æqualia B G, D F, crunt tota AG, C F, 2. pron æqualia. Quare constat, totam quantitatem A E, superare totam C F, codem excessu G E, quo magnitudo D F, adiuncta à magnitudine adiuncta B E, superatur. Quodest propositum.
XVI. SI inæqualibus æqualia adiungantur, erit totorum excessus, excessui eorum, quæ à principio erant, æqualis.
IN eadem figura, inæqualibus quantitatibus B E, D F, addantur æquales A B, C D. Et ex maiore B E, auferatur B G, æqualis ipsi D F, vt G E, sit excessus; quo quantitas B E, quantitatem D F, superat. Quoniam igitur æqualibus B G, D F, addita sunt æqualia A B, C D, erunt tota A G, C F, æqualia. Quamobrem tota quantitas 2 pron A E, superabit totam C F, eodem excessu G E, quo maior quantitas proposita B E, minorem D F, superat. Quod est propositum.
XVII. SI ab æqualibus inæqualia demantur, erit residuorum excessus, excessui ablatorum æqualis.
AB æqualibus A B, C D, auferantur inæqualia B E, D F. Sitqúe E G, excessus, quo quantitas B E superat quantitatem D F, ita vt B G, æqualis sit ipsi D F. Quia igitur ab æqualibus A B, C D, ablata sunt æqualia B G, D F, remanebunt A G, C F, æqualia. Perspicuum 3. pro. ergo est, residuum A E, superari à residuo C F, eodem excessu E G, quo magnitudo ablata B E, ablatam magnitudinem D F, superat. Quod est propositum.
XVIII. SI ab inæqualibus æqualia demantur, erit residuorum excessus excessui totorum æqualis.
AB inæqualibus A B, C D, aufer antur æqualia A E, C F. Sitqúe B G, excessus, quo tota quantitas A B, superat totã quantitatem C D, ita vt A G, æqualis sit ipsi C D. Quoniam igitur ab æqualibus A G. C D, ablata sunt æqualia A E, C F, remanebune E G, F D, æqualia. Quare residuum E B, super abit residuum F D, eodem excessu B G, quo tota quantitas A B, superat totam quantitatem C D. Quod est propositum. IN his quoque quatuor proximè positis pronunciatis, nomine quãtitatum æqualium intelligenda est vna etiam sola quantitas multis communis. Si enim eidem communi inæqualia adijciantur, erit totorum excessus adiunctorum excessui æqualis. Et si inæqualibus idem commune adiungatur, erit totorum excessus, excessui eorum, quæ à principio erant, æqualis. Et si ab eodem communi inæqualia demantur, eritresiduorum excessus excessui ablatorum æqualis. Et si ab inæqualibus idem commune dematur, erit residuorum excessus excessui totorum æqualis. Nam in numeris, si ad 6. addas 5. & 3. fiunt 11. & 9. quorum excessus est 2. idem qui ipsorum 5. & 3. Rursus, si ad 5. & 3. addas 6. fiunt 11. & 9. quorum excessus 2. idem est, qui ipsorum 5. & 3. Item si ex 8. demas 5. & 2. relinquuntur 3. & 6. quorum excessus 3. idem est, qui ipsorum 5. & 2. Denique si ex 10. & 7. demas 3. relinquuntur 7. & 4. quorum excessus 3. idem est, qui ipsorum 10. & 7.
XIX. OMNE totum æquale est omnibus suis partibus simul sumptis.
QVONIAM omnes partes simul sumptæ constituunt totum, cuius sunt partes, manifesta est veritas huius axiomatis.
XX. SI totum totius est duplum, & ablatum ablati; erit & reliquum reliqui duplum.
VT quia totus numerus 20. duplus est totius numeri 10; Et ablatus ex illo 6. ablati ex hoc 3. propterea reliquus illius 14. duplus etiam est reliqui huius 7. In vniuersum autem hoc demon strabitur propositio 5. liber 5. nimirum. Si magnitudo magnitudinis æquè multiplex sit, atque ablata ablatæ, vt decupla, velcentupla, &c. & reliqua reliquæ æquè multiplex erit, atque tota totius.
COLLIGI potest ex dictis cum Proclo, & Gemino hoc discrimen inter postulata, & axiomata, quòd cùm vtraque sint per se nota, & indemonstrabilia, illanaturam sapiunt Problematum, propterea quòd aliquid fieri exposcant; hæc verò, Theoremata imitantur, cùm nihil fieri petant, sed solùm sententiam aliquam notissimam proponant. Differt autem postulatum à problemate, quòd constructio postulati non indigeat vlla demonstratione, problematis autem constructionem concedat nemo sine demonstratione, eo quòd difficile aliquod nobis exhibeat construendum. Idem discrimen inter Axioma, & Theorema reperitur; Illud enim demonstrari non debet, hoc verò concedendum nulla est ratione, nisi demonstretur. Nam nemo huius propositionis demonstrationem, vel etiam probationem requiret. Quæ eidem æqualia, inter se quoque æqualia sunt, Huius autem statim demonstrationem desider abit quis. Omnis trianguli tres anguli interni æquales sunt duobus rectis. Idem iudicium habeto de reliquis axiomatis, atque Theorematis, nec non de postulatis, problematisqúo.
CONSTAT quoque, Postulatorum alia propria esse Geometriæ, qualia sunt illa tria, quæ Euclides nobis proposuit; quædam verò communia & Geometriæ, & Arithmeticæ, cuiusmodi est hoc, Quantitatem posse infinitè augeri. Tam enim numerus, quàm magnitudo, per additionem augeri potest, ita vt nunquam huius incrementi finis reperiatur. Idem dices de axiomatis, siuepronunciatis. Nam octauum, decimum, vndecimum, duodecimum, tertiumdecimum, & quartumdecimum soli Geometriæ conueniunt; Reliqua verò omnia adhibentur & ad demonstrationes Geometricas & ad Arithmeticas. Quemadmodum enim magnitudines æquales ablatæ à magnitudinibus æqualibus, relinquunt magnitudines æquales, siue hæ magnitudines lineæ sint, siue superficies, siue corpora; ita quoque numeri æquales detractic numeris æqualibus relinquunt numeros æquales, &c.
HAEC dicta à nobis sint de triplici hoc genere principiorum, nune ad demonstrationes accedamus, ex quibus pleniùs perfectiúsque principiorum omnium natura percipietur. Sunt enim plurima principia Mathematicorum eiusmodi, vt planè non intelligantur, nisi priùs eorum vsus appareat in demonstrationibus; id quod satis te experientia docebit.
ANTEQVAM porrò ad propositiones Euclidis interpretandas veniamus, paucis explicandum est, quémnam ordinem, ac modum in ipsis demonstrationibus simus secuti. Primum cuilibet propositioni du s numeros affiximus, quorum alter in margine depictus significat ordinem, quem Campanus ex traditione Arabum est secutus in Euclidis propositionibus, alter verò in ipsa propositionum serie descriptus refert dispositionem propositionum ex traditione Theonis, & quam adhuc obseruari cernimus in codicibus Græcis. Id verò eo consilio à nobis est factum: quoniam cùm à quibusdam Geometris propositiones Euclidis iuxta ordinem Campani, ab alijs verò iuxta Theonis seriem citentur, maximeqúe interdum duo hi interpretes inter se discrepent, inserie, atque ordine propositionum, id quod maximè in 6. 7. & 10 liber perspicitur; necessarium esse duximus, vt vtriusque interpretis numerus apponeretur. Ita enim fiet, vt si quando numerus propositionum à Geometra quopiam citatus non respondet alteri interpreti, alteri saltem conueniat. Deinde ne cursus demonstrationum interrumperetur, citauimus principia, & propositiones Euclidis in margine, præfixa euilibet citationi semper literula aliqua alphabeti, vel alio quouis signo, cui similis literula, seu signum respondet in demonstratione, vt faciliùs cognoscatur, ad quem locum quælibet citatio sit referenda. Porrò citationes intelligendæ sunt hoc modo.
1. def. Prima definitio. & sie dealijs numeris , vt 4. def. 23. def. &c.
1. pet. Prima petitio, vel primum Postulatum.
1. Pron. Primum pronunciatum, seu axioma, & ita de reliquis numeris, vt priùs.
1. primi. Prima propositio primi libri.
23. Vndec. Vigesimatertia propositio vndecimi libri
6. tertijd. Sexta tertijdecimi libri.
9. sextid. Nona sextidecimi libri.
13. duod. Decimatertia libri duodecimi.
7. quind. Septima libri quindecimi.
5. quartid. Quinta libri quartidecimi, &c.
Ex his aliæ citationes à quolibet facilè poterunt intelligi. Eadem enim in omnibus est ratio.
 
第十五論
有幾何度不等。若所加之度等。則合幷所贏之度。與元所贏之度等。
如下圖反說之。戊乙、己丁、線不等。於戊乙加乙甲。於己丁加丁丙。則戊甲大於己丙者。戊庚線也。而戊乙大於己丁。亦如之。(p. 一九)
第十六論
有幾何度等若所減之度不等。則餘度所贏之度。與減去所贏之度等。
甲乙丙丁、線等。於甲乙減戊乙。於丙丁減己丁。則乙戊大於丁己者。庚戊也。而丙己大於甲戊。亦如之。
第十七論(p. 二〇)
有幾何度不等。若所減之度等。則餘度所贏之度。與元所贏之度等。
如十四論反說之。甲戊、丙己、線不等。於甲戊減甲乙。於丙己減丙丁。則乙戊長於丁己者。亦庚戊也。與甲戊長於丙己者等矣。
第十八論
全與諸分之井等。
第十九論
有二全度。此全倍於彼全。若此全所減之度。倍於彼全所減之度。則此較亦倍於彼較。相減之餘曰較。
如此度二十。彼度十。於二十減六。於十減三。則此較十四彼較七。(p. 二一) 
Proposition 1. 
PROBLEMA 1. PROPOSITIO 1. 
幾何原本第一卷本篇論三角形計四十八題
第一題 
On a given finite straight line to construct an equilateral triangle. 
SVPER data recta linea terminata triangulum æquilaterum constituere.
IN omni problemate duo potissimùm sunt consideranda, constructio illius, quod proponitur, & demonstratio, quâ ostenditur, constructionem rectè esse institutam. Vt quoniam primum hoc problema iubet constituere triangulum æquilaterum super data recta linea terminata quacunque, ita vt linea recta proposita sit vnum latus trianguli. (Tunc enim figura dicitur constitui super recta linea, quando ipsa linea efficitur vnum figuræ latus) idcirco primum oportet construere ex principiis concessis triangulum aliquod, deinde demonstrare, ipsum eâ ratione constructum, esse æquilaterum, hoc est, habere omnia tria latera inter se æqualia. Quod idem in alijs problematibus perspici potest. Hæc etiam duo reperiuntur ferè in omni Theoremate. Sæpenumerò enim vt demonstretur id, quod proponitur, construendum est, atque efficiendum prius aliquid, ceu manifestum erit in sequentibus. Pauca veró admodum sunt theoremata, quæ nullam requirant constructionem. 
於有界直線上。求立平邊三角形。1  
Let AB be the given finite straight line. 
Sit igitur proposita recta linea terminata A B, 
法曰。甲乙直線上。 
Thus it is required to construct an equilateral triangle on the straight line AB. 
super quam constituere iubemur triangulum æquilaterum. 
求立平邊三角形。 
With centre A and distance AB let the circle BCD be described; [Post. 3]  again, with centre B and distance BA let the circle ACE be described; [Post. 3]  and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1] 
Centro A, & interuallo rectæ A B, 22 describatur circulus C B D: 
22. 3. pet. 
Item centro B, & interuallo eiusdem rectæ B A, alius circulus describatur C A D, secans priorem in punctis C, & D.  Ex quorum vtrouis, nempe ex C, 23 ducantur duæ rectæ lineæ C A, C B, ad puncta A, & B;
Eritque super rectam A B, constitutum triangulum A B C, hoc est, figura rectilinea contenta tribus rectis lineis. Dico, hoc triangulum ita constructum necessariò esse æquilaterum. 
先以甲為心。乙為界。作丙乙丁圜。  次以乙為心。甲為界。作丙甲丁圜。兩圜相交於丙於丁。  末自甲至丙。丙至乙。各作直線。卽甲乙丙為平邊三角形。 
Now, since the point A is the centre of the circle CDB, AC is equal to AB. [Def. 15]  Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15]  But CA was also proved equal to AB;  therefore each of the straight lines CA, CB is equal to AB.  And things which are equal to the same thing are also equal to one another; [C.N. 1]  therefore CA is also equal to CB.  Therefore the three straight lines CA, AB, BC are equal to one another. 
Quoniam rectæ A B, A C, ducuntur ex centro A, ad circumferentiam circuli C B D, 24 erit recta A C, rectæ A B, æqualis: 
23. 1. pet.  24. 15. def. 
Rursus quia rectæ B C, B A, ducuntur ex centro B, ad circumferentiam circuli C A D, erit recta B C, rectæ B A æqualis.  Tam igitur A C, quàm B C,  æqualis est rectæ A B.    25 Quare & A C, B C, inter se æquales erunt,   
論曰。以甲為心。至圜之界。其甲乙線。與甲丙、甲丁、線等。  以乙為心。則乙甲線。與乙丙、乙丁、線亦等。何者。凡為圜。自心至界。各線俱等故。界說十五 旣乙丙等於乙甲。  而甲丙亦等於甲乙。  卽甲丙亦等於乙丙。  公論一    三邊等。 
Therefore the triangle ABC is equilateral;  and it has been constructed on the given finite straight line AB. 
atque idcirco triangulum A B C, erit æquilaterum. 
25. 1. pron. 
Super data ergo recta linea terminata, &c. 
   
  (Being) what it was required to do. 
  Quod faciendum erat.
SCOLIUM
VT autem videas, plures demonstrationes in vna propositione contineri, placuit primam hanc propositionem resoluere in prima sua principia, initio facto ab vltimo syllogismo demonstratiuo. Si quis igitur probare velit, triangulum A B C, constructum methodo prædicta, esse æquilaterum, vtetur hoc syllogismo demonstrante.
Omne triangulum habens tria latera æqualia, 26 est æquilaterum.
Triangulum A B C, tria habet æqualia latera.
Triangulum igitur A B C, est æquilaterum. Minorem confirmabit hoc alio syllogismo. Quæ eidem æqualia sunt, 27 inter se quoque sunt æqualia.
Duo latera A C, B C, æqualia sunt eidem lateri A B.
Igitur & duo latera A C, B C, inter se æqualia sunt.
Ac propterea omnia tria latera A B, B C, A C, æqualia existunt.
Minorem verò huius syllogismi hac ratione colliget.
Lineæ rectæ à centro ductæ ad circumferentiam circuli, 28 inter se sunt æquales.
Lineæ A B, A C, sunt ductæ à centro A, ad circumferentiam C B D.
Sunt igitur lineæ A B, A C, æquales inter se. Eademqúe ratione erunt lineæ A B, B C, æquales, cùm ducantur à centro B, ad circumferentiam C A D. Quamobrem minor præcedentis syllogismi tota confirmata erit.
Non aliter resolui poterunt omnes aliæ propositiones non solùm Euclidis, verùm etiam cæterorum Mathematicorum.
Negligunt tamen Mathematici resolutionem istam in suis demonstrationibus, eò quòd breuiùs ac faciliùs sine ea demonstrent id, quod proponitur, vt perspicuum esse potest ex superiori demonstratione.
SI quis autem super data recta desideret constituere triangulum quoque Isosceles, & scalenum, id cum Proclo in hunc modum efficiet. Sit recta linea A B, circa quam ex centris A, & B, describantur duo circuli, vti prius. 29 Deinde producatur A B, in tramque partem ad circumferentias vsque ad puncta C, & D. Atque centro A, interuallo vero A D, 30 describatur circulus E D F. Item centro B, interuallo vero B C, circulus E C F, secans priorem in punctis E, & F. Ex quorum vtrolibet, nempe ex E, 31 ducantur ad puncta A, & B, duæ rectæ E A, E B. Factumque erit super recta A B, 32 triangulum A B E; quod dico esse Isosceles, nimirum duo later a A E, B E, esse & æqualia inter se, & maiora latere A B. Cùm enim rectæ A E, A D, ducantur è centro A, ad circumferentiam E D F, 33 erit A E, æqualis rectæ A D. Item cùm rectæ B E, B C, ducantur è centro B, ad circumferentiam E C F, 34 erit B E, æqualis rectæ B C. Sunt autem rectæ A D, B C, æquales inter se (vtraque enim A C, & B D, æqualis est rectæ A B; cum A B, A C, ex eodem centro A, ad circumferentiam ducantur; Item B A, B D, ex eodem centro B, ad circumferentiã quoque egrediantur. 35 Quare A C, B D, æquales inter se erunt. Addito igitur communi recta A B, 36 erit tota A D, toti B C, æqualis.) 37 Igitur A E, 2 B E æquales quoque inter se erunt. Quòd verò vtraque A E, B E, 38 maior sit quàm A B, perspicuum est, cum A D, æqualis ostensa ipsi A E, maior sit quàm A B; Item B C, æqualis demonstrata ipsi B E, 39 maior quoque sit, quàm A B. Constitutum igitur est super recta A B, Isosce. 9. pron les A B E, babens duo latera A E, B E, æqualia inter se, & maiora listere A B, quod faciendum erat. Atque hæc est demonstratio Procli, aliorumque interpretum Euclidis.
BREVIVS tamen videtur mihi posse demonstrari, triangulum A B E, esse Isosceles, hac ratione. 40 Quoniam A E, æqualis est rectæ A D, & recta A D, est dupla rectæ A B, propterea quòd B A, B D, æquales inter se sunt; erit & A E, dupla rectæ A B. Rursus quia B E. 15. def æqualis est rectæ B C, & B C, dupla est ipsius A B, propterea quòd A B, 41 A C, æquales sunt inter se, erit & B E, dupla ipsius A B. Cùm igitur vtraque 42 A E, B E, dupla sit eiusdem A B, erunt A E, B E, inter se æquales, maioresque, propterea recta A B. Isosceles ergo est triangulum A B E.
IAM verò, si ex puncto A, 43 ducatur linea recta A G, ad circumferentiam E G F, quæ non sit eadem quæ A E, vel A D, secans circumferentiam E H D, in puncto H, & ex G, ad B 44 ducatur aliæ recta G B, constitutum erit triangulum A B G, super recta A B, 20. def quod dico esse scalenum. Quoniam 45 A G, maior est quàm A H: 46 Sunt autem A H, A E, ex centro A, ductæ, inter se æquales; erit & A G, maior quàm A E, hoc est, quàm B E, quæ ostensa est æqualis ipsi A E, igitur & maior erit A G, quàm B G, cùm B G, sit æqualis ipsi B E. Est autem & B G, maior quàm A B. , propterea quòd tota B C, æqualis ipsi B G, 47 maior est quàm A B, pars. Omnia ergo tria 9. pron later a trianguli A B G, inæqualia sunt, ideoque scalenum est ex definitione; quoderat faciendum.
BREVIVS quoque ostendemus, triangulum A G B, esse scalenum, hac ratione. Quoniam tam 48 A H, A D, ex centro A, ductæ sunt æquales, quàm B G, B C, ex centro B, ductæ: Sunt autem A D. B C, ipsius A B, duplæ, quod A B, vtrique B D, A C, æqualis sit; erunt quoque A H, B G, ipsius A B, duplæ, ac propterea maiores, quàm A B. Cùm ergo 49 A G, maior sit, quàm A H, siue quàm B G, scalenum 9 pron erit triangulum A G B, habens latus A G, maximum, B G, medium, & A B, minimum.
PRAXIS.
CONABIMVR in singulis ferè problematibus Euclidis tradere praxin quandam facilem, & breuem, qua effici possit id, quod Euclides pluribus verbis, atque lineis contendit construere; ldqúe in ijs præsertim obseruabimus, quæ frequentiorem vsum habent apud Mathematicos, & in quibus praxis compendium aliquodsecum videtur afferre.
ITAQVE triangulum æquilaterum ita facilè construetur super data recta A B. Ex centris A, & B, interuallo vero datæ rectæ A B, describantur duo arcus circulorum seintersecantes in puncto C, siue hoc infra lineam contingat, siue suprà. Post bæc ducantur duæ rectæ A C, B C, ex puncto C, ad puncta A, & B, factumqúe erit, quod proponitur. Cuius recadem est demonstrasio cum superiori, simodò circuli essent integri, acperfecti. Transirent enim necessariò per puncta A, & B.
ISOSCELES ita conficietur. Ex centris A, & B, interuallo vero maiore quám A B, si datam rectam esse velimus minus latus; veminore, si eandem in latus maius cligamus, describantur duo arcus secantes se in C. Postea ducantur recta A C, & B C, constructumqúe erit Isosceles: quoniam A C, B C, æquales erunt, propter æquale interu allum assumptum, maius scilicet, aut minus quàm recta A B.
SCALENVM denique hoc modo fabricabitur super data recta A B. Ex centro B, interuallo vero maiore, quam B A, describatur arcus aliquis. Item ex centro A, interuallo vero adhuc maiore, quàm prius assumptum, describatur alter arcus priorem secans in C. Deinde ducantur rectæ A C, B C; constitutumqúe erit Scalenum, vt constat ex inæqualitate interuallorum, quæ assumpta fuerunt in constructione.
CAETERVM quo pacto triangulum constitui debeat habens tria latera æqualia tribus datis lineis quibuscunque, singula singulis, latrùs explicabimus propositio 22. huius libri.
 
  如所求。凡論有二種。此以是為論者。正論也。下倣此。
其用法。不必作兩圜。但以甲為心。乙為界。作近丙一短界線。乙為心。甲為界。亦如之。兩短界線交處。卽得丙。
諸三角形。俱推前用法作之。詳本篇廿二。 
Proposition 2. 
PROBL. 2. PROPOS. 2. 
26. 23. def.  27. 1. pron.  28. 15. def.  29. 2. pet.  30. 3. pet.  31. 1 pet.  32. 20 def.  33. 15. def  34. 15. def  35. 1. pron  36. 2. pron  37. 1. pron  38. 1. pron  39. 9. pron  40. 15. def.  41. 15. def.  42. 6. pro.  43. 1. petit.  44. 1. pet.  45. 9. pro.  46. 15. def.  47. 9. pro.  48. 15. def.  49. 9. pron. 
第二題 
To place at a given point (as an extremity) a straight line equal to a given straight line. 
AD datum punctum, datæ rectæ lineæ æqualem rectam lineam ponere. 
一直線。線或內、或外、有一點。求以點為界。作直線。與元線等。 
Let A be the given point, and BC the given straight line.  Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC. 
SIT punctum datum A, & data recta linea B C,  cui aliam rectam æqualem ponere oportet ad punctum A. Facto alterutro extremo lineæ B C, nempe C, centro 50 describatur circulus B E, interuallo rectæ B C. 
法曰。有甲點。及乙丙線。  求以甲為界。作一線。與乙丙等。先以丙為心。乙為界。乙為心丙為界亦可作。作丙乙圜。第三求 
From the point A to the point B let the straight line AB be joined; [Post. 1]  and on it let the equilateral triangle DAB be constructed. [I. 1]  Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2]  with centre B and distance BC let the circle CGH be described; [Post. 3]  and again, with centre D and distance DG let the circle GKL be described. [Post. 3] 
Et ex A, ad centrum C, 51 recta ducatur A C; (nisi punctum A, intra rectam B C, fuerit: Tunc enim pro linea ducta sumetur A C, vt secunda figura indicat.
Super recta verò A C, 52 construatur triangulum æquilaterum A C D, sursum, aut deorsum versus, vt libuerit;  cuius duo latera modò constituta D A, D C, versus rectam A C, 53 extendantur; D C, quidem oppositum puncto dato A, vsque ad circumferentiam in E; D A, vero oppositum centro C, quantumlibet in F.  Clavius made this circle already in the third sentence of the proposition.  Deinde centro D, interuallo vero rectæ D E, per C, centrum transeuntis, 54 alter circulus describatur E G, secans rectam D F, in G. 
次觀甲點、若在丙乙之外。則自甲至丙。作甲丙線。第一求如上前圖。或甲在丙乙之內。則截取甲至丙一分線。如上後圖。兩法俱以甲丙線為底。  任於上下作甲丁丙平邊三角形。本篇一。  次自三角形兩腰線引長之第二求其丁丙、引至丙乙圜界而止。為丙戊線。其丁甲、引之出丙乙圜外、稍長。為甲己線。  As did Clavius  末以丁為心。戊為界。作丁戊圜。其甲己線、與丁戊圜、相交於庚。 
Then, since the point B is the centre of the circle CGH,BC is equal to BG.  Again, since the point D is the centre of the circle GKL, DL is equal to DG. And in these DA is equal to DB;  therefore the remainder AL is equal to the remainder BG. [C.N. 3]  But BC was also proved equal to BG;  therefore each of the straight lines AL, BC is equal to BG.  And things which are equal to the same thing are also equal to one another; [C.N. 1]  therefore AL is also equal to BC. 
Dico rectam A G, quæ posita est ad punctum datum A, æqualem esse datæ rectæ B C. 
52. 1. primi.  53. 2. pet.  54. 3. pet. 
Quoniam D E, D G, ductæ sunt ex centro D, ad circumferentiam E G, 55 ipsæ inter se æquales erunt: Ablatis igitur D A, D C, æqualibus lateribus trianguli æquilateri A C D, 56 remanebit A G, æqualis rectæ C E.  Sed eidem C E, 57 æqualis est recta B C. (cùm ambæ rectæ C B, C E, cadant ex centro C, ad circumferentiam B E.)  Igitur rectæ A G, B C, quandoquidem vtraque æqualis est ostensa rectæ C E,    inter se 58 æquales erunt.  See above 
卽甲庚線、與乙丙線等。  論曰。丁戊、丁庚線。同以丁為心。戊、庚、為界。故等。界說十五於丁戊線減丁丙。丁庚線減丁甲。其所減兩腰線等。則所存亦等。公論三  夫丙戊、與丙乙。同以丙為心。戊、乙、為界。亦等。界說十五  卽甲庚、與丙乙等。公論一。 (p. 二四)    See previous Chinese sentence  See above 
Therefore at the given point A the straight line AL is placed equal to the given straight line BC.  (Being) what it was required to do. 
Ad datum igitur punctum. &c. 
55. 15. def.  56. 3. pron.  57. 15. def.  58. 1. pron. 
quod erat faciendum.
QVOD si punctum datum fuerit in extremo datæ lineæ, quale est C, facilè absoluetur problema. Si enim centro C, & interuallo C B, 59 describatur circulus, ad cuius circumferentiam recta 60 ducatur vtcunque C E, erit hæc posita ad punctum datum C, 61 æqualis datæ rectæ B C, cum vtraque & B C, & C E, ex eodem centro egrediatur ad circumferentiam B E. 
  若所設甲點。卽在丙乙線之一界。其法尤易。假如點在丙。卽以丙為心。作乙戊圜。從丙至戊、卽所求。 
Proposition 3. 
PROBL. 3. PROPOS. 3. 
59. 3. petit  60. 1. pet.  61. 15. def. 
第三題 
Given two unequal straight lines, to cut off from the greater a straight line equal to the less. 
DVABVS datis rectis lineis inæqualibus, de maiore æqualem minori rectam lineam detrahere. 
兩直線。一長一短。求於長線、減去短線之度。 
Let AB, C be the two given unequal straight lines, and let AB be the greater of them.  Thus it is required to cut off from AB the greater a straight line equal to C the less. 
SINT duæ rectæ inæquales A, minor, & B C, maior,  oporteat que ex maiore B C, detrahere lineam æqualem minori A. 
法曰。甲短線。乙丙長線。  求於乙丙、減甲。 
At the point A let AD be placed equal to the straight line C; [I. 2]  and with centre A and distance AD let the circle DEF be described. [Post. 3] 
Ad alterutrum extremorum lineæ maioris B C, nempe ad punctum B, 62 ponatur aliqua linea, quæ sit B D, æqualis minori A. 
Deinde centro B, interuallo autem B D, circulus 63 describatur secans B C, in E. 
先以甲為度。從乙引至別界。作乙丁線。本篇二  次以乙為心。丁為界。作圜。第三求圜界與乙丙、交於戊。 
Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15]  But C is also equal to AD.  Therefore each of the straight lines AE, C is equal to AD;  so that AE is also equal to C. [C.N. 1] 
Dico B E, detractam esse æqualem ipsi A. Quoniam B E, 64 . æqualis est rectæ B D, & eidem B D, æqualis est recta A, per constructionem; 
65 erunt A, & B E, inter se æquales.    See previous 
卽乙戊、與等甲之乙丁等。蓋乙丁、乙戊。同心、同圜故。界說十五。  See previous  See previous  See previous 
Therefore, given the two straight lines AB, C, from AB the greater AE has been cut off equal to C the less.  (Being) what it was required to do. 
Duabus igitur datis rectis, &c. 
quod erat faciendum.
QVOD si duæ rectæ datæ coniungantur in vno extremo, quales sunt B D, & B C, coniunctæ in extremo vtriusque B; describendus erit circulus ex B, ad interuallum minoris B D. Hic enim auferet B E, æqualem ipsi B D, vt constat ex definitione circuli.
SCHOLIUM
VARIOS etiam posse casus esse in hoc problemate, nemo ignorat, cum duæ lineæ inæquales datæ vel inter se distent, ita vt neutra alteram contingat, vel non; sed vel coniungantur ad vnum extremum, vel se mutuo secent, vel certè altera alteram suo extremo tangat duntaxat, &c. de quare lege Proclum hoc in loco.
 
  (p. 二五) 
Proposition 4. 
THEOREMA 1. PROPOS. 4. 
第四題 
If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. 
SI duo triangula duo latera duobus lateribus æqualia habeant, vtrumque vtrique; habeant verò & angulum angulo æqualem sub æqualibus rectis lineis contentum: Et basim basi æqualem habebunt: eritque triangulum triangulo æquale; ac reliqui anguli reliquis angulis æquales erunt, vterque vtrique, sub quibus æqualia latera subtenduntur. 
兩三角形。若相當之兩腰線各等。各兩腰線間之角等。則兩底線必等。而兩形亦等。其餘各兩角相當者俱等。 
Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF, and the angle BAC equal to the angle EDF.  I say that the base BC is also equal to the base EF,  the triangle ABC will be equal to the triangle DEF,  and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend,  that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. 
SINT duo triangula A B C, D E F, & vnius vtrumque latus A B, A C, æquale sit alterius vtrique lateri D E, D F, hoc est, A B, ipsi D E, & A C, ipsi D F; angulusqúe A, contentus lateribus A B, A C, æqualis angulo D, contento lateribus D E, D F.  Dico basim B C, æqualem quoque esse basi E F;  & triangulum A B C, triangulo D E F;    & vtrumque angulum B, & C, vtrique angulo E, & F, id est, angulos B, & E, qui opponuntur lateribus æqualibus A C, D F, inter se; & angulos C, & F, qui opponuntur æqualibus lateribus A B, D E, inter se quoque esse æquales. 
解曰甲乙丙、丁戊己、兩三角形之甲、與丁、兩角等。甲丙、與丁己、兩線。甲乙、與丁戊、兩線。各等。  題言乙丙、與戊己、兩底線必等。  而兩三角形亦等。    甲乙丙、與丁戊己、兩角。甲丙乙、與丁己戊、兩角。俱等。 
For, if the triangle ABC be applied to the triangle DEF, and if the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE.  Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF;  hence the point C will also coincide with the point F, because AC is again equal to DF.  But B also coincided with E; hence the base BC will coincide with the base EF.  [For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible.  Therefore the base BC will coincide with EF] and will be equal to it. [C.N. 4]  Thus the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it.  And the remaining angles will also coincide with the remaining angles and will be equal to them,  the angle ABC to the angle DEF, and the angle ACB to the angle DFE. 
Quoniam enim recta A B, rectæ D E, ponitur æqualis, fit, vt si altera alteri superponi intelligatur, collocato puncto A, in puncto D, 66 ipsæ sibi mutuo congruant, punctumque B, in punctum E, cadat. Neque enim dicere quis poterit, partem rectæ A B, rectæ D E, congruere, & partem non, quia tunc duæ rectæ haberentidem segmentum commune, 67 quod est impossibile. Quod si quis dicat, posito puncto A, in D, cadere quidem punctum B, in E, sed rectam A B, cadere vel ad dextram, vel ad sinistram D E, claudent duæ rectæ lineæ superficiem, 68 quod fieri non potest. 
Quare recta A B, rectæ D E, congruet, vt dictum est. Cum ergo angulus A, angulo D, ponatur æqualis, congruet quoque alter 69 alteri, hoc est, recta A C, rectæ D F, congruet,  punctumqúe C, in punctum F, cadet, ob æqualitatem rectarum A C, D F.  Basis igitur B C, basi E F, congruet quoque:  alias si supra caderet, aut infra, vt efficeretur recta E G F, vel E H F, clauderent duæ rectæ E F, E G F; vel E F, E H F, superficiem, (negare enim nemo poterit, tam E G F, quàm E H F, rectam esse, cum vtraque ponatur esse eadem, quæ recta B C.) quod est absurdum. Duæ enim rectæ superficiem 70 claudere non possunt.  Quocirca 71 basis B C, basi E F, æqualis erit, cum neutra alteram excedat;  & triangulum A B C, triangulo D E F;  & angulus B, angulo E; & angulus C, angulo F, æqualis, ob eandem causam, existet.   
論曰。如云乙丙、與戊己、不等。卽令將甲角置丁角之上。  兩角必相合、無大小。甲丙、與丁己。甲乙、與丁戊。亦必相合,無大小。公論八 此二俱等。      而云乙丙、與戊己、不等。必乙丙底或在戊己之上、為庚。或在其下、為辛矣。戊己旣為直線而戊庚己又為直線則兩線當別作一形是兩線能相合為形也辛倣此。公論十二此以非為論者。駁論也。下倣此。         
Therefore etc.  (Being) what it was required to prove. 
Quare si duo triangula duo latera duobus lateribus æqualia habeant, &c. 
Quod demonstrandum erat.
RECTE Euclides duas conditiones posuit in antecedente huius theorematis, quarum prima est, vt duo latera vnius trianguli æqualia sint duobus lateribus alterius trianguli, vtrumque vtrique; Secunda, vt angulus etiam vnius contentus illis lateribus æqualis sit angulo alterius contento lateribus, quæ is iis sunt æqualia. Deficiente enim alterutra harum conditionum, neque bases, neque reliqui anguli poterunt vnquam esse æquales, vt probe hoc loco à Proclo demonstratur: Triangula verò ipsalicet possint esse æqualia, posteriore duntaxat conditione deficiente, vt ex scholio propos, 37. huius liber constabit, tamen rarò admodum illud continget. Sint enim triangulorum A B C, D E F, anguli A, & D, æquales, nempe recti, & latera A B, A C, æqualia lateribus D E, D F, non quidem vtrumque vtrique, sed illa simul sumpta hisce simul sumptis, sitque A B, 3. A C, 4. vt ambo simul efficiant 7. At verò D E, sit 2. & D E, 5. vt ambo quoque simul 7. constituant. Quibus posisitis, erit basis B C, 5, & b isis E F, radix quadrata huius numeri 29. quæ maior quidem est quam 5. minor autem, quam 6. Item area trianguli A B C, erit 6. area verò trianguli D E F, 5. Anguli denique super basim B C, inæquales erunt angulis super basim E F. Quæ quidem omnia ita esse, hic ostenderemus, nisi adeorum demonstrationem requirerentur multa, quæ nondum sunt confirmata. Vides igitur omnia inæqualia esse, propterea quod non vtrumque latus vtrique lateri æquale existit in dictis triangulis A B C, D E F.
RVRSVS triangulorum A B C, D E F, latera A B, A C, æqualia sint lateribus D E, D F, vtrumque vtrique, sitqúe vnumquodque 5. anguli verò A, & D, contenti dictis lateribus inæquales, sitqúe A, maior quam D. Quibus concessis, erit basis B C, maior base E F, vt propositio 24. huius libri ostendetur. Quod si basim B C. ponamus esse 8. basim autem E F, 4. erit area trianguli A B C, 12. area verò trianguli D E F, radix quadrata huius numeri 84. quæ maior quidem est quam 9. minor verò, quam 10. id quod notissimum est Geometris. Vt igitur duorum triangulorum & bases, & anguli, nec non triangula ipsa æqualia inter se sint, necesse est, vt vtrumque latus vnius æquale sit vtrique lateri alterius, & anguli quoque dictis lateribus contenti æquales existant, vt optimè dixit Euclides.
 
   
Proposition 5. 
THEOR. 2. PROPOS. 5. 
第五題 
In isosceles triangles the angles at the base are equal to one another,  and, if the equal straight lines be produced further, the angles under the base will be equal to one another. 
ISOSCELIVM triangulorum, qui ad basim sunt, anguli inter se sunt æquales:  Et productis æqualibus rectis lineis, qui sub basi sunt, anguli inter se æquales erunt. 
三角形。若兩腰等。則底線兩端之兩角等。  而兩腰引出之。其底之外兩角亦等。 
Let ABC be an isosceles triangle having the side AB equal to the side AC;  and let the straight lines BD, CE be produced further in a straight line with AB, AC. [Post. 2]  I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE. 
SIT triangulum Isosceles A B C, in quo latera A B, A C, inter se sint æqualia.    Dico angulos A B C, A C B, supra basim B C, æquales inter se esse: Item si latera æqualia A B, A C, producantur quantum libuerit, vsque ad puncta D, & E, angulos quoque D B C, E C B, infra basim eandem B C, esse æquales. 
解曰。甲乙丙三角形。其甲丙、與甲乙、兩腰等。    題言甲丙乙與甲乙丙兩角等, 又自甲丙線任引至戊, 甲乙線任引至丁, 其乙丙戊與丙乙丁兩外角亦等。 
Let a point F be taken at random on BD; from AE the greater let AG be cut off equal to AF the less; [I. 3] and let the straight lines FC, GB be joined. [Post. 1] 
Ex linea enim A E, producta infinite abscindatur A F, æqualis ipsi 72 A D, & ducantur rectæ B F, C D. Confiderentur deinde duo triangula A B F, A C D. 
72. 3 primi. 
論曰。試如甲戊線稍長。卽從甲戊截取一分。與甲丁等。為甲己。本篇三 次自丙至丁乙至己。各作直線。第一求卽甲己乙、甲丁丙、兩三角形必等。 
Then, since AF is equal to AG and AB to AC, the two sides FA, AC are equal to the two sides GA, AB, respectively;  and they contain a common angle, the angle FAG.  Therefore the base FC is equal to the base GB,  and the triangle AFC is equal to the triangle AGB,  and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend,  that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4]  And, since the whole AF is equal to the whole AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG.  But FC was also proved equal to GB;  therefore the two sides BF, FC are equal to the two sides CG, GB respectively;  and the angle BFC is equal to the angle CGB, while the base BC is common to them;  therefore the triangle BFC is also equal to the triangle CGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend;  therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG.  Accordingly, since the whole angle ABG was proved equal to the angle ACF,  and in these the angle CBG is equal to the angle BCF,  the remaining angle ABC is equal to the remaining angle ACB;  and they are at the base of the triangle ABC.  But the angle FBC was also proved equal to the angle GCB;  and they are under the base. 
Quia ergo duo latera A B, A F, trianguli A B F, æqualia sunt duobus lateribus A C, A D, trianguli A C D, utrumque utrique, nempe A B, ipsi A C, ex hypothesi, & A F, ipsi A D, ex constructione; angulusque A, contentus lateribus A B, A F, æqualis est angulo A, contento lateribus A C, A D,  immo angulus A, communis est utrique triangulo:   Erit basis B F, æqualis basi C D;  & angulus F, angulo D; & angulus A B F, angulo A C D; cum & priores duo, & posteriores opponantur æqualibus lateribus in dictis triangulis, ut patet. Rursus considerentur duo triangula B D C, C F B.      Quoniam vero rectæ A D, A F, æquales sunt, per constructionem, fit ut, si auferantur ex ipsis æquales A B, A C, & reliquæ B D, & C F, sint æquales.    Quare duo latera B D, D C, trianguli B D C, æqualia sunt duobus lateribus C F, F B, trianguli C F B, utrumque utrique, videlicet B D, ipsi C F, & D C, ipsi F B, ut probatum est.  Sunt autem & anguli D, & F, contenti dictis lateribus æqualibus æquales, ut ostensum etiam fuit. Igitur erit angulus D B C, angulo F C B, æqualis; & angulus B C D, angulo C B F. Tam enim priores duo, quam posteriores, æqualibus opponunturlateribus, existuntque supra communem basim B C, utriusque trianguli B D C, C F B.          Quod si ex totis angulis æqualibus A B F, A C D, (quos æquales esse iam demonstrauimus in prioribus triangulis) detrahantur anguli æquales C B F, B C D, (quos itidem in posterioribus triangulis modo probauimus esse æquales) remanebunt anguli A B C, A C B, supra basim B C, æquales.    Ostensum est autem in posterioribus triangulis, & angulos D B C, F C B, qui quidem sunt infra eandem basim B C, esse æquales.    
論曰。試如甲戊線稍長。卽從甲戊截取一分。與甲丁等。為甲己。本篇三 次自丙至丁乙至己。各作直線。第一求卽甲己乙、甲丁丙、兩三角形必等。  何者此兩形之甲角同。 
Therefore etc.    Q. E. D. 
Igitur & anguli supra basim inter se, & anguli infra eandem inter se sunt æquales;   Ac propterea isoscelium triangulorum, qui ad basim sunt anguli, &c.   Quod erat demonstrandum.  
 
SCHOLION
HÆC propositio vera etiam est in triangulis æquilateris, cum in quolibet reperiantur duo laterainter se æqualia, licet eam Euclides solis isoscelibus triangulis videatur accommodasse. Existentibus enim duobus lateribus A B, A C, trianguli A B C, æqualibus; siue reliquum latus B C, ipsis quoque sit æquale, ut contingit in triangulo æquilatero, siue inæquale, ut in isoscele accidit, necessario consequitur, & angulos supra basim inter se, & angulos infra eandem inter se quoque esse æquales, vt constat ex demonstratione prædicta. Solet autem theorema hoc tyronibus subdifficile, & obscuriusculum videri, propter multitudinem linearum, & angulorum, quibus nondum sunt assueti. Verum tamen, si diligenter theorematis præcedentis uis ac demonstratio ponderetur, non multo labore boc, quod præ manibus babemus, a quolibet percipietur, si modo memor sit, illos angulos triangulorum probari æquales esse in antecedenti theoremate, qui æqualibus lateribus opponuntur. Quod quidem, quoniam Campanus non apposuit, causa fuit, ut confusa esse videatur, & subobscura eius demonstratio.
COROLLARIUM
Ex hac propofitione quinta liquet, omne triangulum æquilaterum esse æquiangulum quoque: Hoc est, tres angulos cuiuslibet trianguli æquilateri esse inter se æquales. Sit enim triangulum æquilaterum A B C. Quoniam igitur duo latera A B, A C, sunt æqualia, erunt duo anguli B, C æquales. Item quia duo latera A B, B C, sunt æqualia, erunt & anguli C, & A, æquales. Quare omnes tres A, B, & C, æquales erunt. Quod ostendendum erat. 
Proposition 6. 
THEOR. 3. PROPOS. 6. 
第六題 
If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another. 
SI trianguli duo anguli æquales inter se fuerint: & sub æqualibus angulis subtensa latera æqualia inter se erunt. 
三角形。若底線兩端之兩角等。則兩腰亦等。 
Let ABC be a triangle having the angle ABC equal to the angle ACB;  I say that the side AB is also equal to the side AC. 
IN triangulo A B C, sint duo anguli A B C, A C B, super latus B C, æquales.  Dico duo latera illis opposita A B, A C, esse quoque æqualia. 
解曰:甲乙丙三角形。其甲乙丙,與甲丙乙,两角等。  題言甲乙,與甲丙,两腰亦等。 
For, if AB is unequal to AC, one of them is greater.  Let AB be greater; and from AB the greater let DB be cut off equal to AC the less; let DC be joined. 
Si enim non credantur æqualia, existentibus nihilominus angulis dictis æqualibus, erit alterum maius altero;  sit igitur A B, maius quam A C, si fieri potest: Et ex A B,73 abscindatur in D, recta B D, æqualis rectæ A C, (quæ minor dicitur esse, quam A B,) ducaturque recta C D. 
論曰。如云两腰線不等,而一長一短。試辯之。  若甲乙為長線。即令比甲丙線截去所長之度,為乙丁線。 
Then, since DB is equal to AC, and BC is common,  the two sides DB, BC are equal to the two sides AC, CB respectively;  and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB,  and the triangle DBC will be equal to the triangle ACB, the less to the greater:  which is absurd.  Therefore AB is not unequal to AC; it is therefore equal to it.   Therefore etc. 
Considerentur iam duo triangula A C B, D B C. In quibus cum duo latera A C, C B, trianguli A C B, 
æqualia sint duobus lateribus D B, B C, trianguli D B C, utrumque utrique, nempe A C, ipsi D B, (abscidimus enim ex A B, ipsi A C, concessu aduersarij, æqualem D B,) & C B, ipsi B C, cum sit unum & idem;   Sint autem & anguli A C B, D B C, contenti dictis lateribus æquales, per hypothesin: 74   Erunt triangula A C B, D B C, æqualia, totum, & pars;  quod fieri non potest.   Non igitur erunt latera A B, A C, inæqualia, si anguli B, & C, super latus B C, æquales sunt, ne totum parti æquale esse concedamus: sed æqualia existent. Quare si trianguli duo anguli, &c.    
而乙丁,與甲丙等本篇三。  次自丁至丙作直線。則本形成两三角形。其一為甲乙丙。其一為丁乙丙。    而甲乙丙全形,與丁乙丙分形同也。  是全與其分等也公論九。  何者。彼言丁乙丙分形之乙丁,與甲乙丙全形之甲丙,两線既等。丁乙丙分形之乙丙,與甲乙丙全形之乙丙,又同線。而元設丁乙丙,與甲丙乙,两角等。則丁乙丙,與甲乙丙,两形亦等也本篇四是全與其分等也。故底線两端之两角等者。两腰必等也。   
Q. E. D. 
Quod demonstrandum erat.

SCHOLION
CONVERTIT hoc theorema primam partem præcedentis. Nam ibi demonstratum est, si duo latera trianguli inter se æqualia fuerint, angulos, qui ad basim sunt, esse quoque æquales: Hic vero, si anguli ad basim sint æquales, latera quoque angulis illis opposita esse æqualia. Non autem mirum alicui debet videri, si Mathematici aliquando conuertunt propositiones, ita ut nunc ex antecedente quopiam concesso colligant per demonstrationem consequens aliquod, nunc vero rursus ex consequente hoc concesso inferant per aliam demonstrationem antecedens illud, ut ab Euclide in hisce duabus proximis propositionibus factum esse conspicimus: Non debet, inquam, videri mirum, quoniam non semper in rebus Mathematicis reciprocantur antecedens & consequens. Nam in propositionibus necessariis, quales sunt propositiones Geometricæ, potest interdum prædicatum esse uniuersalius subiecto, ut cum Dialecticis loquamur. Quare tunc non poterit conuerti propositio. Nam necessaria est hæc propositio; (Omnis homo est animal.) non tamen conuerti potest uniuersaliter, cum non omne animal sit homo. Ita quoque fieri potest in propositionibus Geometricis necessariis: Cuius ego rei vnum duntaxat nunc exemplum tale in medium proferam. Demonstrat Euclides propos. 16. huius lib. Si trianguli cuiusuis vnum latus producatur, angulum externum maiorem esse duobus internis sibi oppositis; In qua quidem propositione nullo modo antecedens, & consequens reciprocantur. Non enim sequitur, si figuræ cuiusuis rectilineæ uno latere producto, angulus externus maior sit singulis internis oppositis, figuram illam esse triangulum, cum possit etiam esse quadrilatera figura, ut ad propositionem 16. huius liber ostendemus. Eodemque modo multæ aliæ propositiones conuerti nequeunt. Quam ob rem necesse est, ut prius demonstret Geometra, propositionem aliquam conuerti, hoc est, antecedens, & consequens illius reciprocari, antequam ex consequente concesso colligat antecedens. Non conuertit autem Euclides omnes propositiones, quæ conuerti possunt, sed eas duntaxat, quarum conuersione maxime indiget: Nos tamen dabimus operam, ut fere omnes illas conuertamus, quæ aliquam videbuntur afferre utilitatem.

COROLLARIUM
SEQVITVR ex hac propositione, omne triangulum æquiangulum, id est, cuius omnes anguli sunt æquales, esse æquilaterum. Quod quidem conuersum est corollarij quintæ propositionis, ut liquet. Sint enim trianguli A B C, tres anguli æquales. Dico ipsum esse æquilaterum. Cum enim duo anguli B, & C, sint æquales, erunt latera A B, A C, æqualia. Rursus cum duo anguli A, & B, sint æquales, erunt quoque latera A C, B C, æqualia, & idcirco omnia tria latera A B, B C, A C, æqualia. Quod ostendendum erat.

EX PROCLO
LICEBIT nobis etiam conuertere secundam partem quintæ propositionis, hoc modo.
SI trianguli cuiuslibet productis duobus lateribus, anguli infra basim fiant æquales, & duo laterailla æqualia inter se erunt.
Trianguli enim A B C, productis lateribus A B, A C, ad D, & E, fiant anguli D B C, E C B, infra basim B C, æquales. Dico latera A B, A C, esse quoque inter se æqualia. Ex C E, quantumlibet producta abscindatur C F, æqualis ipsi B D, & ducantur rectæ B F, F D, D C. Considerentur deinde triangula D B C, F C B. In quibus cum latera D B, B C, æqualia sint lateribus F C, C B, utrumque utrique, nempe D B, ipsi F C, per constructionem, & B C, ipsi C B, quod sit unum & idem: sint autem & anguli D B C, F C B, dictis lateribus contenti æquales, per hypothesim: erunt & bases C D, B F, & anguli B C D, C B F, super has bases, cum opponantur æqualibus lateribus B D, C F, æquales. Ablatis igitur hisce angulis æqualibus B C D, C B F, ex angulis F C B, D B C, per bypothesin, æqualibus; remanebunt anguli F C D, D B F, æquales. Considerentur rursus triangula D B F, F C D. In quibus quoniam latera D B, B F, æqualia sunt lateribus F C, C D, utrumque utrique, nempe D B, ipsi F C, per constructionem, & B F, ipsi C D, ut modo ostensum est; Sunt autem & anguli contenti dictis lateribus D B F, F C D, æquales, ut eiam fuit nuper demonstratum: Erit angulus B D F, super basim D F, trianguli D B F, æqualis angula C F D, super eandem basim F D, trianguli F C D. Hi enim æqualibus lateribus opponuntur. Cum igitur in triangulo A D F, duo anguli A D F, A F D, sint æquales, ut nunc ostendimus, erunt latera A D, A F, æqualia. A quibus si rectæ B D, C F, per constructionem, æquales demantur, remanebunt A B, A C, latera trianguli A B C, æqualia.
Quod erat ostendendum.  
 
Proposition 7. 
THEOR. 4. PROPOS. 7. 
第七題 
Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it. 
SVPER eadem recta linea, duabus eisdem rectis lineis aliæ duæ rectæ lineæ æquales, vtraque vtrique, non constituentur, ad aliud atque aliud punctum, ad easdem partes, eosdemque terminos cum duabus initio ductis rectis lineis habentes. 
一線為底。出兩腰線。其相遇止有一點。不得別有腰線與元腰線等。而於此點外相遇。 
For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it,  so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined. 
SVPER recta A B, constituantur ad punctum quoduis C, duæ rectæ lineæ A C, B C. Dico super eandem rectam A B, versus partem eandem C, non posse ad aliud punctum, vt ad D, constitui duas alias rectas lineas,   quæ sint æquales lineis A C, B C, vtraque vtrique, nempe A C, ipsi A D, quæ eundem habent terminum A; & B C, ipsi B D, quæ eundem etiam terminum possident B. 
Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5]  therefore the angle ADC is greater than the angle DCB;  therefore the angle CDB is much greater than the angle DCB.  Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB.  But it was also proved much greater than it: which is impossible. 
Sint enim, si fieri potest, rectæ A C, A D, inter se, & rectæ B C, B D, inter se etiam æquales. Aut igitur punctum D, erit in alterutra rectarum A C, B C, ita vt recta A D, in ipsam rectam A C vel B D, in ipsam B C, cadat; aut intra triangulum A B C; aut extra. Sit primo punctum D, in altera rectarum A C, B C, nempe in A C, vt A D, sit pars ipsius A C. Quoniam igitur rectæ A C, A D, eundem terminum A, habentes dicuntur æquales, erit pars A D, toti A C, æqualis. Quod fieri non potest. Sit deinde punctum D, intra triangulum A B C, & ducta recta C D, producantur rectæ B C, B D, vsque ad E, & F. Quoniam igitur in triangulo A C D, ponuntur latera A C, A D, æqualia erunt anguli A C D, A D C, super basim C D, æquales;   Est autem angulus A C D, minor angulo D C E; nempe pars toto: Igitur & angulus A D C, minor erit eodem angulo D C E.   Quare angulus C D F, pars ipsius A D C, multo minor erit eodem angulo D C E.  Rursus, quia in triangulo B C D, latera B C, B D, ponuntur æqualia, erunt anguli C D F, D C E, sub basi C D, æquales.  Ostensum autem fuit, quod idem angulus C D F, multo sit minor angulo D C E. Idem ergo angulus C D F, & minor est angulo D C E, & eidem æqualis, quod est absurdum. Sit postremo punctum D, extra triangulum A B C. Aut igitur in tali erit loco, vt vna linea super alteram cadat, vt in priori figura, dummodo loco D, intelligas C, & loco C, ipsum D; ex quo rursus colligetur pars æqualis toti, quod est absurdum. Aut in tali erit loco, vt posteriores duæ lineæ ambiant priores duas, ceu in posteriori figura, si modo loco D, iterum intelligas C, & D, loco C. Quo posito, inidem absurdum incidemus, nempe angulum D C F, & minorem esse angulo C D E, & eidem æqualem, vt perspicuum est. Aut denique punctum D, ita erit extra triangulum A B C, vt altera linearum posteriorum, nempe A D, secet alteram priorum, vt ipsam B C. Ducta igitur recta C D, cum in triangulo A C D, latera A C, A D, ponantur æqualia, erunt anguli A C D, A D C, supra basim C D, æquales: Ac proinde cum angulus A D C, minor sit angulo B D C, pars toto, erit & angulus A C D, minor eodem angulo B D C. Quare multo minor erit angulus B C D, pars anguli A C D, angulo eodem B D C. Rursus, cum in triangulo B D C, latera B C, B D, ponantur æqualia, erunt anguli B C D, B D C, super basim C D, æquales: Est aut etiam ostensum, angulum B C D, multo esse minorem angulo B D C. Idem igitur angulus B C D, & minor est angulo B C D, & eidem æqualis, quod est absurdum. Non ergo æquales sunt inter se A C, A D, & inter se quoque B C, B D.  
Therefore etc.  Q. E. D. 
Quare super eadem recta linea, duabus eisdem rectis lineis, &c.   Quod erat demonstrandum.

SCHOLION
FIERI potest, vt duæ lineæ A D, B D, æquales sint duabus A C, B C, vtraque vtrique, vt A D, ipsi B C, & B D, ipsi A C, vt vltima figura indicat. Verum hoc modo non egrediuntur ab eodem puncto lineæ illæ, quæ sunt æquales inter se, vt constat. Solæ enim A C, A D, eundem limitem possident A; Item B C, B D, eundem B; optimeque demonstratum fuit ab Euclide, fieri non posse, vt A C, A D, inter se sint æquales, ita vt B C, B D, quoque inter se æquales existant. Recte igitur in propositione apposita sunt hæc verba: eosdemque terminos cum duabus initio ductis rectis lineis habentes. Rursus possunt esse duæ lineæ simul sumptæ A D, B D, æquales duabus lineis A C, B C, simul sumptis, vt in eadem figura perspici potest: Sed hoc non ostendit Euclides fieri non posse. Dixit enim non posse vtramque vtrique esse æqualem, &c.
Eadem ratione possunt ex A, & B, infra A B, basim trianguli A B C, hoc est, ad contrarias partes, duci duæ lineæ rectæ A D, B D, conuententes ad aliquod punctum, ita vt A D, exiens e puncto A, æqualis sit ipsi A C; & B D, egrediens ex B, æqualis ipsi B C, vt perspicuum est in apposita figura. Non igitur sine causa adiecit Euclides: ad easdem partes. Denique esse poterunt duæ lineæ A C, A D, æquales inter se eundem terminum A, possidentes; Sed hoc posite, fieri nulla ratione poterit, vt reliquæ duæ B C, B D, terminum habentes eundem B, inter se quoque sint æquales, vt in hac figura apparet, & ab Euclide est demonstratum. Apposite igitur dictum est in propositione: duabus eisdem rectis lineis aliæ duæ rectæ lineæ æquales, vtraque vtrique, &c. Quare vt plane scopus Euclidi in hac propositione propositus intelligatur, diligenter singula verba propositionis sunt ponderanda. 
Proposition 8. 
THEOR. 5. PROPOS. 8. 
第八題 
If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines. 
SI duo triangula duo latera habuerint duobus lateribus; vtrumque vtrique æqualia, habuerint verò & basim basi æqualem: Angulum quoque sub æqualibus rectis lineis contentum angulo æqualem habebunt. 
兩三角形。若相當之兩腰各等。兩底亦等。則兩腰間角必等。 
Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF;  and let them have the base BC equal to the base EF;  I say that the angle BAC is also equal to the angle EDF. 
SINT duo latera A B, A C, trianguli A B C, duobus lateribus D E, D F, trianguli D E F, æqualia, vtrumque vtrique, nempe A B, ipsi D E, & A C, ipsi D F;   sit autem & basis B C, basi E F, æqualis.   Dico angulum A, æqualem esse angulo D, quorum videlicet vterque dictis lateribus continetur.  
For, if the triangle ABC be applied to the triangle DEF, and if the point B be placed on the point E and the straight line BC on EF, the point C will also coincide with F, because BC is equal to EF.  Then, BC coinciding with EF, BA, AC will also coincide with ED, DF;  for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF,  then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.  But they cannot be so constructed. [I. 7]  Therefore it is not possible that, if the base BC be applied to the base EF, the sides BA, AC should not coincide with ED, DF;  they will therefore coincide, so that the angle BAC will also coincide with the angle EDF, and will be equal to it. 
Nam si mente intelligatur basis B C, superponi basi E F, neutra excedet alteram, sed punctum B, congruet puncto E, & punctum C, puncto F, cum hæ bases ponantur æquales inter se.     Deinde si triangulum A B C, cogitetur cadere super triangulum D E F, cadet punctum A, aut in ipsum punctum D, aut alio. Si punctum A, in ipsum punctum D, cadat, congruent sibi mutuo triangulorum latera, cum ponantur æqualia; Ac propterea angulus A, æqualis erit angulo D, cum neuter alterum excedat.  Quod si punctum A, alio dicatur cadere, ut ad G, quomodocunque id contingat, hoc est, sive in latus E D, sive intra triangulum E D F, sive extra, ut in figuris apparet; erit perpetuo E G, (quæ eadem est, quæ B A,) æqualis ipsi E D; & F G, (quæ eadem est, quæ C A) æqualis ipsi F D, propterea quod latera unius trianguli æqualia ponuntur lateribus alterius.    Hoc autem fieri non posse, iamdudum demonstratum est, cum tam rectæ E G, E D, terminum eundem E, quam rectæ F G, F D, eundem limitem F, possideant.   Non igitur punctum A, cadet alio quam in punctum D: ac propterea angulus A, angulo D, æqualis erit. 
If therefore etc.  Q. E. D. 
Quare si duo triangula duo latera habuerint duobus lateribus, &c.   Quod erat demonstrandum.

SCHOLION
VT vides, hæc propositio conuertit primam partem propositionis quartæ. Sicut enim ibi ex æqualitate angulorum, qui lateribus æqualibus continentur, collecta fuit basium æqualitas; ita hic ex æqualitate basium concludit Euclides æqualitatem angulorum, qui lateribus æqualibus comprehenduntur. Possumus eodem modo ex prima, & tertia parte conclusionis quartæ propositionis inferre totum antecedens eiusdem, ita ut theorema proponatur in hanc formam.
SI duo triangula bases habuerint æquales, & angulos super bases constitutos æquales, utrumque utrique: Habebunt quoque reliqua latera æqualia, utrumque utrique, quæ videlicet æqualibus angulis subtenduntur, angulosque reliquos hisce lateribus inclusos æquales.
SIT enim basis B C, æqualis basi E F, & angulus B, angulo E, angulusque C, angulo D F E. Dico latus quoque A B, lateri D E, & latus A C, lateri D F, æquale esse, angulumque A, angulo D. Nam si basis basi superponatur, congruent sibi mutuo extrema earum, nec non & lineæ angulorum æqualium. Quare omnia sibi congruent, proptereaque omnia inter sese æqualia erunt. Verum hoc idem theorema a nobis propositum, quod quidem magis proprie conuertere videtur quartam propositionem, quam illud Euclidis, aliter demonstrabit Euclides in prima parte propositionis 26. ut eo loco monebimus.

COROLLARIUM
PORRO ex antecedente huius octauæ propositionis non solum colligi potest, angulos lateribus æqualibus contentos æquales esse, verum etiam reliquos angulos, qui ad bases constituuntur, utrumque utrique, ut angulum B, angulo E, & angulum C, angulo F; imo totum triangulum toti triangulo, ut constat ex eadem superpositione unius trianguli super alterum. Nam sibi mutuo congruent & dicti anguli, & tota triangula, ut perspicuum est. Quod etiam ex quarta propositio colligi poterit, postquam demonstratum fuerit, angulos æqualibus comprehensos lateribus æquales esse. Inde enim fiet, cum latera quoque sint æqualia, & reliquos angulos, & tota triangula esse æqualia, ut in propositio 4. demonstratum est.

EX PROCLO
PHILONIS familiares conantur hoc idem theorem a octavum ostenduns demonstratione affirmativa, hac ratione. Posito enim eodem antecedente, superponi intelligatur basis B C, basi E F, ita ut triangulum A B C, cadat in diuersas partes, & non super triangulum D E F, quare est triangulum A E F. Aut igitur duo latera, nempe D F, F A, constituunt unam lineam rectam, quod quidem continget, si duo anguli C, & F, recti extiterint; aut non. Si constituant unam lineam rectam, veluti D A, ita propositum concludetur. Quoniam in triangulo A E D, duo latera A E, D E, ponuntur æqualia (est enim nunc A E, recta eadem, quæ A B, quæ per hypothesin rectæ D E, æqualis est) erunt anguli A, & D, super basin A D, æquales, quod erat ostendendum. Si vero neque D E, F A, neque D E, E A, lineam rectam conficiant, ducatur ex D, ad A, linea recta D A, quæ vel cadet intra triangula, vel extra. Cadat primum intra, quod quidem accidet, quando anguli ad E, & F, sunt acuti. Quoniam igitur in triangulo A E D, duo latera A E, D E, æqualia ponuntur, erunt duo anguli E A D, E D A, æquales ad basim D A. Eadem ratione, cum duo latera A F, D F, æqualia sint per hypothesin, erunt duo anguli F A D, F D A, super basin D A, æquales. Si igitur hi æquales illis æqualibus addantur, fient toti anguli E A F, E D F, æquales. Quod erat ostendendum. Cadat deinde recta D A, extra triangula, quod demum fiet, quando anguli ad F, fuerint obtusi. Quoniam igitur in triangulo A E D, duo latera A E, D E, ponuntur æqualia, erunt anguli E A D, E D A, æquales super basin D A. Eadem ratione, cum duo latera A F, D F, in triangulo A F D, sint per hypothesin æqualia, erunt anguli F A D, F D A, super basin D A, æquales. His ergo a prioribus ablatis, remanebunt anguli E A F, E D F, æquales; Quod demonstrandum proponebatur.
 
Proposition 9. 
PROBL. 4. PROPOS. 9. 
第九題 
To bisect a given rectilineal angle. 
DATVM angulum rectilineum bifariam secare. 
有直線角, 求兩平分之。 
Let the angle BAC be the given rectilineal angle.  Thus it is required to bisect it. 
SIT diuidendus rectilineus angulus B A C,  bifariam, hoc est, in duos angulos æquales.  
法曰: 乙甲丙角,  求兩平分之。 
Let a point D be taken at random on AB; let AE be cut off from AC equal to AD; [I. 3] let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined.  I say that the angle BAC has been bisected by the straight line AF. 
In recta A B, sumatur quodcunque punctum D, &rectæ A D, secetur ex A C, recta A E, 75 æqualis, ducaturque recta D E. Deinde super D E, constituatur triangulum æquilaterum D F E, & 76 ducatur recta A F, diuidens angulum B A C, in angulos B A F, C A F.  
75. 3 primi.  76. 1 primi. 
Dico hos angulos inter se esse æquales.  
先於甲乙線任截一分為甲丁, 本篇三。 次於甲丙亦截甲戊與甲丁等, 次自丁至戊作直線, 次以丁戊為底, 立平邊三角形, 本篇一。為丁戊己形, 末自己至甲作直線,   卽乙甲丙角為兩平分。 
For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively.  And the base DF is equal to the base EF;  therefore the angle DAF is equal to the angle EAF. [I. 8] 
Cum enim latera D A, A F, trianguli D A F, æqualia sint lateribus E A, A F, trianguli E A F, utrumque utrique, quod D A, ipsi E A, per constructionem, sit æquale, & A F, commune;  Sit autem & basis D F, basi E F, æqualis, propterea quod triangulum D F E, constructum est æquilaterum.  Erit angulus D A E, angulo E A F, æqualis,  
Therefore the given rectilineal angle BAC has been bisected by the straight line AF.  Q. E. F. 
ideoque angulus B A C, diuisus bifariam,  quod erat faciendum.

SCHOLION
QUOD si loco trianguli æquilateri construamus triangulum Isosceles, nihilo minus idem demonstrabimus. Id quod etiam in proximis tribus propositionibus, quæ sequuntur, fieri potest.

PRAXIS
DICTO citius angulus quilibet rectilineus, ut B A C, bifariam secabitur, hoc modo. Ex centro A, circino aliquo abscindantur rectæ æquales A D, A E, cuiuscunque magnitudinis. Et circino non variato (posses tamen ipsum variare, si velles) ex centris D, & E, describantur duo arcus secantes sese in F. Recta igitur ducta A F, secabit angulum B A C, bifariam. Si enim ducerentur rectæ D F, E F, essent hæ æquales, nempe semidiametri circulorum æqualium. Unde ut prius demonstrabitur, angulum D A F, æqualem esse angulo E A F. Non descripsimus autem dictas lineas, ut nudæ praxis haberetur. Id quod in aliis quoque praxibus, quoad eius fieri poterit, obseruabimus, ne linearum multitudo tenebras nobis offundat, pariatque confusionem.

SCHOLION
HINC aperte colligitur, angulum rectilineum quemvis diuidi posse etiam in 4. angulos æquales, in 8. in 16. in 32. in 64. & ita deinceps, semper procedendo per augmentum duplex. Nam postquam angulus quilibet rectilineus in duos æquales angulos fuerit diuisus, si horum uterque iterum bifariam secetur, habebimus 4. angulos æquales; Quod si singuli rursus diuidantur bifariam, obtinebimus 8. angulos æquales, & sic deinceps; Non docuit autem Euclides usquam, quanam ratione angulus rectilineus in quotvis partes æquales possit diuidi, quia id a nemine usque ad illum diem fuerat demonstratum. Ex Pappo tamen Alexandrino nos id docebimus, beneficio cuiusdam lineæ curvæ, vel inflexæ, ad finem lib. 6. Interim vero, si quis angulum rectilineum quemcunque propositum in quotvis partes æquales diuidere desideret, rudi, ut dicitur, Minerva, uti eum necesse erit circino, ut quasi attentando, & sæpius repetendo praxim ipsam ad finem desideratum perveniat; hac nimirum ratione. Sit angulus rectilineus B A C, diuidendus in 5. angulos æquales. Ex A, centro describatur arcus circuli B C, ad quodcunque intervallum, secans rectas A B, A C, in B, & C. Deinde hic arcus beneficio circini (eius crura modo dilatando magis, modo restringendo, donec debitam habeant distantiam) diuidatur in tot partes æquales, in quot angulus propositus est diuidendus, ut in exemplo proposito in quinque partes in punctis D, E, F, G. Si namque ad hæc puncta ex A, rectæ ducantur lineæ, diuisus erit angulus B A C, in quinque æquales angulos. Cum enim circino sumpta sint æqualia interualla B D, D E, &c. si ducantur rectæ B D, D E, &c. erunt hæ omnes inter se æquales. Quare erunt duo latera B A, A D, trianguli B A D, æqualia duobus lateribus E A, A D, trianguli E A D, utrumque utrique, cum omnia ex centro egrediantur ad circumferentiam usque. Basis autem B D, basi quoque D E, ut dictum fuit, æqualis est: Angulus igitur B A D, angulo E A D, æqualis existet; Eademque ratione demonstrabitur, angulum E A D, angulo E A F, æqualem esse, & sic de cæteris. Brevius autem colligetur, omnes angulos ad A, esse inter se æquales, ex 27. propositio tertii lib. propterea quod circumferentiæ B D, D E, &c. acceptæ sunt omnes æquales inter sese. Nemo vero miretur, quod praxes exhibeamus interdum, quarum demonstrationes ex sequentibus propositionibus pendent. Hoc enim, ut supra ad defin. 10. diximus, eo consilio facimus, ut quoad eius fieri potest, singula propriis in locis tractentur, diuisio nimirum anguli rectilinei cuiusvis in quotlibet partes æquales eo in loco, in quo Euclides docet diuisionem eiusdem anguli in duas partes æquales. Et diuisio lineæ rectæ in quotuis partes æquales, ubi eandem diuidit Euclides bifariam, & ita de singulis. Neque enim ad praxes huiusmodi requiruntur semper sequentes demonstrationes, sed solum, ut probetur recte esse per ipsas effectum, quod imperabatur. Quamobrem is, qui non contentus nuda praxi demonstrationem requirit, poterit regredi ad praxin quamlibet, postquam demonstrationes ad eam necessarias diligenter perceperit. Nam semper propositiones illas, quæ ad hanc rem debent adhiberi, citabimus in demonstrationibus nostrarum praxium; quemadmodum & in proxima praxi citavimus propositionem 27, tertii libri.
 
Proposition 10. 
PROBL. 5. PROPOS. 10. 
第十題 
To bisect a given finite straight line. 
DATAM rectam lineam finitam bifariam secare. 
一有界線。求兩平分之。 
Let AB be the given finite straight line.  Thus it is required to bisect the finite straight line AB. 
SIT recta finita A B,   diuidenda bifariam, id est, in duas partes æquales. 
Let the equilateral triangle ABC be constructed on it, [I. 1]  and let the angle ACB be bisected by the straight line CD; [I. 9]  I say that the straight line AB has been bisected at the point D. 
Describatur super A B, triangulum æquilaterum A B C,   cuius angulus C, per rectam C D, diuidatur bifariam, rectaque C D, rectam A B, secet in D.   Dico rectam A B, bifariam esse diuisam in D.  
For, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD respectively;  and the angle ACD is equal to the angle BCD;  therefore the base AD is equal to the base BD. [I. 4] 
Quoniam duo latera A C, C D, trianguli A C D, æqualia sunt duobus lateribus B C, C D, trianguli B C D, utrumque utrique, nempe A C, ipsi B C, cum sint ambo latera trianguli æquilateri, & C D, est commune;  Est autem & angulus A C D, angulo B C D, æqualis, per constructionem:   Erit basis A D, basi B D, æqualis.  
Therefore the given finite straight line AB has been bisected at D.  Q. E. F. 
Datam ergo rectam A B, bifariam secuimus in D,   quod facere oportebat.

PRAXIS
EX centro A, ad quoduis intervallum, quod tamen dimidium linea A B, excedat, describantur duo arcus, unus superne, alter inferne; Et ex centro B, ad idem intervallum omnino alii duo arcus delinquentur, qui priores secent in C, & D. Recta enim ducta C D, secabit rectam A B, in E, bifariam. Si enim ex A, & B, ad C, & D, ducantur quatuor rectæ, erunt hæ omnes inter se æquales, cum ex centris ad circumferentias æqualium circulorum cadant; Nam arcus circulorum descripti sunt eodem intervallo. Quoniam igitur latera A B, C D, æqualia sunt lateribus B C, C D, utrumque utrique, & basis A D, basi B D, erit angulus A C D, angulo B C D, æqualis. Rursus quia latera A C, C E, æqualia sunt lateribus B C, C E, utrumque utrique, & angulus A C E, angulo B C E, ut ostensum fuit; erit basis A E, basi B E, æqualis.

SCHOLION
PERSPICVVM est, eodem modo dividi posse eandem lineam rectam A B, in 4. partes æquales, & in 8. in 16. in 32. &c. sicuti in propositione præcedenti diximus de diuisione trianguli rectilinei. Qua vero ratione quævis recta linea proposita diuidenda sit in quotcunque partes æquales, uberrime trademus ad propositionem 10. liber 6. ubi varias, & non iniucundas praxes in medium adducemus. Ibi enim videtur esse proprius huic rei locus, cum huiusmodi praxes fere omnes per linearum proportiones facilius demonstrentur. Neque vero unquam indigebimus diuisione lineæ in plures, quam in duas partes æquales, ad eum locum usque.
 
Proposition 11. 
PROBL. 6. PROPOS. 11. 
第十一題 
To draw a straight line at right angles to a given straight line from a given point on it. 
DATA recta linea, a puncto in ea dato, rectam lineam ad angulos rectos excitare. 
一直線。任於一點上求作垂線。 
Let AB be the given straight line, and C the given point on it.  Thus it is required to draw from the point C a straight line at right angles to the straight line AB. 
RECTA linea data sit A B; & in ea punctum C,   a quo iubemur erigere super A B, lineam ad angulos rectos, seu perpendicularem. 
法曰:甲乙直線,任指一點於丙。 
Let a point D be taken at random on AC; let CE be made equal to CD; [I. 3]  on DE let the equilateral triangle FDE be constructed, [I. 1]  and let FC be joined;  I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it. 
A puncto C, sumatur recta C D, cui æqualis auferatur C E.   Deinde super D E, constituatur triangulum æquilaterum D E F,  atque ex F, ad C, ducatur recta F C,  quam dico esse perpendicularem ad A B.  
For, since DC is equal to CE, and CF is common,  the two sides DC, CF are equal to the two sides EC, CF respectively;  and the base DF is equal to the base FE;  therefore the angle DCF is equal to the angle ECF; [I. 8]  and they are adjacent angles.  But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [Def. 10]  therefore each of the angles DCF, FCE is right. 
Quoniam latera D C, C F, trianguli D C F, æqualia sunt lateribus E C, C F, trianguli E C F, utrumque utrique, nempe D C, ipsi E C, per constructionem, & C F, commune;    Est vero & basis D F, basi E F, æqualis, ob triangulum æquilaterum:   Erunt anguli ad C, contenti dictis lateribus, æquales.       Quare dicetur uterque rectus, 
Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it.  Q. E. F. 
atque adeo F C, recta, ad A B, perpendicularis. Data igitur recta linea a puncto in ea dato, &c.   Quod faciendum erat.

PRAXIS
EX puncto C, abscindantur utrinque lineæ æquales C D, C E, & ex D, & E, describantur duo arcus secantes sese in F. Recta namque ducta F C, erit perpendicularis. Demonstratio eadem est, quæ Euclidis, si modo ducantur rectæ D F, E F, quæ æquales erunt, propter æquales circulos ex D, & E, descriptos, qui se intersecant in puncto F. Quod si punctum datum in linea recta fuerit extremum, producenda erit linea in rectum & continuum, ad partes puncti dati, ut ex illo erigatur secundum praxim datam linea perpendicularis. Ut si linea data fuerit A C, & punctum datum C, extremum; protrahenda erit A C, in B, & sumendæ æquales C D, C E, &c. Si vero ad aliquam lineam constituenda sit linea perpendicularis, non quidem in puncto assignato, sed utcunque, id efficietur hac methodo. Ex duobus punctis A, & B, quibuscunque lineæ propositæ describantur tam superne, quam inferne duo arcus sese intersecantes in C, & D. Nam recta ducta C D, erit perpendicularis ad A B, hoc est, faciet duos angulos ad E, rectos, seu æquales. Quod non aliter probabis, quam supra praxim, qua lineam in duas æquales diuisimus partes, demonstrauimus. Nam per 4. propos. erunt anguli ad E, æquales, quippe qui super æquales bases A E, B E, consistant, opponanturque æqualibus lateribus A C, B C, quæ ex C, ad puncta A, & B, ducerentur.

EX PROCLO
SI punctum in linea datum fuerit extremum, & linea commods produci nequiuerit, poterimus ex puncto dato educere lineam perpendicularem, linea non producta, hac ratione. Sit recta A B, & punctum A. Ex C, puncto quolibet intra lineam educatur perpendicularis C D, ut docuit Euclides; & abscindatur C E, æqualis ipsi A C: Deinde diuidatur angulus C, bifariam, ducta recta C F: Et ex E, rursus, ut docuit Euclides, educatur E G, perpendicularis ad C D, secans rectam C F, in G. Ducta enim recta G A, perpendicularis erit ad A B. Quoniam eum latera A C, C G, triangula A C G, æqualia sint lateribus E C, C G, trianguli E C G, utrumque utrique, & anguli hisce lateribus contenti æquales quoque, per constructionem: Erunt anguli A, & E, oppositi communi lateri C G, æquales; Sed E, est rectus per constructionem; igitur & A, rectus erit, ideoque A C, ad A B, perpendicularis.

SCHOLION
BREVIVS lineam perpendicularem erigemus ex puncto dato, sive extremum illud sit, sive non, hoc modo. Sit data linea A B, punctumque in ea A. Ex centro C, extra lineam assumpto, ubi libuerit (dummodo recta A B, producta cum ipso non conueniat) interuallo vero accepto usque ad A, describatur arcus circuli secans A B, in D. Et ex D, per C, recta ducatur secans arcum in E. Recta enim ducta E A, erit perpendicularis ad A B. Nam angulus A, est rectus, cum sit in semicirculo D A E, ut ostendemus propositione 31. lib. 3.
 
Proposition 12. 
PROBL. 7. PROPOS. 12. 
第十二題 
To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line. 
SVPER datam rectam lineam infinitam, a dato puncto, quod in ea non est, perpendicularem rectam deducere. 
有無界直線。線外有一點。求於點上作垂線。至直線上。 
Let AB be the given infinite straight line, and C the given point which is not on it;  thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line. 
Sit recta A B, interminatæ quantitatis, & extra ipsam punctum C,  a quo oporteat lineam perpendicularem deducere ad rectam A B. 
 
For let a point D be taken at random on the other side of the straight line AB,  and with centre C and distance CD let the circle EFG be described; [Post. 3]  let the straight line EG be bisected at H, [I. 10]  and let the straight lines CG, CH, CE be joined. [Post. 1]  I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. 
77  
Centro C, interuallo vero quolibet circulus describatur secans A B, in D, & E. (quoniam interuallum assumptum tantum esse debet, ut transcendat rectam A B; alias eam non secaret. Diuisa autem recta D E, bifariam in F,   ducatur recta C F,   quam dico perpendicularem esse ad A B. 
For, since GH is equal to HE, and HC is common,  the two sides GH, HC are equal to the two sides EH, HC respectively;  and the base CG is equal to the base CE;  therefore the angle CHG is equal to the angle EHC. [I. 8]  And they are adjacent angles.  But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right,  and the straight line standing on the other is called a perpendicular to that on which it stands. [Def. 10] 
Si enim ducantur C D, C E,   erunt duo latera D F, F C, trianguli D F C, æqualia duo bus lateribus E F, F C, trianguli E F C, utrumque utrique, per constructionem;  est autem & basis C D, basi C E, æqualis, cum hæ sint ex centro C, ad circumferentiam.   Quare erit angulus D F C, angulo E F C, æqualis,   & propterea uterque rectus.     
Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.  Q. E. F. 
Ducta est igitur C F, perpendicularis,  quod faciendum erat.

SCHOLION
PROBE apposuit Euclides hanc particulam: infinitam. Si enim linea esset infinita, non posset semper a puncto dato extra ipsam perpendicularis ad eam deduci. Ut si linea finita esset B E, & punctum C, non posset ex C, describi circulus secans B E, in duobus punctis, quare neque ex C, perpendicularis duci ad B E. Hac igitur de causa vult Euclides, rectam datam esse infinitam, hoc est, non habere magnitudinem determinatam, ut saltem ad ipsam productam perpendicularis possit deduci. Ita enim fiet hic, si B E, producatur, donec circulus ex C, descriptus secet totam B A, productam in D, & E, &c.

PRAXIS
CENTRO facto C, & interuallo quovis eodem, describantur duo arcus secantes rectam datam in A, & B. Deinde ex A, & B, eodemque interuallo, vel alio, si placuerit, alii duo arcus describantur secantes se in D. Nam ducta recta C D secans A B, in E, erit perpendicularis ad A B. Demonstratio huius operationis non differt a demonstratione tradita in praxi propositionis 10. Nam anguli ad E, erunt recti, nempe inter se æquales.
IDEM officiemus hoc modo. Ex quovis puncto A, in linea data, & interuallo quolibet usque ad C, assumpto, arcus circuli describatur: Deinde ex quolibet alio puncto B, interualloque usque ad idem C, alius arcus describatur priorem secans in C, & D; Eritque recta C D, secans A E, in E, perpendicularis ad A B. Demonstratio eadem est, quæ prior. Non est autem necesse, ut intervallum B C, æquale sit intervallo A C, ut in hac figura apparet. Facilior tamen erit, & breuior operatio, si idem semper intervallum accipiatur.
 
Proposition 13. 
THEOR. 6. PROPOS. 13. 
If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles. 
CVM recta linea super rectam consistens lineam angulos facit, aut duos rectos, aut duobus rectis æquales efficiet. 
一直線。至他直線上所作兩角。非直角。卽等於兩直角。 
For let any straight line AB set up on the straight line CD make the angles CBA, ABD;  I say that the angles CBA, ABD are either two right angles or equal to two right angles. 
RECTA linea AB, consistens super rectam CD, faciat duos angulos A B C, A B D.   
Now, if the angle CBA is equal to the angle ABD, they are two right angles. [Def. 10]  But, if not, let BE be drawn from the point B at right angles to CD; [I. 11]  therefore the angles CBE, EBD are two right angles.  Then, since the angle CBE is equal to the two angles CBA, ABE, let the angle EBD be added to each;  therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C. N. 2]  Again, since the angle DBA is equal to the two angles DBE, EBA, let the angle ABC be added to each;  therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. [C. N. 2]  But the angles CBE, EBD were also proved equal to the same three angles;  and things which are equal to the same thing are also equal to one another; [C. N. 1]  therefore the angles CBE, EBD are also equal to the angles DBA, ABC.  But the angles CBE, EBD are two right angles;  therefore the angles DBA, ABC are also equal to two right angles. 
Si igitur A B, fuerit perpendicularis ad C D, erunt dicti 10. def anguli duo recti.   Si vero A B, non fuerit perpendicularis, faciet unum quidem angulum obtusum, alterum vero acutum. Dico igitur ipsos duobus esse rectis æquales. Educatur enim B E, ex B, perpendicularis ad C D,   ut sint duo anguli E B C, E B D, recti.  Quoniam vero angulus rectus E B D, æqualis est duobus angulis D B A, A B E; erunt, apposito communi angulo recto E B C,  duo recti E B D, E B C, tribus angulis D B A, A B E, E B C, æquales.   Rursus quia angulus A B C, duobus angulis A B E, E B C, æqualis est; apposito communi angulo A B D,  erunt duo anguli A B C, A B D, tribus angulis D B A, A B E, E B C, æquales.   Sed illis tribus ostensum fuit esse etiam æquales duos rectos E B D, E B C;  quæ autem eidem æqualia, inter se sunt æqualia.  Duo igitur anguli A B C, A B D, æquales sunt duobus rectis E B D, E B C.     
Therefore etc.  Q. E. D. 
Cum ergo recta linea super rectam consistens lineam, &c.   Quod ostendere oportebat.

SCHOLION
VIDETVR hæc propositio pendere ex communi quadam animi notione. Quo enim angulus A B C, superat rectum angulum E B C, eo reliquus angulus A B D, superatur ab angulo recto E B D. Nam sicut ibi excessus est angulus A B E, ita hic defectus est idem angulus A B E. Quocirca anguli A B C, A B D, duobus rectis æquales esse convincuntur, siquidem tantum unus eorum supra rectum acquirit, quantum alter deperdit.
 
Proposition 14. 
THEOR. 7. PROPOS. 14. 
第十四題 
If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another. 
SI ad aliquam rectam lineam, atque ad eius punctum, duæ rectæ lineæ non ad easdem partes ductæ eos, qui sunt deinceps, angulos duobus rectis æquales fecerint; in directum erunt inter se ipsæ rectæ lineæ. 
一直線。於線上一點。出不同方兩直線。偕元線、每旁作兩角。若每旁兩角、與兩直角等卽後出兩線、為一直線。 
For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles;  I say that BD is in a straight line with CB. 
Ad punctum C, lineæ rectæ A B, in diuersas partes eductæ sint duæ rectæ C D, C E, facientes cum A B, duos angulos A C D, A C E, vel rectos, vel duobus rectis æquales.  Dico ipsas C D, C E, inter se esse constitutas in directum, ita ut D C E, sit una linea recta. 
For, if BD is not in a straight line with BC, let BE be in a straight line with CB.  Then, since the straight line AB stands on the straight line CBE, the angles ABC, ABE are equal to two right angles. [I. 13]  But the angles ABC, ABD are also equal to two right angles;  therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1]  Let the angle CBA be subtracted from each;  therefore the remaining angle ABE is equal to the remaining angle ABD, [C.N. 3] the less to the greater:  which is impossible.  Therefore BE is not in a straight line with CB.  Similarly we can prove that neither is any other straight line except BD.  Therefore CB is in a straight line with BD. 
Si enim non est recta D C E; producta D C, ad partes C, in directum, & continuum cadet aut supra C E, ut sit recta D C F, aut infra C E, ut sit recta D C G.  Si cadit supra, cum A C, consistat super rectam D C F, fient duo anguli A C D, A C F, duobus rectis æquales;  Ponuntur autem & duo anguli A C D, A C E, æquales duobus rectis;  & omnes recti sunt inter se æquales. Quare duo anguli A C D, A C F, duobusangulis A C D, A C E, erunt æquales.   Ablato igitur cmmuni angulo A C D,   remanebunt anguli A C F, A C E, inter se æquales, pars & totum,  quod est absurdum.  Non igitur recta D C, producta cadet supra C E;   Sed neque infra cadet; Eadem enim ratione probarentur anguli A C E, A C G, æquales.  Igitur D C, producta eadem efficietur, quæ C E; 
Therefore etc.  Q. E. D. 
proptereaque, si ad aliquam rectam lineam, atque ad eius punctum, &c.  Quod demonstrandum erat.

SCHOLION
EST hæc propositio præcedentis conuersa. In ea enim probatum fuit, si D C E, sit recta, angulos A C D, A C E, duobus esse rectis æquales. In hac vero demonstratum est, si dicti anguli sint duobus rectis æquales, rectas D C, E C, esse unam lineam rectam.

EX PROCLO
RECTE Euclides addidit in propositione hac (& non ad easdem partes.) Quoniam, ut ait Porphyrius, fieri potest, ut ad punctum aliquod lineæ datæ ad easdem partes duæ lineæ ducantur, facientes cum data duos angulos duobus rectis æquales, quæ tamen non constituant unam lineam, eo quod non ad diuersas sint ductæ partes. Sit enim punctum C, in linea A B, datum. Ducatur C D, perpendicularis ad A B, diuidaturque rectus angulus A C D, bifariam per rectam C E. Deinde ex D, quolibet puncto rectæ C D, ducatur D E, perpendicularis ad C D, secans rectam C D, in E. Producta autem E D, ad partes D, sumatur D F, æqualis rectæ D E, & ducatur recta F C. Quoniam igitur latera E D, D C, trianguli E D C, æqualia sunt lateribus F D, D C, trianguli F D C, utrumque utrique, & anguli D, ipsis contenti æquales, nempe recti; erit basis E C, basi C F, æqualis, & angulus E C D, angulo F C D. Sed angulus E C D, dimidium est recti. (Est enim rectus A C D, divisus bifariam.) Igitur & F C D, dimidium erit recti. Quare C F, cum A C, faciet angulum A C F, constantem ex recto, & dimidio recti; Facit autem C E, cum eadem A C, angulum A C E, dimidium etiam recti. Duo igitur anguli A C F, A C E, quos ad easdem partes faciunt rectæ C F, C E, cum A B; æquales sunt duobus rectis; Et tamen C F, C E, non sunt una linea recta, propterea quod non sunt ducta ad diuersas partes, sed ad easdem.
 
Proposition 15. 
THEOR. 8. PROPOS. 15. 
第十五題 
If two straight lines cut one another, they make the vertical angles equal to one another. 
SI duæ rectæ lineæ se mutuo secuerint, angulos ad verticem æquales inter se efficient. 
凡兩直線相交作四角。每兩交角必等。 
For let the straight lines AB, CD cut one another at the point E;  I say that the angle AEC is equal to the angle DEB, and the angle CEB to the angle AED. 
Secent se duæ rectæ A B, C D, in puncto E, utcunque.   Dico angulos, quos faciunt ad verticem E, inter se esse æquales, angulum videlicet A E D, angulo B E C, & angulum A E C, angulo B E D.  
For, since the straight line AE stands on the straight line CD, making the angles CEA, AED, the angles CEA, AED are equal to two right angles [I. 13]  Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB,  the angles AED, DEB are equal to two right angles. [I. 13]  But the angles CEA, AED were also proved equal to two right angles;  therefore the angles CEA, AED are equal to the angles AED, DEB. [Post. 4 and C. N. 1]  Let the angle AED be subtracted from each;  therefore the remaining angle CEA is equal to the remaining angle BED. [C. N. 3]  Similarly it can be proved that the angles CEB, DEA are also equal. 
Quoniam recta D E, consistit super rectam A B, erunt duo anguli A E D, D E B, æquales duobus rectis.  Rursus quia recta B E, super rectam C D, consistit,  erunt eadem ratione duo anguli C E B, B E D, duobus rectis æquales.   Cum igitur omnes recti anguli inter se sint æquales;  erunt duo anguli AED, D E B, duobus angulis D E B, B E C, æquales.  Dempto igitur communi angulo D E B,  remanebit angulus A E D, angulo B E C, æqualis.  Eadem ratione confirmabitur, angulos A E C, B E D, inter se æquales esse. Nam duo anguli A E C, C E B, qui duobus sunt rectis æquales, æquales erunt duobus quoque angulis D E B, B E C, qui duobus rectis sunt æquales. Ablato igitur angulo communi B E C, remanebunt anguli A E C, B E D, æquales inter se. 
Therefore etc.  Q. E. D.  [Porism. From this it is manifest that, if two straight lines cut one another, they will make the angles at the point of section equal to four right angles.] 
Si igitur duæ rectæ lineæ se mutuo secuerint, &c.   Quod ostendere oportebat.

COROLLARIUM. I.
EUCLIDES colligit ex demonstratione huius theorematis (ex sententia Procli, quoniam alia exemplaria hoc corollarium non habent) duas lineas rectas se mutuo secantes efficere ad punctum sectionis quatuor angulos quatuor rectis angulis æquales. Nam in demonstratione ostensum fuit, tam duos angulos A E D, D E B, quam duos A E C, C E B, duobus esse rectis æquales, per 13. propos. Omnes igitur quatuor anguli ad E, constituti æquipollent bis duobus rectis angulis. Quare quatuor rectis æquales existunt.

COROLLARIUM. II.
EADEM ratione colligemus, omnes angulos circa unum & idem punctum constitutos, quotcunque fuerint, quatuor duntaxat rectis angulis æquales esse. Si enim ex E, aliæ lineæ quotlibet educantur, diuidentur solummodo illi quatuor anguli ad E, constituti in plurimas partes, quæ omnes simul sumptæ totis suis adæquantur. Cum ergo illi quatuor anguli æquales sint quatuor rectis, ex 1. corollario, erunt quoque omnes alii simul sumpti quatuor tantum rectis æquales. Ex quo perspicuum est, omne spatium punctum aliquod in plano circumstans, æquivalere quatuor rectis angulis, ut multi auctores afferunt: quia omnes anguli, qui circa illud punctum constitui possunt, quatuor sunt rectis angulis æquales. Simili modo constat, quotlibet lineas rectas se inuicem secantes, facere ad punctum sectionis angulos æquales quatuor rectis.

EX PROCLO
SI ad aliquam rectam lineam, ad eiusque signum, duæ rectæ lineæ non ad easdem partes sumptæ, angulos ad verticem æquales fecerint; ipsæ rectæ lineæ in directum sibi inuicem erunt.
EX puncto C, rectæ A B, in diuersas partes egrediantur duæ rectæ C D, C E, facientes angulos A C E, B C D, inter se æquales: Vel etiam duos A C D, B C E. Dico duas C D, C E, efficere unam lineam rectam. Quoniam enim angulus A C E, æqualis est angulo B C D; addito communi angulo B C E, erunt duo anguli A C E, E C B, duobus angulis D C B, B C E, æquales: Sed anguli A C E, E C B, sunt æquales duobus rectis. Igitur & duo D C B, B C E, duobus erunt rectis æquales. Quamobrem C D, C E, erunt linea una recta. Hoc autem, ut vides, conuersum est propositionis decimæquintæ.

EX PELETARIO
SI quatuor rectæ lineæ ab uno puncto exeuntes binos angulos oppositos inter se æquales fecerint, erunt quælibet duæ lineæ aduersæ in rectum sibi, & continuum coniunctæ.
EX puncto A, quatuor lineæ eductæ A B, A C, A D, A E, faciant duos angulos oppositos B A E, C A D, inter se æquales: Item duos B A C, D A E, inter se æquales. Dico tam B A, A D, facere unam lineam rectam, quam C A, A E. Quoniam æquales sunt anguli B A E, C A D, si æquales illis addantur anguli B A C, D A E, erunt duo anguli B A E, B A C, æquales duobus angulis C A D, D A E. Tam ergo illi, quam hi, dimidium sunt quatuor angulorum circa punctum A, consistentium: At hi quatuor æquales sunt quatuor rectis, per 2. coroll. præcedentis propos. Igitur duo anguli B A E, B A C, æquales sunt duobus rectis; atque adeo C A, A E, unam efficient lineam rectam. Eodem pacto ostendetur, duas B A, A D, unam rectam efficere lineam. Nam eadem ratione erunt duo anguli B A E, E A D, æquales duobus angulis D A C, C A B. Quare, ut prius, concludetur propositum. Peletarius autem demonstrat hoc idem ratione ducente ad id, quod fieri nequit. Nos tamen demonstrationem nostram ostensiuam eius demonstrationi iure optimo præposuimus.
 
 
一系。推顯兩直線相交。於中點上作四角。與四直角等。
二系一點之上。兩直線相交不論幾許線、幾許角。定與四直角等。公論 \\ 十八。
增題一直線內、出不同方兩直線、而所作兩交角等。卽後出兩線、為一直線。 
Proposition 16. 
THEOR. 9. PROPOS. 16. 
第十六題 
In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles. 
CVIVSCVNQVE trianguli vno latere producto, externus angulus vtrolibet interno, & opposito, maior est. 
凡三角形之外角。必大于相對之各角。 
Let ABC be a triangle,  and let one side of it BC be produced to D;  I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC. 
TRIANGVLI A B C,   latus B A, producatur ad D.  Dico angulum externum D A C, maiorem esse interno, & opposito A C B, itemque maiorem interno, & opposito A B C. 
解曰甲乙丙角形。  自乙甲線,引之至丁。  題言外角丁甲丙。必大于相對之内角甲乙丙甲丙乙。 
Let AC be bisected at E [I. 10],  and let BE be joined and produced in a straight line to F;  let EF be made equal to BE[I. 3],  let FC be joined [Post. 1],  and let AC be drawn through to G [Post. 2]. 
Dividatur 78 enim A C, bifariam in E;  
& ex B, per E, extendatur recta B E F,  ita ut E F, 79 abscissa sit æqualis rectæ E B;   ducaturque recta F A.   
論曰。欲顯丁甲丙角,大于甲丙乙角。試以甲丙線兩平分于戊本篇十  自乙至戊。作直線,引長之。  從戊外截取戊己,與乙戊等本篇三  次自甲至己,作直線。   
Then, since AE is equal to EC, and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively;  and the angle AEB is equal to the angle FEC,  for they are vertical angles. [I. 15]  Therefore the base AB is equal to the base FC,  and the triangle ABE is equal to the triangle CFE,  and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]  therefore the angle BAE is equal to the angle ECF.  But the angle ECD is greater than the angle ECF; [C. N. 5]  therefore the angle ACD is greater than the angle BAE.  Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15], can be proved greater than the angle ABC as well. 
Quoniam igitur latera C E, E B, trianguli C E B, æqualia sunt lateribus A E, E F, trianguli A E F, utrumque utrique, per constructionem;  
Sunt autem & anguli ad E, dictis lateribus comprehensi, 80 inter se æquales,  cum sint circa verticem E, & oppositi:  Erit 81 basis C B, æqualis basi A F,       & angulus E C B, angulo E A F;  Est autem angulus D A C, externus maior angulo E A F, totum videlicet parte.   Igitur & externus angulus D A C, maior erit interno, & opposito angulo A C B.  Quod si latus C A, producatur ad G; & A B, diuidatur bifariam in H; extendaturque recta C H I, ut H I, æqualis sit rectæ H C, & ducatur recta I A: demonstrabitur eadem prorsus ratione, angulum externum G A B, maiorem esse interno angulo, & opposito A B C; Est autem 82 angulus D A C, angulo G A B, æqualis, cum lineæ B D, C G, se mutuo secent in A. Igitur & angulus D A C, maior erit interno & opposito angulo A B C. Est autem idem angulus D A C, maior quoque ostensus angulo interno & opposito A C B. 
即甲戊己,戊乙丙,兩角形之戊己,與戊乙,兩線等。戊甲與戊丙兩線等。  甲戊己,乙戊丙,兩交角又等本篇十五。    則甲己與乙丙兩底亦等本篇四    兩形之各邊,各角,俱等。  而己甲戊,與戊丙乙,兩角亦等矣。  夫己甲戊。乃丁甲丙之分。  則丁甲丙,大于己甲戊,亦大于相等之戊丙乙。 而丁甲丙外角。不大于相對之甲丙乙内角乎。次顯丁甲丙,大于甲乙丙。  試自丙甲線,引長之,至庚。次以甲乙線兩平分于辛本篇十。自丙至辛。作直線,引長之。從辛外截取辛壬,與丙辛等本篇三。次自甲至壬,作直線。依前論,推顯甲辛壬,辛丙乙,兩角形之各邊,各角,俱等。則壬甲辛,與辛乙丙,兩角亦等矣。夫壬甲辛。乃庚甲乙之分。必小于庚甲乙也。庚甲乙,又與丁甲丙,兩交角等本篇十五。則甲乙丙内角。不小于丁甲丙外角乎。其餘乙丙上作外角,俱大于相對之内角。 
Therefore etc.  Q. E. D. 
Cuiuscunque ergo trianguli uno latere producto, &c.  
Quod demonstrandum erat.

SCHOLION
NON dixit Euclides, angulum externum D A C, maiorem esse angulo B A C, interno, qui sibi est deinceps; sed solum magnitudine superare utrumlibet A C B, A B C, internorum, sibique oppositorum: quoniam externus angulus æqualis potest esse angulo interno sibi deinceps, quando scilicet externus rectus est. Tunc enim necessario is, qui sibi est deinceps, rectus quoque erit: Potest & esse minor, quando nimirum est acutus. Hoc enim posito, angulus illi deinceps obtusus erit. Solum ergo, quando obtusus erit externus, superabit internum sibi deinceps; Hic enim necessario acutus existet. Quæ omnia facile colliguntur ex propos. 13. per quam angulus externus, & internus illi deinceps, æquales sunt duobus rectis.
ID vero, quod in scholio propos. 6. huius libri nos demonstraturos recepimus, nimirum hanc propositionem non posse conuerti; cum & uno latere figuræ quadrilateræ producto, externus angulus quolibet interno, & opposito possit esse maior; hac ratione absolvemus.
SIT figura quadrilatera A B C D, cuius angulus B A D, obtusus, & A B C, rectus constituatur, hac tamen lege, ut rectæ A B, D C, productæ ad partes B, & C, in puncto E, nec non & rectæ D A, C B, ad partes A & B, in puncto F, coeant. Dico, si A D, producatur ad G, angulum externum C D G, maiorem esse quolibet tribus internis B A D, A B C, B C D, sibi oppositis. Cum enim A D E, triangulum sit, erit angulus externus E D G, maior interno opposito D A E. Rursus cum D A B, obtusus maior sit recto A B C, maior quoque multo erit E D G, ipso A B C. Postremo, quia & in triangulo C D F, angulus externus C D G, maior est interno, & opposito F C D, manifestum est, in quadrilatero A B C D, externum angulum C D G, maiorem esse internis, & oppositis B A D, A B C, B C D. Quam ob rem propositio hæc 16. conuerti nequit, quippe cum eius antecedens, & consequens non reciprocentur, ut demonstratum est.

EX PROCLO
SEQVITVR ex hac propositione, ab eodem puncto ad unam eandemque lineam rectam non posse duci plures lineas rectas, quam duas inter se æquales. Si enim fieri potest, ducantur ex A, ad lineam B C, tres lineæ rectæ æquales A B, A C, A D. Quoniam igitur latera A B, A C, sunt æqualia, erunt anguli A C B, & A B C, æquales super basim B C. Rursus quia latera A B, A D, sunt æqualia, erunt anguli A D B, & A B D, super basim B D, æquales. Quare cum uterque angulus A C D, & A D B, æqualis sit angulo A B C, erit angulus A D B, æqualis angulo A C D, externus interno opposito, quod est absurdum, cum per hanc 16. propos. externus interno maior sit. Non ergo plures lineæ rectæ, quam duæ, inter se æquales, ex A, ad B C, possunt duci. Quod est propositum.
 
依此推顯。   
Proposition 17. 
THEOR. 10. PROPOS. 17. 
第十七題 
In any triangle two angles taken together in any manner are less than two right angles. 
CVIVSCVNQVE trianguli duo anguli duobus rectis sunt minores, omnifariam sumpti. 
凡三角形之每兩角。必小於兩直角。 
Let ABC be a triangle;  I say that two angles of the triangle ABC taken together in any manner are less than two right angles. 
Sit triangulum A B C;  Dico duos angulos A B C, & A C B, minores esse duobus rectis; Item duos C B A, & C A B; Itemque duos B A C, & B C A. 
For let BC be produced to D. [Post. 2]  Then, since the angle ACD is an exterior angle of the triangle ABC, it is greater than the interior and opposite angle ABC. [I. 16]  Let the angle ACB be added to each;  therefore the angles ACD, ACB are greater than the angles ABC, BCA.  But the angles ACD, ACB are equal to two right angles. [I. 13]  Therefore the angles ABC, BCA are less than two right angles.  Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well. 
Producantur enim duo quæuis latera, nempe C B, C A, ad D, & E.   Quoniam igitur angulus A B D, externus maior est interno & opposio angulo A C B;  si addatur communis angulus A B C,   erunt duo anguli A B D, A B C, maiores duobus angulis A B C, A C B:  Sed A B D, A B C, æquales sunt duobus rectis.   Igitur A B C, A C B, minores sunt duobus rectis.   Eadem ratione erunt anguli C B A, & C A B, minores duobus rectis. Item duo B A C, & B C A. 
Therefore etc.  Q. E. D. 
Cuiuscunque igitur trianguli, &c.   Quod erat demonstrandum.

EX PROCLO
HINC perspicuum est, ab eodem puncto ad eandem rectam lineam non posse deduci plures lineas perpendiculares, quam unam. Si enim fieri potest, ducantur ex A, ad rectam B C, duæ perpendiculares A B, A C. Erunt igitur in triangulo A B C, duo anguli interni B, & C, duobus rectis æquales, cum sint duo recti, quod est absurdum. Sunt enim quilibet duo anguli in triangulo quocunque ostensi minores duobus rectis. Non ergo plures perpendiculares, quam vna, ex A, ad B C, deduci possunt. Quod est propositum.

COROLLARIVM
CONSTAT etiam ex his, In omni triangulo, cuius unus angulus fuerit rectus, vel obtusus, reliquos esse acutos, ceu monuimus defin. 26. huius lib. Cum enim per hanc propositio duo quilibet anguli sint duobus rectis minores, necesse est, si unus fuerit rectus, vel obtusus, quemcunque reliquorum esse acutum, ne duos angulos in triangulo rectos, aut duobus rectis maiores esse fateamur.
 
Proposition 18. 
THEOR. 11. PROPOS. 18. 
第十八題 
In any triangle the greater side subtends the greater angle.  For let ABC be a triangle having the side AC greater than AB;  I say that the angle ABC is also greater than the angle BCA. 
OMNIS trianguli maius latus maiorem angulum subtendit.  IN triangulo A B C, sit latus A C, maius latere A B.   Dico angulum A B C, subtensum a maiori latere A C, maiorem esse angulo A C B, qui a minori latere A B, subtenditur.  
凡三角形。大邊對大角。小邊對小角。 
For, since AC is greater than AB, let AD be made equal to AB [I. 3], and let BD be joined. 
Nam ex A C, auferatur A D, æqualis ipsi A B, & ducatur recta B D.  
Then, since the angle ADB is an exterior angle of the triangle BCD, it is greater than the interior and opposite angle DCB. [I. 16]  But the angle ADB is equal to the angle ABD, since the side AB is equal to AD;  therefore the angle ABD is also greater than the angle ACB;  therefore the angle ABC is much greater than the angle ACB. 
Quonim igitur duo latera A B, A D, æqualia sunt per constructionem, erunt anguli A B D, A D B, æquales: Est autem angulus A D B, maior angulo A C B.   .  Igitur & angulus A B D, maior erit angulo A C B.83   Quamobrem cum angulus totus A B C, maior adhuc sit angulo A B D; erit angulus A B C, multo maior angulo A C B. Eadem ratione, si latus A C, maius ponatur latere B C, ostendens angulum A B C, maiorem esse angulo B A C; si nimirum ex C A, abscindatur linea æqualis ipsi C B, &c.  
Therefore etc.  Q. E. D. 
Quare omnis trianguli maius latus maiorem angulum subtendit;  
Quod demonstrandum erat.

COROLLARIVM
EX hoc sequitur, omnes tres angulos trianguli Scaleni esse inæquales, ut monuimus defin. 15. huius lib. Sit enim triangulum Scalenum A B C, cuius maximum quidem latus A C, minimum autem B C, & medium locum habens A B. Dico eiusdem omnes angulos inæquales esse. Cum enim latus A C, ponatur maius latere A B, erit per hanc propositio angulus B, angulo C, maior. Eadem ratione maior erit angulus C, angulo A, quandoquidem & latus A B, latere B C, maius ponitur. Sunt igitur omnes tres anguli inæquales, maximus quidem B, minimus vero A, & C, medium locum inter utrumque tenens.
 
Proposition 19. 
THEOR. 12. PROPOS. 19. 
第十九題 
In any triangle the greater angle is subtended by the greater side. 
OMNIS trianguli maior angulus maiori lateri subtenditur. 
凡三角形。大角對大邊。小角對小邊。 
Let ABC be a triangle having the angle ABC greater than the angle BCA;  I say that the side AC is also greater than the side AB. 
IN triangulo A B C, angulus B, maior sit angulo C.   Dico latus A C, subtendens maiorem angulum B, maius esse latere A B, quod angulum minorem C, subtendit. 
For, if not, AC is either equal to AB or less.  Now AC is not equal to AB;  for then the angle ABC would also have been equal to the angle ACB; [I. 5]  but it is not; therefore AC is not equal to AB.  Neither is AC less than AB,  for then the angle ABC would also have been less than the angle ACB; [I. 18]  but it is not; therefore AC is not less than AB.  And it was proved that it is not equal either.  Therefore AC is greater than AB. 
Si enim latus A C, maius non est latere A B, erit vel æquale illi, vel minus.  Si dicatur A C, æquale esse ipsi A B,   erit angulus B, æqualis angulo C;  Est autem & maior per hypothesin, quod est absurdum.   Si vero A C, minus esse dicatur latere A B,   erit angulus B, subtensus a minore latere A C, minor angulo C, subtenso a maiore latere A B;   Ponitur autem maior, quod magis est absurdum.   Cum igitur A C, latus neque æquale sit lateri A B, neque minus eo,  erit maius. Eadem ratione probabitur, latus A C, maius esse latere B C, si angulus B, maior esse concedatur angulo A. 
Therefore etc.  Q. E. D. 
Omnis ergo trianguli maior angulus maiori lateri subtenditur;  Quod demonstrandum proponebatur.

EX PROCLO
POSSVMVS hoc idem theorema ostendere affirmatiua demonstratione, sine adminiculo præcedentis, si tamen prius demonstretur hoc sequens theorema.
SI trianguli angulus bifariam sectus fuerit, secansque angulum recta linea ad basin ducta in partes inæquales ipsam diuidat; Latera illum angulum continentia inæqualia erunt, & maius quidem illud, quod cum maiori basis segmento coincidit, minus vero, quod cum minori.
TRIANGVLI A B C, angulus B A C, diuidatur bifariam per rectam A D, quæ secet basin B C, in partes inæquales, maiusque segmentum sit D C. Dico latus A C, maius esse latere A B. Producatur enim A D, ad E, ut sit D E, æqualis ipsi A D. Deinde ex maiori segmento D C, auferatur recta D F, æqualis minori segmento D B, & per F, ex E, extendatur recta E F G. Quoniam igitur latera A D, D B, trianguli A D B, æqualia sunt lateribus E D, D F, trianguli E D F, utrumque utrique, per constructionem; sunt autem & anguli A D B, E D F, dictis lateribus contenti æquales: Erunt bases A B, & E F, æquales, & angulo B A D, angulus E F D, æqualis: Est vero & angulus C A D, angulo B A D, æqualis, per hypothesin: Igitur anguli C A D, G E A, trianguli A G E, æquales erunt, ideoque latera A G, E G, æqualia erunt. Est autem recta A C, maior quam A G: quare & A C, maior erit, quam E G. Et quia E G, maior est, quam E F, erit & A C, multo maior, quam E F. Cum igitur demonstratum sit rectam E F; æqualem esse rectæ A B, erit A C, latus maius latere A B, quod erat ostendendum.
HOC ostenso theoremate, ita propositio 19. demonstrabitur. In triangulo A B C, angulus A B C, maior sit angulo C. Dico latus A C, maius esse latere A B. Divia enim recta B C (super quam constituti sunt dicti anguli inæquales) bifariam in D, ex A per D, extendatur recta A D F, ut sit D E, æqualis ipsi A D: ducaturque recta B E. Quoniam igitur latera A D, D C, trianguli A D C, æqualia sunt lateribus E D, D B, trianguli E D B, utrumque utrique, per constructionem, sunt autem & anguli A D C, E D B, dictis comprehensi lateribus æquales: Erunt bases A C, & B E, æquales, angulusque A C D, angulo E B D, æqualis: Et quia angulus A C D, ponitur esse minor angulo A B C, erit & angulus E B D, minor eodem angulo A B C. Ideoque angulus A B E, per rectam B D, diuidetur in partes inæquales. Si igitur bifariam secetur per rectam B F, cadet B F, supra E D, eo quod angulus A B D, maior sit angulo E B D. Quia vero E F, maior est quam E D, & E D, posita est aqualis ipsi A D, erit E F maior quam A D. Sed adhuc A D, maior est quam A F. Multo igitur maior erit E F, quam A F. Itaque quia recta B F, diuidens angulum A B E, bifariam, secat basin A E, inæqualiter in F, estque maius segmentum E F, minus autem A F; erit per theorema a Proclo proxime demonstratum, latus B E, maius latere A B. Ostensum est autem B E, æquale esse lateri A C. Igitur & A C, latus latere A B, maius erit. Quod erat demonstrandum.

SCHOLION
HAEC propositio 19. conuersa est propositionis 18. ut perspicuum est. Campanus autem duarum istarum propositionum ordinem prorsus inuertit, ita ut ea, quæ apud nos est 18. apud ipsum sit 19. & contra. Quarum utramque ostendit ducendo ad id, quod fieri nequit, cum tamen Euclides propositionem 18. directe & ostensiue confirmauerit, ut ex dictis liquido constat.

POTERIMVS quoque theorema a Proclo demonstratum conuertere, hocmodo.
SI trianguli duo latera inæqualia fuerint, linea recta bifariam diuidens angulum ipsis contentum, secabit basin in partes inæquales, maiusque segmentum erit prope maius latus.
DVO latera A B, A C, trianguli A B C, sint inæqualia; A C, maius, & A B, minus. Recta autem A D, diuidens angulum B A C, bifariam, secet basin B C, in D. Dico segmentum D C, maius esse segmento D B. Si enim non est maius, erit vel æquale, vel minus. Si dicatur esse æquale; producatur A D, ad E, ut D E, æqualis sit ipsi D A, ducaturque recta E C. Quoniam igitur latera A D, D B, æqualia sunt lateribus E D, D C, utrumque utrique; A D, videlicet ipsi E D, per constructionem, & D B, ipsi D C, per hypothesin aduersarii. Sunt autem & anguli ad D, dictis lateribus contenti æquales: erit basis A B, basi E C, æqualis & angulo B A D, angulus C E D. Positus autem est & angulo B A D, angulus C A D, æqualis. Igitur & anguli C E D, C A D, æquales erunt. Ideoque latus A C, lateri E C, æquale. Cum igitur ostensum sit, lateri E C, æquale esse quoque latus A B, erunt latera A C, A B, æqualia; quod est absurdum, quia A C, maius ponebatur, quam A B. Non erit igitur segmentum D C, segmento D B, æquale. Quod si D C, dicatur esse minus, & D B, maius; erit per theorema Procli, latus A B, maius latere A C. Ponebatur autem minus, quod multo magis est absurdum. Non igitur minus erit D C, quam D B. Quare erit necessario maius.

EODEM modo demonstrari poterit hoc theorema.
SI trianguli angulum recta linea bifariam diuidens, basin bifariam quoque secet, erunt duo latera angulum continentia inter se æqualia: Quod si latera æqualia fuerint, basin etiam bifariam secabit linea recta, quæ angulum bifariam diuidit.
PRIMO recta A D, secans angulum B A C, bifariam diuidat quoque basin B C, in D, bifariam. Dico latera A B, A C, inter se æqualia esse. Hoc autem demonstrabimus eadem ratione, qua in præcedenti theoremate ostensum fuit, latus A C, æquale esse lateri A B, si D C, segmentum segmento D B, æquale ponatur, dummodo figuram eodem modo construas. Cum enim latera A D, D B, æqualia sint lateribus E D, D C, & anguli ad D, dictis lateribus contenti æquales; erunt bases A B, E C, æquales, & angulus C E D, angulo B A D, hoc est, angulo C A D, æqualis. Quare A C, æquale erit ipsi E C, hoc est, ipsi A B.
SECVNDO sint latera A B, A C, æqualia, & recta A D, secans basin B C, in D, diuidat angulum B A C, bifariam. Dico segmentum D C, æquale esse segmento D B. Cum enim latera A D, A B, æqualia sint lateribus, A D, A C, utrumque utrique, & anguli quoque ad A, contenti dictis lateribus æquales per hypothesim, erunt bases B D, D C, æquales.
 
Proposition 20. 
THEOR. 13. PROPOS. 20. 
第二十題 
In any triangle two sides taken together in any manner are greater than the remaining one. 
OMNIS trianguli duo latera reliquo sunt maiora, quomodocunque assumpta. 
凡三角形之兩邊。并之必大於一邊。 
For let ABC be a triangle;  I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one,  namely BA, AC greater than BC, AB, BC greater than AC, BC, CA greater than AB. 
SIT triangulum A B C.   Dico quælibet eius duo latera,84   nempe A B, A C, simul maiora esse reliquo latere B C. 
For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined. 
Producatur unum ex illis, ut C A, usque ad D, sitque recta A D, æqualis alteri lateri non producto A B, & ducatur recta D B. 
Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5]  therefore the angle BCD is greater than the angle ADC. [C.N. 5]  And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19] therefore DB is greater than BC.  But DA is equal to AC;  therefore BA, AC are greater than BC.  Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB. 
Quoniam igitur duo latera A B, A D, æqualia inter se sunt, per hypothesin, erunt anguli A B D, A D B, æquales inter se:  Est autem angulo A B D, maior angulus C B D.  Igitur & angulus C B D, maior erit angulo A D B. In triangulo ergo C B D, latus C D, oppositum maiori angulo C D B, maius erit latere B C, quod minori angulo C D B, opponitur.  Cum igitur duo latera A B, A C, simul æqualia sint ipsi C D, (si enim æqualibus A B, A D, commune addatur A C, fient tota æqualia; nimirum linea composita ex A B, A C, & linea composita ex A D, A C)  erunt quoque latera A B, A C, simul maiora latere B C.  Eodem modo demonstrabitur, quælibet alia duo latera maiora esse reliquo. 
Therefore etc.  Q. E. D. 
Quare omnis trianguli duo latera reliquo sunt maiora, &c.  Quod demonstrandum erat.

EX PROCLO
ALITER hoc theorema a familiaribus Heronis, & Porphyrii demonstratur, nullo latere producto, hac ratione. Sit probandum duo latera A B, A C, trianguli A B C, maiora esse latere B C. Diuidatur angulus B A C, illis lateribus contentus bifariam per rectam A D. Quoniam igitur trianguli C D A, latus C D, protractum est ad B, erit angulus externus B D A, maior interno & opposito C A D. Igitur & maior angulo B A D. Quare in triangulo A B D, latus A B, maiori angulo A D B, oppositum maius erit latere B D, quod minori angulo B A D, opponitur. Eadem ratione ostendetur, latus A C, maius esse quam C D, quia angulus C D A, maior est angulo B A D, hoc est, angulo C A D, &c. Quamobrem duo latera A B, A C, maiora erunt latere B C. Eademque est ratio quorumcunque duorum laterum, si angulus ipsis comprebensus bifariam secetur.
 
Proposition 21. 
THEOR. 14. PROPOS. 21. 
第二十一題 
If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle. 
SI super trianguli uno latere, ab extremitatibus duæ rectæ lineæ interius constitutæ fuerint; hæ constitutæ reliquis trianguli duobus lateribus minores quidem erunt, maiorem vero angulum continebunt. 
凡三角形。於一邊之兩界、出兩線。復作一三角形、在其內。則內形兩腰。幷之必小於相對兩腰。而後兩線所作角。必大於相對角。 
On BC, one of the sides of the triangle ABC, from its extremities B, C, let the two straight lines BD, DC be constructed meeting within the triangle;  I say that BD, DC are less than the remaining two sides of the triangle BA, AC, but contain an angle BDC greater than the angle BAC. 
IN triangulo A B C, super extremitates B, & C, lateris B C, intra triangulum constituantur duæ rectæ lineæ B D, C D, in puncto D, concurrentes.  Dico B D, C D, simul minores esse duobus lateribus B A, C A, simul; At vero angulum B D C, maiorem angulo B A C. 
For let BD be drawn through to E.  Then, since in any triangle two sides are greater than the remaining one, [I. 20]  therefore, in the triangle ABE, the two sides AB, AE are greater than BE.  Let EC be added to each;  therefore BA, AC are greater than BE, EC.  Again, since, in the triangle CED, the two sides CE, ED are greater than CD, let DB be added to each;  therefore CE, EB are greater than CD, DB.  But BA, AC were proved greater than BE, EC;  therefore BA, AC are much greater than BD, DC. 
Producatur enim altera linearum interiorum, nempe B D, ad punctum E, lateris C A.    Quoniam igitur in triangulo B A E, duo latera B A, A F, maiora sunt latere B E,  si addatur commune E C,  erunt B A, A C, maiora quam B E, E C.  Rursus quia in triangulo C E D, duo latera C E, E D, maiora sunt latere C D; si commune apponatur D B,  erunt C E, E B, maiora quam C D, D B.  Ostensum vero iam fuit, A B, C A, maiora esse quam B E, E C.  Multo igitur maiora erunt B A, C A, quam B D, C D, quod primo proponebatur. 
Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16]  therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED.  For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC.  But the angle BDC was proved greater than the angle CEB;  therefore the angle BDC is much greater than the angle BAC. 
  Præterea, quoniam angulus B D C, maior est angulo D E C, externus interno;  & angulus D E C, angulo B A C, maior quoque est, eandem ob causam:    Erit angulus B D C, multo maior angulo B A C; quod secundo proponebatur. 
Therefore etc.  Q. E. D. 
Si igitur super trianguli uno latere, ab extremitatibus, &c.  Quod erat ostendendum.

SCHOLION
QVAM recte Euclides dixerit, duas illas lineas intra triangulum constitutas, duci debere ab extremitatibus unius lateris, aperte intelligi potest ex eo, quod mox ex Proclo demonstrabimus; in triangulis videlicet rectangulis, vel etiam amblygoniis, intra triangulum constitui posse duas lineas super unum latus circa angulum rectum, vel obtusum, quarum quidem una ab extremitate dicti lateris, altera vero a quouis puncto prope aliud extremum lateris eiusdem educitur, quæ maiores sint reliquis duobus trianguli lateribus. Item in triangulis scalenis eodem modo super maximum latus duas rectas intra triangulum constitui posse, quæ minorem comprehendant angulum, &c.

EX PROCLO
SIT triangulum habens exempli gratia angulum A B C, obtusum. Dico ab extremo C, & a quouis puncto, nempe a D, prope aliud extremum B, lateris B C, duci posse duas lineas intra triangulum ad aliquod punctum, quæ maiores sint duobus lateribus B A, A C. Ducatur enim recta D A: Et quoniam in triangulo A B D, duo anguli A B D, A D B, minores sunt duobus rectis. Ponitur autem A B D, maior recto, nempe obtusus; erit A D B minor recto, ideoque minor angulo A B D. Quare latus A D, maius erit latere A B. Ex D A, abscindatur recta D E, æqualis rectæ A B. Et reliqua linea A E, bifariam diuidatur in F. Si igitur ab extremo C, ad F, recta dueatur C F, erunt duæ lineæ rectæ constitutæ C F, D F, intra triangulum maiores duobus lateribus B A, A C. Quoniam enim in triangulo A F C, duo latera A F, F C, maiora sunt latere A C. Est autem recta A F, ipsi F E, æqualis, per constructionem; erunt C F, F E, maiores quoque latere C A. Si igitur æqualia addantur E D, & A B, fient rectæ C F, F D, maiores lateribus C A, A B. Quod est propositum. Quod si ad F, ex B, extremo recta duceretur, essent duæ rectæ constitutæ C F, B F, minores duobus lateribus C A, A B, ut Euclides demontrauit.
RVRSVS sit triangulum scalenum A B C, cuius latus maximum B C, minimum A B. Ex B C, auferatur B D, æqualis rectæ A B, & ducatur A D, recta, ad cuius punctum quodlibet, ut ad E, ab extremo C, recta ducatur C E. Constitutæ igitur erunt intra triangulum duæ lineæ C E, D E, quæ minorem angulum comprebendunt eo, quem efficiunt duo latera A B, A C. Cum enim duo tatera B A, B D, æqua lia sint, erunt duo anguli B A D, B D A, æquales: Sed B D A, angulus maior est angulo C E D. Maior igitur erit & angulus B A D, angulo C E D. Quare multo maior erit totus angulus B A C, angulo C E D. Quod est propositum. Recte igitur Euclides monuit, duas lineas intra triangulum constitutas educi debere, ab extremis punctis unius lateris, ut minores quidem sint duob us reliquis trianguli lateribus, maiorem vero complectantur angulum. Alias enim propositio vera non esset, ut iam est demonstratum.
 
Proposition 22. 
PROBL. 8. PROPOS. 22. 
第二十二題 
Out of three straight lines, which are equal to three given straight lines, to construct a triangle: 
EX tribus rectis lineis, quæ sint tribus datis rectis lineis æquales, triangulum constituere.  
三直線。求作三角形。 
thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [I. 20] 
Oportet autem duas reliqua esse maiores omnifariam sumptas: quoniam vniuscuiusque trianguli duo latera omnifariam sumpta reliquo sunt maiora. 
其每兩線幷。大於一線也。 
Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one,  namely A, B greater than C, A, C greater than B, and B, C greater than A;  thus it is required to construct a triangle out of straight lines equal to A, B, C. 
TRES lineæ rectæ datæ sint A, B, & C, quarum quælibet duæ reliqua sint maiores, (Alias ex ipsis non posset constitui triangulum, ut constat ex propositio 20. in qua ostensum fuit, duo quæuis latera trianguli reliquo esse maiora.)    oporteatque construere triangulum habens tria latera tribus datis lineis æqualia. 
Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E,  and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3]  With centre F and distance FD let the circle DKL be described;  again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined;  I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C. 
Ex assumpta recta quauis D E, infinitæ magnitudinis abscindatur recta D F, æqualis rectæ A.  Et ex reliqua F E, recta F G, æqualis rectæ B; & ex reliqua G E, recta G H, æqualis rectæ C.85   Deinde centro F, interuallo vero F D, circulus describatur D I K.  Item centro G, interuallo autem G H, alius circulus describatur HIK, qui necessario priorem secabit in punctis I, & K, (cum enim duæ F D, G H, maiores ponantur recta F G; Si ex F E, sumatur recta F L, æqualis ipsi F D: & ex G D, recta G M, æqualis ipsi G H, cadet punctum M, inter L, & D. Si namque M, caderet in L, punctum, essent G L, F L, hoc est, G H, & F D, æquales rectæ F G: Si vero M, caderet inter G, & L, essent eædem duæ F L, G M, hoc est, D F, G H, minores recta F G, quorum utrumque est contra hypothesin. Id quod ex appositis figuris apparet) ex quorum quolibet, nimirum ex K, ducantur ad puncta, F, G, rectæ K F, K G,  factumque erit triangulum F G K, cuius latera dico æqualia esse datis rectis A B, & C. 
For, since the point F is the centre of the circle DKL, FD is equal to FK.  But FD is equal to A;  therefore KF is also equal to A.  Again, since the point G is the centre of the circle LKH, GH is equal to GK.  But GH is equal to C;  therefore KG is also equal to C.  And FG is also equal to B;  therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C. 
Cum enim recta F K, æqualis sit rectæ F D, 
& recta A, per constructionem eidem F D, æqualis;  erit latus F K, rectæ A, æquale.  Rursus quia G K, æqualis est ipsi G H,  & recta C, eidem G H;  erit quoque latus G K, rectæ C, æquale:  Positum autem fuit per constructionem, reliquum latus F G, reliquæ rectæ B, æquale.  Omnia igitur tria latera F K, F G, G K, tribus datis rectis A, B, C, æqualia sunt. 
Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed.  Q. E. F. 
Constituimus ergo ex tribus rectis lineis, quæ sunt tribus datis rectis lineis æquales, triangulum:  Quod faciendum erat.

PRAXIS
SVMATVR recta D E, æqualis cuicunque rectarum datarum, nempe ipsi B, quam nunc volumus esse basin. Deinde ex D, ad interuallum rectæ A, arcus describatur: Item ex E, ad interuallum rectæ C, alter arcus secans priorem in F. Si igitur ducantur rectæ D F, E F, factum erit triangulum habens tria latera æqualia tribus datis lineis. Erit enim latus D F, æquale rectæ A, propter interuallum ipsius A, assumptum: & latus E F, ipsi C, propter assumptum interuallum C: D E, vero latus, acceptum est rectæ B, æquale, ab initio.

SCHOLION
HAC arte cuicunque triangulo proposito alterum prorsus æquale & quoad latera, angulosque & quoad aream ipsius, constituemus. Sit namque triangulum quodcunque A B C, cui æquale omni ex parte est construendum. Intelligo eius latera, tanquam tres lineas rectas datas A B, B C, C A, quarum quælibet duæ maiores sunt reliqua. Deinde sumo rectam D E, æqualem uni lateri, nempe B C; & ex D, interuallo lateris A B, arcum describo, item alium ex E, interuallo reliqui lateris C A, qui priorem secet in F, &c.
 
Proposition 23. 
PROBL. 9. PROPOS. 23. 
第二十三題 
On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle. 
AD datam rectam lineam, datumque; in ea punctum, dato angulo rectilineo æqualem angulum rectilineum constituere. 
一直線。任於一點上、求作一角。與所設角等。 
Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle;  thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE. 
DATA recta sit A B, datumque in ea punctum C, & datus angulus D E F.  Oportet igitur ad rectam A B, in puncto C, angulum constituere æqualem angulo E. 
On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined,  and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed  in such a way that CD is equal to AF, CE to AG, and further DE to FG. 
Sumantur in rectis E D, E F, duo puncta utcunque G, H, quæ recta G H, connectantur:  Deinde constituatur triangulum C I K, habens tria latera æqualia tribus rectis E G, G H, H E,  ita ut C I, æquale sit ipsi E G; & C K, ipsi E H; & I K, ipsi G H. (Quod facile fiet, si C I, sumatur æqualis ipsi E G, & C L, ipsi E H, & I M, ipsi G H. Deinde ex centris C, & I, interuallis vero C L, & I M, circuli describantur secantes sese in K, &c.) Dico angulum C, æqualem esse angulo E. 
Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG. [I. 8] 
Quoniam enim duo latera C I, C K, æqualia sunt duobus lateribus E G, E H, utrumque utrique, & basis I K, basi G H, per constructionem; erit angulus C, angulo E, æqualis: 
Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE.  Q. E. F. 
Effecimus igitur angulum ad C, æqualem angulo F, &c.  Quod facere oportebat.

PRAXIS
NON differt huius problematis praxis ab illa, quam in præcedente problemate tradidimus; propterea quod triangulum constituere oportet æquale alteri triangulo, ut angulus dato angulo æqualis exhibeatur, ut perspicuum est. Facilius tamen hac arte problema efficies. Sit linea data A B, punctumque in ea C, & angulus datus E. Centro igitur E, & interuallo quouis arcus describatur G H: Eodemque interuallo ex centro C, arcus describatur I K, sumaturque beneficio circini arcus I K, arcui G H, æqualis. Recta enim ducta C K, faciet angulum ad C, æqualem angulo E. Nam si ducerentur rectæ I K, G H, essent ipsæ æquales, propterea quod circino non variato utramque distantiam I K, G H, accepimus. Cum ergo & duo latera I C, C K, æqualia sint duobus G E, E H, ob æqualia interualla, quibus arcus sunt descripti; erunt anguli I C K, G E H, æquales.
 
Proposition 24. 
THEOR. 15. PROPOS. 24. 
第二十四題 
If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base. 
SI duo triangula duo latera duobus lateribus æqualia habuerint, utrumque utrique, angulum vero angulo maiorem sub æqualibus rectis lineis contentum: Et basin basi maiorem habebunt. 
兩三角形。相當之兩腰、各等。若一形之腰間角大。則底亦大。 
Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively,  namely AB to DE, and AC to DF, and let the angle at A be greater than the angle at D;  I say that the base BC is also greater than the base EF. 
DVO latera A B, A C, trianguli A B C, æqualia sint duobus lateribus D E, D F, utrumque utrique,  nempe A B, ipsi D E, & A C, ipsi D F; Angulus vero A, maior sit angulo E D F.  Dico basin B C, maiorem esse base E F. 
For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG equal to the angle BAC; [I. 23]  let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined. 
Ad lineam enim D E, ad eiusque punctum D, constituatur angulus E D G, æqualis angulo A; (cadetque recta D G, extra triangulum D E F, cum angulus E D F, minor ponatur angulo A)  ponaturque D G, æqualis ipsi D F, hoc est, ipsi A C. Ducta deinde recta E G, cadet ea aut supra rectam E F; aut in ipsam, aut infra ipsam. Cadat primum supra E F, ducaturque recta F G. 
Then, since AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two sides ED, DG, respectively;  and the angle BAC is equal to the angle EDG;  therefore the base BC is equal to the base EG. [I. 4]  Again, since DF is equal to DG, the angle DGF is also equal to the angle DFG; [I. 5]  therefore the angle DFG is greater than the angle EGF.  Therefore the angle EFG is much greater than the angle EGF.  And, since EFG is a triangle having the angle EFG greater than the angle EGF,  and the greater angle is subtended by the greater side, [I. 19]  the side EG is also greater than EF.  But EG is equal to BC.  Therefore BC is also greater than EF. 
Quia ergo latera A B, A C, æqualia sunt lateribus D E, D G, utrumque utrique,  & angulus A, æqualis angulo E D G, per constructionem:  Erit basis B C, basi E G, æqualis.  Rursus quia duo latera D F, D G, inter se sunt æqualia; erunt anguli D F G, D G F, æquales: Est autem angulus D G F, maior angulo E G F.  Igitur & angulus D F G, eodem angulo E G F, maior erit.  Quare multo maior erit totus angulus E F G, eodem angulo E G F.       In triangulo igitur E F G, maius erit latus E G, latere E F.  Est autem ostensum E G, æquale esse ipsi B C.  Maior igitur erit quoque B C, quam E F. 
Therefore etc.  Q. E. D. 
  Quod est propositum.
CADAT deinde E G, in ipsam E F. Et quia rursus, ut prius basis E G, æqualis est basi B C: & E G, maior quam E F, quod est propositum.
CADAT tertio E G, infra E F, producanturque rectæ D F, D G, usque ad H, & I, & ducatur recta F G. Erit autem rursus, ut prius basis E G, basi B C, æqualis. Deinde quia duo latera D F, D G, æqualia sunt inter se, per constructionem, erunt anguli G F H, F G I, infra basin F G, æquales: Est autem angulus F G I, maior angulo F G E. Igitur & angulus G F H, eodem angulo F G E, maior erit. Quare multo maior erit totus angulus E F G, eodem angulo F G E. In triangulo ergo E F G, maius erit latus E G, latere E F. Est autem ostensum E G, æquale esse ipsi B C. Maior igitur erit quoque B C, basis basi E F. Si igitur duo triangula duo latera duobus lateribus, &c. Quod erat ostendendum.

SCHOLION
SI quis forte roget, cur in 4. propositione Euclides ex eo, quod duo latera unius trianguli æqualia sint duobus lateribus alterius trianguli, utrumque utrique, & anguli contenti dictis lateribus æquales, concluserit non solum æqualitatem basium, verum etiam triangulorum, & reliquorum angulorum; hic autem ex eo, quod duo latera unius trianguli æqualia sint duobus lateribus alterius trianguli, utrumque utrique, anguli vero lateribus illis comprehensi inæquales, colligat tantum inæqualitatem basium, non autem triangulorum, & reliquorum angulorum: Huic respondendum est, necessario id ab Euclide peritissimo Geometra esse factum. Nam ex antecedente huius theorematis semper consequitur basium inæqualitas, ita ut basis illius trianguli, cuius angulus contentus lateribus assumptis est maior, superet basin alterius, cuius angulus minor existit, ut demonstratum est; non autem necesse est, triangulum illud maius hoc esse. Vt enim clarissime ex Proclo demonstrabimus ad propos. 37. huius primi libri, Triangulum maiorem habens angulum aliquando æquale est triangulo minorem habenti angulum, aliquando vero minus eodem, & aliquando maius. Non igitur potuit in uniuersum inferri, ex eo, quod angulus unius trianguli maior est angulo alterius, triangulum etiam maius esse, cum modo æquale sit, modo minus, & modo maius. Idem dici potest de angulis reliquis. Nam in prima figura huius theorematis angulus A B C, minor est semper angulo D E F; cum angulus D E G, (qui æqualis est per 4. propos. angulo A B C,) minor sit eodem angulo D E F, pars toto. In secunda autem figura, existit quidem angulus A B C, angulo D E F, æqualis, per 4. propos. At vero angulus A C B, minor est angulo D F E, cum angulus D F E, maior sit angulo D G F, externus interno, & opposito; & angulus D G F, æqualis sit angulo A C B. In tertia denique figura angulus A B C, maior quidem est angulo D E F, propterea quod angulus D E G. (æqualis existens per 4. propos. angulo A B C,) maior est eodem angulo D E F, totum parte. Sed angulus A C B, minor est angulo D F E. Nam si recta E F, producatur secans rectam D G, in K, fiet angulus D F E, maior angulo D K E, externus interno; Est autem & angulus D K E, maior adhuc angulo D G E, externus quoque interno, & opposito. Multo igitur mator erit angulus D F E, angulo D G E, qui per 4. propos. æqualis est angulo A C B. Quare neque certi quicquam colligi potuit de inæqualitate reliquorum angulorum, cum modo unus altero sit maior, modo minor, & modo æqualis.
 
Proposition 25. 
THEOR. 16. PROPOS. 25. 
第二十五題 
If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other. 
SI duo triangula duo latera duobus lateribus æqualia habuerint, utrumque utrique, basin vero basi maiorem: Et angulum sub æqualibus rectis lineis contentum angulo maiorem habebunt. 
兩三角形。相當之兩腰各等。若一形之底大、則腰間角亦大。 
Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively,  namely AB to DE, and AC to DF;  and let the base BC be greater than the base EF;  I say that the angle BAC is also greater than the angle EDF. 
DVO latera A B, A C, trianguli A B C, æqualia sint duobus lateribus D E, D F, trianguli D E F, utrumque utrique,  hoc eft, A B, ipsi D E, & A C, ipsi D F;  Basis autem B C, maior sit base E F.  Dico angulum A, maiorem esse angulo D. 
For, if not, it is either equal to it or less.  Now the angle BAC is not equal to the angle EDF;  for then the base BC would also have been equal to the base EF, [I. 4]  but it is not;  therefore the angle BAC is not equal to the angle EDF.  Neither again is the angle BAC less than the angle EDF;  for then the base BC would also have been less than the base EF, [I. 24]  but it is not;  therefore the angle BAC is not less than the angle EDF.  But it was proved that it is not equal either;  therefore the angle BAC is greater than the angle EDF. 
Si enim non est angulus A, maior angulo D, erit vel æqualis, vel minor.  Si dicatur esse æqualis, cum etiam duo latera circa A, æqualia sint duobus circa D, utrumque utrique, per hypothesin;  erit & basis B C, æqualis basi B F;  quod est absurdum;  Ponitur enim basis B C, base E F, maior:  Si vero angulus A, dicatur esse minor angulo D; erit, propter æqualitatem laterum circa istos angulos,  basis E F, maior base B C;  quod magis est absurdum,  cum E F, ponatur esse minor quam B C.    Quare angulus A, cum neque possit æqualis esse angulo D, neque minor, erit maior. 
Therefore etc.  Q. E. D. 
Si igitur duo triangula duo latera duobus lateribus æqualia habuerint, &c.  Quod erat ostendendum.

SCHOLION
THEOREMA hoc conuersum est præcedentis. In eo enim ex maiori angulo demonstratum est, basin illi respondentem esse maiorem: In hoc autem ex maiori basi ostensum fuit, angulum illi respondentem maiorem esse. Differunt autem plurimum hæc duo theoremata, nempe 24. & 25. ab illis, quæ explicata sunt propos. 18, & 19. Nam in 19 demonstratum est, in uno eodemque triangulo maiori angulo maius latus respondere: At in 24. idem ostensum fuit in duobus diuersis triangulis, quorum duo latera unius æqualia sunt duobus lateribus alterius &c. Idemque discrimen reperies inter propos. 18. & 25.
MENELAVS Alexandrinus, ut ait Proclus, demonstrat hoc idem theorema ostensiue, hac ratione. Positis eisdem triangulis, ex base maiore B C, abscindatur recta B G, æqualis basi minori E F. Fiat quoque angulus G B H, æqualis angulo D E F, & sit B H, æqualis ipsi B A, atque adeo ipsi D E. Ducta autem recta linea A H, ducatur quoque recta per G, ex H, secans A C, in I. Quoniam igitur duo latera B A, B H, æqualia sunt, erunt anguli B A H, B H A, æquales. Rursus quia latera B G, B H, æqualia sunt lateribus E F, E D, utrumque utrique, & angulus G B H, æqualis, angulo D E F, per constructionem: erit basis H G, basi D F, atque adeo ipsi A C, æqualis, angulusque G H B, angulo E D F. Et quoniam recta H I, maior est quam H G, quæ est ostensa æqualis ipsi A C, erit quoque maior H I, quam A C: Sed A C, maior est adhuc, quam A I. Multo ergo maior erit H I, quam A I. Quare angulus I A H, maior erit angulo I H A. Additis igitur duobus angulis B A H, B, H A, qui ostensi sunt æquales, fiet totus angulus B A C, toto angulo B H G, maior: Sed angulus B H G, demonstratus fuit æqualis angulo D. Maior igitur etiam erit angulus B A C, angulo D, quod est propositum.
HERON autem idem ex eodem Proclo hoc modo demonstrat. Positis eisdem triangulis, producatur basis minor E F, ad G, ut sit E G, æqualis basi maiori B C. Deinde centro D, interuallo autem D F, describatur circulus, producaturque E D, ad H, in circunferentiam. Quoniam igitur D H, est æqualis ipsi D F, erit quoque D H, æqualis ipsi A C. Additis igitur æqualibus D E, A B, fient A C, A B, simul æquales toti H E: Sed A C, A B, simul maiores sunt, quam B C, atque adeo quam E G. Igitur & H E, maior erit, quam E G; Quare circulus descriptus ex centro E, & interuallo E G, intersecabit rectam E H, atque adeo circumferentiam prioris circuli in I, & K, punctis: ad K, autem ducantur rectæ D K, E K. Et quoniam duo latera A B, A C, æqualia sunt duobus lateribus D E, D K, utrumque utrique, (est enim D K, æquale ipsi D F, per definitionem circuli: D F, autem positum est æquale lateri A C.) & basis B C, basi E K, æqualis: (cum E K, æqualis sit ipsi E G, per definitionem circuli: E G, vero recta per constructionem facta sit æqualis basi B C.) Erit angulus B A C, angulo E D K, æqualis: Sed angulus E D K, maior est angulo E D F. Quare & angulus A, angulo E D F, maior existet. Quod est propositum.
 
Proposition 26. 
THEOR. 17. PROPOS. 26. 
第二十六題 
If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle. 
SI duo triangula duos angulos duobus angulis æquales habuerint, utrumque utrique, unumque latus uni lateri æquale, siue quod æqualibus adiacet angulis, seu quod uni æqualium angulorum subtenditur: & reliqua latera reliquis lateribus æqualia, utrumque utrique, & reliquum angulum reliquo angulo æqualem habebunt. 
二支
兩三角形。有相當之兩角等、及相當之一邊等。則餘兩邊必等。餘一角亦等。其一邊。不論在兩角之內、及一角之對。 
Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively,  namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD;  and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF;  I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF,  and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF. 
SINT duo anguli B, & C, trianguli A B C, æquales duobus angulis E, & E F D, trianguli D E F, uterque utrique,  hoc est, B, ipsi E, & C, ipsi E F D;  Sitque primo latus B C, quod angulis B, & C, adiacet, lateri E F, quod angulis E, & E F D, adiacet, æquale.  Dico, reliqua quoque latera A B, A C, reliquis lateribus D E, D F, æqualia esse, utrumque utrique, hoc est, A B, ipsi D E, & A C, ipsi D F, ea nimirum, quæ æqualibus angulis subtenduntur;  reliquumque angulum A, reliquo angulo D. 
For, if AB is unequal to DE, one of them is greater.  Let AB be greater, and let BG be made equal to DE; and let GC be joined. 
Si enim latus A B, non est æquale lateri D E,  sit D E, maius, a quo abscindatur recta linea E G, æqualis rectæ lineæ A B, ducaturque recta G F. 
Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively;  and the angle GBC is equal to the angle DEF;  therefore the base GC is equal to the base DF,  and the triangle GBC is equal to the triangle DEF,  and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]  therefore the angle GCB is equal to the angle DFE.  But the angle DFE is by hypothesis equal to the angle BCA;  therefore the angle BCG is equal to the angle BCA, the less to the greater: which is impossible.  Therefore AB is not unequal to DE, and is therefore equal to it.  But BC is also equal to EF; therefore the two sides AB, BC are equal to the two sides DE, EF respectively,  and the angle ABC is equal to the angle DEF;  therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I. 4] 
Quoniam igitur latera A B, B C, æqualia sunt lateribus G E, E F, utrumque utrique,  & anguli B, & E, æquales per hypothesin: Erit angulus C, æqualis angulo E F G.          Ponitur autem angulus C, æqualis angulo E F D.  Quare & angulus E F G, eidem angulo E F D, æqualis erit, pars toti; Quod est absurdum.  Non est igitur latus A B, inæquale lateri D E, sed æquale.  Quamobrem, cum latera A B, B C, æqualia sint lateribus D E, E F, utrumque utrique,  & anguli contenti B, & E, æquales;  erunt & bases A C, D F, & anguli reliqui A, & D, æquales. Quod est propositum. 
Again, let sides subtending equal angles be equal, as AB to DE;  I say again that the remaining sides will be equal to the remaining sides,  namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF. 
SINT secundo latera A B, D E, subtendentia æquales angulos C, & E F D, inter se æqualia.  Dico rursus reliqua latera B C, C A, reliquis lateribus E F, F D, esse æqualia, utrumque utrique,  hoc est, B C, ipsi E F, & C A, ipsi F D; reliquumque angulum A, reliquo angulo D, æqualem. 
For, if BC is unequal to EF, one of them is greater.  Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.  Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively,  and they contain equal angles;  therefore the base AH is equal to the base DF,  and the triangle ABH is equal to the triangle DEF,  and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]  therefore the angle BHA is equal to the angle EFD.  But the angle EFD is equal to the angle BCA;  therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA: which is impossible. [I. 16]  Therefore BC is not unequal to EF, and is therefore equal to it.  But AB is also equal to DE;  therefore the two sides AB, BC are equal to the two sides DE, EF respectively,  and they contain equal angles;  therefore the base AC is equal to the base DF, the triangle ABC equal to the triangle DEF, and the remaining angle BAC equal to the remaining angle EDF. [I. 4] 
Si enim latus B C, non est æquale lateri E F,  sit E F, maius; ex quo sumatur recta E G, æqualis ipsi B C, ducaturque recta D G.  Quoniam igitur latera A B, B C, æqualia sunt lateribus D E, E G, utrumque utrique,  & anguli contenti B, & E æquales, per hypothesin;        Erit angulus C, angulo E G D, æqualis:   Ponitur autem angulus C, angulo E F D, æqualis;  Igitur & angulus E G D, angulo eidem E F D, æqualis erit, externus interno, & opposito, quod est absurdum. Est enim maior.  Non ergo est latus B C, lateri E F, inæquale. Quocirca, ut prius, colligetur institutum ex 4. propos. huius libri.         
Therefore etc.  Q. E. D. 
Si duo igitur triangula duos angulos duobus angulis æquales habuerint, &c.  Quod demonstrandum erat.

SCHOLION
PRIOR huius theorematis pars conuersa est 4. propositionis, quoad eam partem, in qua ex æqualitate laterum, & angulorum ipsis contenterum, collecta fuit æqualitas basium, & angulorum super bases. Nam in priori parte huius theorematis ex æqualitate basium B C, E F, & angulorum super has bases, demonstratum est, reliqua latera unius trianguli reliquis lateribus æqualia esse, reliquumque angulum reliquo angulo, &c. Quod quidem alia nos ratione iam demonstrauimus ad propositionem octauam huius liber quem modum eo loco monuimus.
 
Proposition 27. 
THEOR. 18. PROPOS. 27. 
第二十七題 
If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another. 
SI in duas rectas lineas recta incidens linea alternatim angulos æquales inter se fecerit: parallelæ erunt inter se illæ rectæ lineæ. 
兩直線。有他直線交加其上若內相對兩角等。卽兩直線必平行。 
For let the straight line EF falling on the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another;  I say that AB is parallel to CD. 
IN duas rectas A B, C D, incidens recta E F, faciat angulos alternatim A G H, D H G, inter se æquales.  Dico lineas A B, C D, esse parallelas. 
For, if not, AB, CD when produced will meet either in the direction of B, D or towards A, C.  Let them be produced and meet, in the direction of B, D, at G.  Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite angle EFG: which is impossible. [I. 16]  Therefore AB, CD when produced will not meet in the direction of B, D.  Similarly it can be proved that neither will they meet towards A, C.  But straight lines which do not meet in either direction are parallel; [Def. 23]  therefore AB is parallel to CD. 
Si enim non sunt parallelæ, coibunt tandem, si producantur infinite. Si namque non coirent unquam, parallelæ essent, ex parallelarum definitione.  Conueniant ergo ad partes B, & D, in puncto I.  Quoniam igitur triangulum est G I H, (cum A B, recta continuata sit, item recta C D, usque ad punctum I,) & angulus A G H, positus est æqualis angulo D H G; erit externus angulus A G H, æqualis interno, & opposito D H G, quod est absurdum; quoniam externus interno maior est.    Quod si A B, C D, coire dicantur ad partes A, & C, in puncto K, erit rursus eadem ratione angulus externus D H G, æqualis interno, & opposito A G H, quod est absurdum. Non igitur coibunt lineæ A B, C D.    Quare parallelæ erunt. Eodem modo, si ponantur anguli alterni B G H, C H G, æquales, demonstrabitur, lineas A B, C D, esse parallelas. 
Therefore etc.  Q. E. D. 
Si igitur in duas rectas lineas recta incidens, &c.  Quod erat ostendendum.

SCHOLION
NECESSE est, ut lineæ, quæ dicuntur parallelæ, in eodem existant plano, ut ex definitione constat: Quare non satis est, duos angulos alternos æquales inter se esse, ut duæ lineæ probentur esse parallelæ, nisi ponatur, eas in uno, eodemque existere plano. Fieri enim potest, ut lineæ rectæ incidens in duas rectas non in eodem plano existentes, faciat alternos angulos æquales. Sit enim C D, perpendicularis ad A B, rectam, quæ in subiecto plano existit; & ex C, in alio plano, ad C D, ducatur alia perpendicularis C E, ita ut punctum E, intelligatur in sublimi. Quo posito, perspicuum est, rectam C D, incidentem in rectas C E, A B, facere duos angulos E C D, A D C, alternos æquales, cum sint recti; & tamen C E, A B, non sunt parallelæ, quod non in eodem existant plano. Non apposuit autem Euclides in propositione hanc conditionem; in eodem plano existentes: sicut neque in subsequentibus; quoniam cum in prioribus sex libris agatur de planis duntaxat, ut supra diximus, omnia intelligenda sunt necessario in eodem plano exsistere. In undecimo vero libro, & aliis, qui ipsum sequuntur, monebit semper, lineas aliquas in eodem esse plano, vel in diuersis planis, quia in illis libris disseritur de solidis, in quibus diuersa plana confiderari possunt. Quod idem dicendum est de punctis exira lineas, & superficies, &c.
 
Proposition 28. 
THEOR. 19. PROPOS. 28. 
第二十八題 
If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another. 
SI in duas rectas lineas recta incidens linea externum angulum interno, & opposito, & ad easdem partes, æqualem fecerit; Aut internos, & ad easdem partes duobus rectis æquales: Parallelæ erunt inter se ipsæ rectæ lineæ. 
二支
兩直線。有他直線交加其上。若外角、與同方相對之內角等。或同方兩內角、與兩直角等。卽兩直線必平行。 
For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same side, namely BGH, GHD, equal to two right angles;  I say that AB is parallel to CD. 
IN duas rectas A B, C D, recta incidens E F, faciat primo externum angulum E G A, æqualem angulo interno, & opposito ad easdem partes G H C.  Dico rectas A B, C D, esse parallelas. 
For, since the angle EGB is equal to the angle GHD, while the angle EGB is equal to the angle AGH, [I. 15]  the angle AGH is also equal to the angle GHD; and they are alternate;  therefore AB is parallel to CD. [I. 27] 
Quoniam enim angulo E G A, æqualis ponitur angulus G H C; & eidem angulo E G A, æqualis est angulus H G B;  erunt anguli alterni G H C, H G B, æquales.  Quare lineæ A B, C D, parallelæ erunt. Idem ostendetur, si angulus externus E G B, æqualis ponatur interno G H D. 
Again, since the angles BGH, GHD are equal to two right angles, and the angles AGH, BGH are also equal to two right angles, [I. 13]  the angles AGH, BGH are equal to the angles BGH, GHD.  Let the angle BGH be subtracted from each;  therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate;  therefore AB is parallel to CD. [I. 27] 
SECVNDO faciat recta E F, angulos internos ex eadem parte, nempe A G H, C H G, duobus rectis æquales. Dico rursus rectas A B, C D, esse parallelas. Quoniam enim anguli A G H, C H G, duobus rectis æquales ponuntur; Sunt autem & anguli A G E, A G H, duobus rectis æquales;  Erunt duo anguli A G H, C H G, duobus angulis A G E, A G H, æquales.  Ablato igitur communi angulo A G H,  remanebit angulus A G E, externus angulo C H G, interno, & opposito ad easdem partes æqualis.  Quare ut iam ostensum est, erunt rectæ A B, C D, parallelæ. Idem ostendetur, si duo anguli B G H, D H G, duobus rectis ponantur æquales. 
Therefore etc.  Q. E. D.   
Si igitur in duas rectas lineas recta incidens linea externum angulum, &c.  Quod erat demonstrandum.

SCHOLION
IAMDVDVM pronuntiatum undecimum a principiorum numero reieciemus. Cum igitur sequens propositio 29. illi innitatur, ita ut absque eo demonstrari non posse, necesse est, ut illud ex hactenus demonstratis theorematibus, qua ex eo nulla ratione dependent, cum Proclo confirmemus, ut antea polliciti sumus. Hoc autem facile præstabimus, si prius duo explicemus, quorum primum hoc sit.

SI ab uno puncto duæ rectæ lineæ angulum facientes infinite producantur, ipsarum distantia omnem finitam magnitudinem excedet.
EXEANT a puncto A, duæ rectæ A B, A C, facientes angulum A. Quoniam igitur puncta D, & E, plus inter se distant, quam F, & G. Item punctæ B, & C, plus quam D, & E, & ita deinceps, si producantur ultra rectæ lineæ A B, A C, perspicuum est, extrema earum puncta infinito spatio inter se distare, si infinite ipsæ producantur. Si enim non infinito spatio distarent, augeri posset eorum distantia; igitur & lineæ ipsæ ultra produci, quod est absurdum, cum ponantur infinite iam esse productæ. Quare si dictæ lineæ A B, A C, producantur infinite, ipsarum distantia excedet omnem finitam distantiam. Hoc pronunciato usus est & Aristoreles liber I. de cœlo, ubi demonstrauit, mundum non esse infinitum. Secundum quod debes explicari, ita se habet.

SI duarum parallelarum rectarum linearum alteram secet quædam recta linea, reliquam quoque productam secabit.
SINT duæ parallelæ A B, C D, & recta E F, secet ipsam A B, in G. Dico rectam E F, si producatur, secturam esse quoque ipsam C D. Quoniam duæ rectæ G B, G F, in puncto G, angulum faciunt, si producantur infinite, excedent omnem finitam distantiam; igitur & distantiam, qua parallelæ A B, a parallela C D, distat, cum hac distantia sit finita, alias enim non essent lineæ parallelæ. Quare quando distantia G B, a G F, maior iam fuerit ea, quæ inter parallelas est, necesse est rectam G F, productam secuisse rectam C D. Nam quamdiu G F, continebitur inter duas parallelas, minori distantia a G B, remouebitur, quam C D, ab eadem G B, ut constat. His igitur ita expositis, facile demonstrabitur hoc theorema, quod est apud Euclidem, tertium decimum pronunciatum.

SI in duas rectas lineas altera recta incidens internos, ad easdemque partes, angulos duobus rectis minores faciat; Duæ illæ rectæ lineæ infinite productæ sibi mutuo incident ad eas partes, ubi sunt anguli duob rectis minores.
IN rectas A B, C D, incidens recta E F, faciat internos angulos ad partes B, & D, vt B G H, D H G, duobus rectis minores. Dico rectas A B, C D, coire ad easdem partes B, & D. Quoniam enim duo anguli B G H, D H G, minores ponuntur esse duobus rectis: Sunt autem duo anguli D H G, D H F, duobus rectis æquales: Erunt duo anguli D H G, D H F, maiores duobus angulis D H G, B G H. Ablato ergo communi angulo D H G, remanebit angulus D H F, maior angulo B G H. Si igitur ad rectam F G, & ad punctum, G, constituatur angulus K G H, æqualis angulo D H F, cadet G K. supra G B, secabitque producta rectam A B. Quoniam igitur in duas rectas I K, C D, recta incidens E F, facit angulum externum D H F, æqualem interno, & opposito K G H. Erunt rectæ I K, C D, parallelæ. Secat autem recta A B, ipsam I K, in G. Producta igitur secabit quoque ipsam C D, ut demonstratum est. Quare A B, cum C D, conueniet ad partes B, & D, nimirum in puncto L. quod est propositum.
QVAMVIS autem, concesso principio nostro, optime a nobis demonstratum sit tertiumdecimum hoc axioma, & a Proclo etiam, si eius principium difficilius quidem, quam nostrum, admittatur, ut iure optimo inter theoremata, & non inter principia possit connumerari; tamen ne ordinem Euclidis in quoquam immutemus, utemur eo in omnibus propositionibus, quarum demonstrationes ex ipso pendent, tanquam pronunciato, praesertim cum facile ei assensus præberi queat, intellecta prius recte propositione 28. Si enim lineæ rectæ propositæ parallelæ sunt, ita ut nunquam coeant, sed semper æquali inter se distantia progrediantur, etiamsi infinite producantur, quando recta in eas incidens facit duos angulos internos, ad easdem partes duobus rectis æquales, ut demonstratum fuit; quis non videt, si eadem incidens in duas rectas faciat anulos internos, ad easdem partes duobus rectis minores, alteram alteri appropinquare, ad eas partes, ad quas sunt interni anguli duobus rectis minores; quandoquidem æquali distantia procederent, si iidem anguli paulo maiores essent, duobus videlicet rectis æquales, ut hæc propositio 28. demonstrauit?
 
 
 
Proposition 29. 
THEOR. 20. PROPOS. 29. 
第二十九題三支 
A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles. 
IN parallelas rectas lineas recta incidens linea; Et alternatim angulos inter se æquales efficit; & externum interno, & opposito, & ad easdem partes æqualem; & internos, & ad easdem partes, duobus rectis æquales facit. 
兩平行線。有他直線交加其上。則內相對兩角、必等。外角與同方相對之內角、亦等。同方兩內角、亦與兩直角等。 
For let the straight line EF fall on the parallel straight lines AB, CD;  I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the same side, namely BGH, GHD, equal to two right angles. 
IN parallelas A B, C D, recta incidat E F.  Dico primum, angulos alternos A G H, D H G, inter se esse æquales. 
先解曰。此反前二題。故同前圖。有甲乙、丙丁、二平行線。加他直線戊己。交於庚、於辛。  題言甲庚辛、與丁辛庚、內相對兩角必等。 
For, if the angle AGH is unequal to the angle GHD, one of them is greater.  Let the angle AGH be greater.  Let the angle BGH be added to each;  therefore the angles AGH, BGH are greater than the angles BGH, GHD.  But the angles AGH, BGH are equal to two right angles; [I. 13]  therefore the angles BGH, GHD are less than two right angles.  But straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]  therefore AB, CD, if produced indefinitely, will meet;  but they do not meet, because they are by hypothesis parallel.  Therefore the angle AGH is not unequal to the angle GHD, and is therefore equal to it.  Again, the angle AGH is equal to the angle EGB; [I. 15]  therefore the angle EGB is also equal to the angle GHD. [C.N. 1]  Let the angle BGH be added to each;  therefore the angles EGB, BGH are equal to the angles BGH, GHD. [C.N. 2]  But the angles EGB, BGH are equal to two right angles; [I. 13]  therefore the angles BGH, GHD are also equal to two right angles. 
Si enim non sunt æquales, sit alter, nempe A G H, maior.  Quoniam igitur angulus A G H, maior est angulo D H G,  si addatur communis angulus BGH;  erunt duo A G H, B G H, maiores duobus D H G, B G H:  At duo A G H, B G H, æquales sint duobus rectis.  duo D H G, B G H, minores sunt duobus rectis.    Quare cum sint interni, & ad easdem partes, B, & D, coibunt lineæ A B, C D, ad eas partes,  quod est absurdum, cum ponantur esse parallelæ.  Non est igitur angulus A G H, maior angulo D H G: Sed neque minor. Eadem enim ratione ostenderetur, rectas coire ad partes A, & C.  Igitur æquales erunt anguli alterni A G H, D H G. Eademque est ratio de angulis alternis B G H, C H G.  DICO secundo, angulum externum A G E, æqualem esse interno, & ad easdem partes opposito C H G. Quoniam enim angulo B G H, æqualis est alternus C H G, ut ostensum est; & eidem B G H, æqualis est angulus A G E. Erunt anguli A G E, C H G, inter se quoque æquales. Eodem modo demonstrabitur, angulum B G E, æqualem esse angulo D H G.  DICO tertio, angulos internos ad easdem partes, A G H, C H G, æquales esse duobus rectis. Quoniam enim ostensum fuit, angulum externum A G E, æqualem esse angulo C H E, interno; si addatur communis A G H,  erunt duo A G E, A G H, duobus C H G, A G H, æquales:  Sed duo A G E, A G H, æquales sunt duobus rectis.  Igitur & duo anguli C H G, A G H, æquales duobus rectis erunr. Eodem modo anguli B G H, D H G, duobus erunt rectis æquales. 
論曰。如云不然、  而甲庚辛、大於丁辛庚。  則丁辛庚、加辛庚乙。  宜小於辛庚甲、加辛庚乙矣。公論四  夫辛庚甲、辛庚乙。元與兩直角等。本篇十三據如彼論。  則丁辛庚、辛庚乙、兩角、小於兩直角。  而甲乙、丙丁、兩直線。向乙丁行。必相遇也。公論十一  可謂平行線乎。        次解曰。戊庚甲外角、與同方相對之庚辛丙內角、等。論曰。乙庚辛、與相對之丙辛庚、兩內角等。本題 則乙庚辛交角相等之戊庚甲、本篇十五 與丙辛庚、必等。公論一。  後解曰。甲庚辛、丙辛庚、兩內角、與兩直角等。
論曰。戊庚甲、與庚辛丙、兩角旣等。本題 而每加一甲庚辛角。 
則庚辛丙、甲庚辛、兩角、與甲庚辛、戊庚甲、兩角、必等。公論二  夫甲庚辛、戊庚甲、本與兩直角等。本篇十三  則甲庚辛、丙辛庚、兩內角、亦與兩直角等。 
Therefore etc.  Q. E. D. 
In parallelas ergo rectas lineas recta incidens linea, & alternatim angulos, & c.  Quod erat demonstrandum.

SCHOLION
CONVERTIT autem hoc præsens theorema duo præcedentia theoremata, ut perspicuum est.
 
   
Proposition 30. 
THEOR. 21. PROPOS. 30. 
第三十題 
Straight lines parallel to the same straight line are also parallel to one another. 
QVÆ eidem rectæ lineæ parallelæ, & inter se sunt parallelæ. 
兩直線。與他直線平行。則元兩線亦平行。 
Let each of the straight lines AB, CD be parallel to EF;  I say that AB is also parallel to CD. 
SINT rectæ A B, C D, eidem rectæ E F, parallelæ.  Dico & ipsas A B, C D, esse inter se parallelas. 
For let the straight line GK fall upon them; 
Quoniam enim omnes hæ lineæ in eodem ponuntur esse plano. (Nam propositio 9. undecimi libri agetur de lineis in diuersis planis) ducta recta G H, secabis omnes, nimirum A B, in I; C D, in K; & E F, in L. 
Then, since the straight line GK has fallen on the parallel straight lines AB, EF, the angle AGK is equal to the angle GHF. [I. 29]  Again, since the straight line GK has fallen on the parallel straight lines EF, CD,the angle GHF is equal to the angle GKD. [I. 29]  But the angle AGK was also proved equal to the angle GHF;  therefore the angle AGK is also equal to the angle GKD; [C.N. 1]  and they are alternate.  Therefore AB is parallel to CD. 
Quia igitur A B, ponitur parallela ipsi E F, erit angulus A I L, alterno F L I, æqualis.  Rursus quia C D, ponitur etiam parallela ipsi E F, erit angulus D K I, eidem angulo F L I, nempe internus externo, vel externus interno, æqualis.    Quare anguli A I L, D K I, æquales inter se quoque erunt.  Cum igitur sint alterni,  erunt rectæ A B, C D, parallelæ inter se. 
  Q. E. D. 
Quæ igitur eidem rectæ lineæ parallelæ, & inter se sunt parallelæ.  Quod demonstrandum erat.

SCHOLION
QVOD si quis dicat, duas rectas A I, B I, parallelas esse rectæ C D, & tamen ipsas non esse parallelas; Occurrendum est, duas A I, B I, non esse duas lineas, sed partes tantum unius lineæ. Concipiendum enim est animo, quaslibet parallelas infinite esse productas; Constat autem A I, productam coincidere cum B I. Quamobrem quæ eidem rectæ lineæ parallelæ, & inter se sunt parallelæ: vel certe, quando inter se coeunt. Quod ita demonstrabitur. Sint duæ rectæ A B, A C, coeuntes in A, parallelæ ipsi D E. Dico illas in rectum esse constitutas. Ex puncto enim A, ducatur recta A F, secans D E, in F, utcunque. Quoniam igitur A B, D E, sunt parallelæ, erunt anguli alterni B A F, A F E, æquales. Addito ergo communi angulo C A F, erunt duo anguli ad A, æquales duobus angulis C A F, A F E. Sed hi duo æquales sunt duobus rectis, cum sint interni inter duas parallelas A C, D E. Igitur & duo anguli ad A, duobus eruntrectis æquales; ac propterea in rectum erunt constitutæ ipsæ AB, AC. Quod est propositum.
 
Proposition 31. 
PROBL. 10. PROPOS. 31. 
第三十一題 
Through a given point to draw a straight line parallel to a given straight line. 
A dato puncto, datæ rectæ lineæ parallelam rectam lineam ducere. 
一點上求作直線、與所設直線平行。 
Let A be the given point, and BC the given straight line;  thus it is required to draw through the point A a straight line parallel to the straight line BC. 
EX puncto A, ducenda sit linea parallela lineæ B C.    
Let a point D be taken at random on BC, and let AD be joined;  on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23];  and let the straight line AF be produced in a straight line with EA. 
Ducatur ex A, ad B C, linea A D,  utcunque, faciens angulum quemcunque A D B; Cui ad A, æqualis constituatur E A D.  Dico rectam E A, extensam ad F, quantumlibet, parallelam esse ipsi B C. 
Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. [I. 27] 
Cum enim anguli alterni A B D, D A E, æquales sint, per constructionem, erunt rectæ B C, E F, parallelæ. 
Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC.  Q. E. F. 
A dato igitur puncto, datæ rectæ lineæ, &c.  Quod erat faciendum.

SCHOLION
DEBET autem punctum datum in tali esse loco situm extra lineam datam, ut hæc producta cum illo non conueniat. Quod quidem aperte colligitur ex ipsa constructione problematis. Nam ex puncto dato ducenda est linea faciens angulum aliquem cum linea data, qui fieri non posset, si punctum in directum iaceret cum ipsa linea data. Quemadmodum autem ab uno, eodemque puncto ad eandem rectam non plures perpendiculares, quam una, ducuntur, ut ostendimus propositio 17. ex Proclo; ita etiam per idem punctum, datæ rectæ plures parallelæ, quam una, duci nequeunt. Si enim duæ ducerentur, conuenirent ipsæ in puncto illo eodem, quod est absurdum, cum sint parallelæ inter se, propterea quod uni & eidem, cui parallelæ dicuntur duci, sunt parallelæ.
EX hoc porro problemate, & illo, quod propositio 23. continetur, facili negotio constituemus parallelogrammum, cuius unus angulorum æqualis sit dato angulo rectilineo, lateraque angulum illum comprehendentia datis duabus rectis lineis æqualia.
SINT enim datæ rectæ A B, oporteatque constituere parallelogrammum habens angulum æqualem dato angulo rectilineo C, lateraque circa illum angulum rectis A B, æqualia. Sumpta recta D E, quæ rectæ A, sit æqualis, fiat angulus E D F, angulo C, & recta D F, rectæ B, æqualis. Deinde per E, agatur recta E G, ipsi D F, parallela, & per F, recta F G, ipsi D E, parallela secans E G, in G. Quoniam ergo & latera D F, E G, & D E, F G, parallela sunt, ex constructione; parallelogrammum erit D E G F. Quod cum ex constructione, habeat angulum D, angulo dato C, æqualem, & latera D E, D F, circa dictum angulum D, datis rectis A, B, æqualia; factum erit, quod proponitur.

PRAXIS
SIT ducenda parallela ipsi B C, per punctum A. Ducatur recta A D, utcunque ad B C, & ex D, & A, ad idem interuallum quodlibet describantur duo arcus ad diuersas partes, unus ad partes B, alter ad partes C. Deinde beneficio circini arcui E F, abscindatur ex arcu altero arcus G H, æqualis. Si igitur ex A, per H, recta ducatur, erit hæc parallela ipsi B C. Nam anguli EDF, HAG, sunt æquales, ut constat ex praxi propos. 23. &c.
ALIO modo ducetur per idem punctum A, datum linea parallela lineæ datæ B C, hac arte. Ex centro A, ad quoduis interuallum describatur arcus secans B C, in puncto D; & eodem interuallo ex D, sumatur punctum E, in eadem recta B C. Deinde eodem interuallo ex A, & E, describantur duo arcus secantes sese in F. Nam ducta recta A F, erit parallela rectæ B C. Quoniam enim propter idem interuallum assumptum recta A F, æqualis est rectæ D E; & recta A D, rectæ E F, si ducerentur hæ lineæ; erit A F, oppositæ D E, parallela, ut postea demonstrabimus propositio 34. Quod si punctum A, vicinum fuerit rectæ B C, commodius hac lege parallela optata ducetur. Ex A, sumatur punctum D, in B C, ad quoduis interuallum; Et ex quouis puncto eiusdem rectæ B C, nempe E, quod tæmen aliquantulum distet a puncto D (Quo enim maior fuerit distantia inter D, & E, eo facilius & accuratius parallela ducetur) eodem interuallo arcus describatur ad partes A. Deinde ex A, interuallo D E, alter arcus descriptus secet priorem arcum in F. Recta namque ducta A F, erit parallela rectæ B C, ut prius; quia recta A F, æqualis est rectæ D E, ob idem interuallum; & recta A D, rectæ E F, si hæ rectæ ductæ essent, &c.
 
Proposition 32. 
THEOR. 22. PROPOS. 32. 
第三十二題 
In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles. 
CVIVSCVNQVE trianguli vno latere producto: Externus angulus duobus internis, & oppositis est æqualis. Et trianguli tres interni anguli duobus sunt rectis æquales. 
凡三角形之外角、與相對之內兩角幷、等。凡三角形之內三角幷、與兩直角等。 
Let ABC be a triangle, and let one side of it BC be produced to D;  I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC,  and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles. 
PRODVCATVR in triangulo A B C, latus B C, ad D.  Dico primo, angulum externum A C D, æqualem esse duobus internis, & oppositis simul A, & B.  86  
Then, since AB is parallel to CE, and AC has fallen upon them, the alternate angles BAC, ACE are equal to one another. [I. 29]  Again, since AB is parallel to CE, and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. [I. 29]  But the angle ACE was also proved equal to the angle BAC;  therefore the whole angle ACD is equal to the two interior and opposite angles BAC, ABC. 
Quoniam igitur recta A C, incidit in parallelas A B, C E, erunt anguli interni A, & A C E, æquales.  Rurus, quia recta B D, in easdem parallelas incidit, erit angulus externus D C E, æqualis interno B.    Additis igitur æqualibus A C E, & A, fiet totus A C D (qui ex duobus D C E, A C E, componitur) duobus A, & B, simul æqualis. Quod est propositum. 
Let the angle ACB be added to each;  therefore the angles ACD, ACB are equal to the three angles ABC, BCA, CAB.  But the angles ACD, ACB are equal to two right angles; [I. 13]  therefore the angles ABC, BCA, CAB are also equal to two right angles. 
DICO secundo, tres angulos internos eiusdem trianguli A, B, & A C B, duobus esse rectis æquales. Cum enim externus angulus A C D, ut ostensum fuit, æqualis sit duobus internis A, & B; si addatur communis A C B,  erunt duo anguli A C D, A C B, æquales tribus A, B, & A C B;  Sed duo A C D, A C B, æquales sunt duobus rectis.  Igitur & tres interni, A, B, A C B, duobus sunt rectis æquales. 
Therefore etc.  Q. E. D. 
Quare cuiuscunque trianguli uno latere producto, &c.  Quod erat demonstrandum.

SCHOLION
CVM demonstratum sit propositio 16. angulum externum cuiusuis trianguli maiorem esse utrolibet interno, & opposito; Hic autem, eundem externum eisdem internis simul esse æqualem, perspicuum est, alterutrum internorum, & oppositorum superari ab externo, reliquo interno angulo opposito. Vt in triangulo proposito angulus A, internus superatur ab angulo externo A C D, angulo B, interno: Et angulus B, internus superatur ab eodem externo angulo A C D, angulo A, interno, quandoquidem angulus A C D, duobus angulis A, & B, est ostensus hoc loco æqualis. Rursum, quia demonstratum est propositio 17. duos angulos cuiuslibet trianguli quomodocunque sumptos, duobus esse rectis minores; Hic vero omnes tres duobus rectis æquales esse; manifestum est, duos a duobus rectis deficere, reliquo angulo trianguli. Vt in eodem triangulo, duo anguli A, & B, a duobus rectis deficiunt angulo A C B, &c.
OMNE porro triangulum habere tres angulos duobus rectis æquales, primi omnium, ut refert Eudemus, Pythagorei demonstrarunt hac ratione. Sit triangulum A B C, & per punctum A, ducatur rectæ B C, parallela D E. Quoniam igitur anguli alterni D A B, & A B C, æquales sunt; si addantur æquales E A C, & A C B (sunt enim & hi alterni) erunt duo anguli D A B, E A C, duobus A B C, A C B, æquales. Addito ergo communi angulo B A C, erunt tres anguli D A B, B A C, C A E, æquales tribus angulis A B C, B A C, A C B. Sed anguli D A B, B A C, C A E, æquales sunt duobus rectis, ut constat ex propositione decimatertia. Igitur & in triangulo A B C, anguli A B C, B A C, A C B, duobus sunt rectis æquales, quod est propositum. Ex hoc autem facile concludemus, angulum externum A C F, si latus B C, sit protractum, æqualem esse duobus internis, & oppositis A B C, B A C. Quoniam enim anguli A B C, B A C, A C B, æquales sunt duobus rectis, ut ostensum fuit. Sunt autem & anguli A C F, A C B, duobus rectis æquales; Erunt anguli A B C, B A C, A C B, angulis A C F, A C B, æquales. Dempto igitur communi angulo A C B, remanebit angulus A C F, duobus angulis A B C, B A C, æqualis.
FACILE etiam conuerti poterit prima pars propositionis Euclidis: Hoc est, si ab uno angulo trianguli linea recta ducatur, ut angulus externus æqualis sit duobus internis, & oppositis, illam lineam esse in directum ipsi lateri constitutam. Ex C, enim ducatur C D, recta, sitque angulus A C D, æqualis duobus angulis A B C, B A C. Dico rectas B C, C D, in directum iacere. Cum enim angulus A C D, æqualis sit angulis A B C, B A C, si addatur communis angulus A C B, erunt anguli A C D, A C B, æquales angulis A B C, B A C, A C B: Sed A B C, B A C, A C B, æquales sunt duobus rectis. Igitur & anguli A C D, A C B, duobus erunt rectis æquales. Quare B C, C D, unam lineam rectam constituunt.

QVOT ANGVLIS RECTIS æquiualeant anguli omnes interni cuiuscunque figuræ rectilineæ.

DVOBVS modis ex hac propos. 32. colligemus, quotnam rectis angulis æquiualeant interni anguli figuræ cuiuslibet rectilineæ, quorum primus hic est.

OMNES anguli figuræ rectilineæ cuiusuis sunt æquales bis tot rectis angulis, quota ipsa est inter figuras rectilineas.

HOC est, omnes anguli primæ figuræ rectilineæ æquales sint bis uni recto, id est, duobus rectis; Anguli vero secundæ figuræ rectilineæ æquales sunt bis duobus rectis, nempe quatuor rectis; Anguli autem tertiæ figuræ rectilineæ æquales sunt bis tribus rectis, sex videlicet rectis. Et sic reliquis. Eum autem locum quælibet figura rectilinea obtinet inter figuras rectilineas, quem indicat numerus laterum, seu angulorum, dempto binario; quoniam duæ lineæ rectæ superficiem non concludunt, unde nec figuram constituunt, sed cum minimum tres rectæ lineæ ad figuræ constitutionem requiruntur. Ex quo fit triangulum, quia habet tria latera, totidemque angulos, esse primam inter rectilineas figuras. Nam binario dempto ex tribus, relinquitur unum. Sic erit figura habens 20. latera seu angulos, inter figuras rectilineas decimaoctaua, cum binarius subtractus ex 20. relinquat 18. Idem iudicium de aliis figuris est habendum. Itaque figura contenta 20. lateribus, cum sit decimaoctaua, habebit 20. angulos æquiualentes 36. rectis angulis, nempe bis 18. angulis rectis, ut dictum est. Ita quoque omnes 10. anguli figuræ 10. lateribus contentæ, æquiualebunt 16. angulis rectis, cum talis figura sit octaua inter rectilineas figuras. Hoc autem hac ratione demonstrabitur. Omnis figura rectilinea in tot triangula diuiditur, quota ipsa est inter figuras, seu quot ipsa habet angulos lateraue, binario dempto. Nam a quouis angulo ipsius ad omnes angulos oppositos duci possunt lineæ rectæ, solum ad duos propinques angulos non possunt duci. Quare in tot triangula distribuetur, quot ipsa habet angulos, demptis duobus illis angulis. Sic vides, triangulum non posse diuidi in alia triangula; quadrangulum vero in duo secari; quinquangulum in tria; sexangulum in quatuor, &c. Cum igitur anguli horum triangulorum constituant omnes angulos rectilineæ figuræ oppositæ, & omnes anguli cuiuslibet trianguli æquales sint duobus rectis; perspicuum est, omnes angulos figuræ cuiusuis rectilineæ æquales esse bis tot rectis, in quot triangula diuiditur, hoc est, quota ipsa est inter rectilineas figuras. Quod quidem manifeste perspicitur in propositis figuris.
Secundus modus, quo scitur valor angulorum cuiuslibet figuræ rectilineæ, hic est.

OMNES anguli figuræ rectilineæ cuiusuis, æquales sunt bis tot rectis angulis, demptis quatuor, quotipsa continet latera, seu angulos.

HOC est, anguli cuiuslibet trianguli æquales sunt bis tribus rectis, demptis quatuor, nempe duobus rectis. Ita etiam anguli figuræ continentis 20. latera æquiualebunt bis 20. angulis rectis, minus quatuor, nimirum 36. rectis angulis, &c. Demonstratio autem huius rei talis est. Si a quouis puncto intra figuram assumpto ad omnes angulos rectæ lineæ ducantur, efficientur tot triangula, quot latera, angulosue figura ipsa continet. Cum igitur anguli cuiuscunque trianguli æquales sint duobus rectis, erunt omnes anguli illorum triangulorum æquales bis tot rectis, quot latera figuram ambiunt. At anguli eorundem triangulorum circa punctum intra figuram assumptum consistentes non pertinent ad angulos figuræ rectilineæ propositæ, ut constat. Quare si hi auferantur, erunt reliqui triangulorum anguli constituentes angulos figuræ propositæ, bis quoque tot rectis æquales, demptis illis circa punctum assumptum constitutis, quot latera, vel angulos continet figura. Sunt autem illi anguli, quotquot sint, circa dictum punctum existentes æquales 4. rectis, ut collegimus ex propos. 15. Quamobrem anguli cuiusque figuræ bis tot rectis sunt æquales, ablatis quatuor, quot ipsa figura continet angulos, seu latera, quod est propositum.
EX hoc porro secundo modo liquet, si singula latera figuræ cuiusuis rectilineæ producantur ordinatim versus eandem partem, omnes angulos externos æquales esse quatuor rectis. Nam quilibet externus, & illi deinceps internus, æquantur duobus rectis; atque adeo omnes externi una cum omnibus internis æquales erunt bis tot rectis, quot latera, angulosue figura continet. Sunt autem & soli interni bis tot rectis æquales, minus quatuor, ut demonstrauimus. Si igitur interni auferantur, remanebunt externi quatuor tantum rectis æquales, qui nimirum desunt internis angulis, ut interni & externi simul bis tot rectos conficiant, quot latera figuram propositam ambiunt. Exemplum. In triangulo quouis, anguli interni & externi simul æquales sunt sex rectis. Cum igitur interni duobus sint rectis æquales, erunt soli externi æquales quatuor duntaxat rectis. In quadrilatero, anguli externi & interni simul æquales sunt octo rectis. Cum igitur interni soli æquales sint quatuor rectis, ut ostendimus, erunt & soli externi quatuor etiam rectis æquales. In pentagono, seu quinquangulo, anguli interni & externi sunt æquales 10. rectis. Quoniam vero interni adæquantur sex rectis, ut demonstrauimus, remanebunt externi æquales quatuor tantum rectis. Quæ omnia in appositis figuris conspiciuntur. Eademque est ratio in aliis omnibus figuris.

EX CAMPANO

SI pentagoni singula latera producantur in partem utramque, ita ut quælibet duo extra pentagonum coeant, efficientur quinque anguli ex lateribus coeuntibus æquales duobus solum rectis.

IN pentagono A B C D E, latera in utramque partem producta coeant in punctis F, G, H, I, K. Dico quinque angulos F, G, H, I, K, æquales tantum esse duobus rectis. In triangulo enim B H K, cum latus H B, sit protractum ad F, erit externus angulus F B K, duobus internis, & oppositis H, K, æqualis. Eadem ratione in triangulo A I G, erit externus angulus F A G, æqualis duobus internis, & oppositis I, G. Quare duo anguli F B A, F A B, æquales sunt quatuor angulis G, H, I, K. Addito igitur communi angulo F, erunt tres anguli A, B, F, trianguli A B F, æquales quinque angulis F, G H I K. Sed anguli A, B, F, trianguli A B F, æquales sunt duobus rectis. Igitur & quinque anguli F, G, H, I, K, duobus sunt rectis æquales. Quod est propositum.

COROLLARIVM I

EX hac propos. 32. colligitur, tres angulos cuiuslibet trianguli simul sumptos æquales esse tribus angulis cuiusque alterius trianguli simul sumptis: Quoniam tam illi tres, quam hi, æquales sunt duobus angulis rectis. Unde si duo anguli unius trianguli fuerint æquales duobus angulis alterius trianguli, erit & reliquus illius reliquo huius æqualis, æquiangulaque erunt ipsa triangula.

COROLLARIVM II

CONSTAT etiam, in omni triangulo Isoscele, cuius angulus lateribus æqualibus comprehensus rectus fuerit, quemlibet reliquorum esse semirectum; Nam reliqui duo simul conficiunt unum rectum, cum omnes tres sint æquales duobus rectis: & tertius ille ponatur rectus. Quare cum duo reliqui inter se sint æquales, erit quilibet eorum semirectus. At vero si angulus æqualibus lateribus contentus fuerit obtusus, quemlibet aliorum esse semirecto minorem. Reliqui enim duo simul minores erunt uno recto, &c. Si denique dictus angulus extiterit acutus, utrumque reliquorum maiorem esse semirecto. Quoniam reliqui duo simul maiores erunt uno recto.

COROLLARIVM III

PERSPICVVM quoque est, quemuis angulum trianguli æquilateri esse duas tertias partes unius recti, vel tertiam partem duorum rectorum. Duo enim anguli recti, quibus æquales sunt tres anguli trianguli æquilateri, diuisi in tres angulos, faciunt duas tertias partes unius recti.

COROLLARIVM IIII

LIQVET etiam, si ab uno angulo trianguli æquilateri perpendicularis ad latus oppositum ducatur, constitui duo triangula scalena, quorum unumquodque habet unum angulum rectum prope perpendicularem; alium duas tertias partes unius recti, illum scilicet, qui est, & angulus trianguli æquilateri; reliquum denique tertiam partem unius recti.

SCHOLION

PORRO ex tertio corollario depromi potest methodus, qua angulus rectus in tres angulos æquales diuidatur. Sit enim angulus rectus A B C. Super rectam A B, constituatur triangulum æquilaterum A B D. Et quia per corollarium 3. angulus A B D, facit duas tertias partes anguli recti A B C; erit angulus C B D, pars tertia eiusdem recti. Diuiso igitur angulo A B D, bifariam, per rectam B E, erit uterque angulus A B E, E B D, tertia quoque pars recti. Quare rectus angulus A B C, diuisus est in tres angulos æquales. Quod est propositum.
 
Proposition 33. 
THEOR. 23. PROPOS. 33. 
第三十三題 
The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel. 
RECTÆ lineæ, quæ æquales, & parallelas lineas ad partes easdem coniungunt; Et ipsæ æquales, & parallelæ sunt. 
兩平行相等線之界。有兩線聯之。其兩線亦平行。亦相等。 
Let AB, CD be equal and parallel, and let the straight lines AC, BD join them (at the extremities which are) in the same directions (respectively);  I say that AC, BD are also equal and parallel. 
SINT rectæ lineæ A B, C D, æquales, & parallelæ; Ipsas autem coniungant ad easdem partes rectæ A C, B D.  Dico A C, B D, æquales quoque esse, & parallelas. 
Let BC be joined.  Then, since AB is parallel to CD, and BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]  And, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB;  and the angle ABC is equal to the angle BCD;  therefore the base AC is equal to the base BD  and the triangle ABC is equal to the triangle DCB,  and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]  therefore the angle ACB is equal to the angle CBD.  And, since the straight line BC falling on the two straight lines AC, BD has made the alternate angles equal to one another, AC is parallel to BD. [I. 27]  And it was also proved equal to it. 
Ducatur enim recta A D.  Quoniam igitur A D, incidit in parallelas A B, C D, erunt anguli alterni B A D, C D A, æquales.  Quare cum duo latera B A, A D, trianguli B A D, æqualia sint duobus lateribus C D, D A, trianguli C D A, utrumque utrique,  & anguli quoque dictis lateribus inclusi æquales;  erunt bases B D, A C, æquales,      & angulus A D B, angulo D A C, æqualis.  Cum igitur hi anguli sint alterni inter rectas A C, B D, erunt A C, B D, parallelæ:  Probatum autem iam fuit, eadem esse æquales. 
Therefore etc.  Q. E. D. 
Rectæ ergo lineæ, quæ æquales, & parallelas lineas, &c.  Quod erat demonstrandum.

SCHOLION
DIXIT Euclides, lineas æquales, & parallelas ad easdem partes debere coniungi, ut ipsæ coniungentes sint & æquales & parallelæ. Nam si ad partes diuersas coniungerentur, ut ad A, & D. Item ad B, & C, neque coniungentes lineæ essent parallelæ unquam, sed perpetuo se mutuo secarent, neque essent æquales, nisi raro admodum, ut ex sequenti propositione constabit.
 
Proposition 34. 
THEOR. 24. PROPOS. 34. 
第三十四題 
In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas. 
PARALLELOGRAMMORVM spatiorum æqualia sunt inter se, quæ ex aduerso & latera, & anguli; atque illa bifariam secat diameter. 
凡平行線方形。每相對兩邊線、各等。每相對兩角、各等。對角線、分本形、兩平分。 
Let ACDB be a parallelogrammic area, and BC its diameter;  I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it. 
SIT parallelogrammum A B C D, quale definiuimus definitione 35.  Dico latera opposita A B, D C, inter se esse æqualia, nec non latera opposita A D, B C. Item angulos oppositos B, & D, æquales inter se esse, nec non & angulos oppositos D A B, & D C B: Denique ducta diametro A C, parallelogrammum ipsum bifariam secari. 
For, since AB is parallel to CD, and the straight line BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]  Again, since AC is parallel to BD, and BChas fallen upon them, the alternate angles ACB, CBD are equal to one another. [I. 29]  Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively,  and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC;  therefore they will also have the remaining sides equal to the remaining sides respectively,  and the remaining angle to the remaining angle; [I. 26]  therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle CDB.  And, since the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. [C.N. 2]  And the angle BAC was also proved equal to the angle CDB. 
Cum enim A B, D C, sint parallelæ erunt anguli alterni B A C, D C A, æquales.  Rursus quia A D, B C, sunt parallelæ, erunt & anguli alterni B C A, D A C, æquales.  Itaque cum anguli B A C, B C A, trianguli A B C, æquales sint duobus angulis D C A, D A C, trianguli A D C, uterque utrique,  & latus A C, dictis angulis adiacens, commune utrique triangulo;  erit recta A B, æqualis oppositæ rectæ D C, & recta B C, oppositæ rectæ A D, quod est primum.  Erit rursus eadem de causa angulus B, angulo D, æqualis.  See the record before the previous.  Et quia si æqualibus angulis B A C, D C A, addantur æquales anguli D A C, B C A, toti quoque anguli B A D, B C D, fiunt æquales;  constat secundum, angulos nimirum oppositos esse æquales. 
Therefore in parallelogrammic areas the opposite sides and angles are equal to one another. 
 
I say, next, that the diameter also bisects the areas.  For, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB respectively;  and the angle ABC is equal to the angle BCD;  therefore the base AC is also equal to DB,  and the triangle ABC is equal to the triangle DCB. [I. 4] 
  Quoniam vero duo latera A B, B C, trianguli A B C, æqualia sunt duobus lateribus C D, D A, trianguli C D A, utrumque utrique,  & angulus B, angulo D, æqualis, ut iam ostendimus;    erunt triangula A B C, C D A, æqualia, 
Therefore the diameter BC bisects the parallelogram ACDB.  Q. E. D. 
ideoque parallelogrammum A B C D, diuisum erit bifariam a diametro A C, quod tertio loco proponebatur. Parallelogrammorum igitur spatiorum æqualia sunt inter se, quæ ex aduerso, &c.  Quod ostendendum erat.

SCHOLION
APPOSITE dixit Euclides, solummodo parallelogramma a diametro diuidi bifariam, non autem & angulos. In Quadrato enim, & Rhombo duntaxat anguli etiam bifariam diuiduntur a diametro; At in figura Altera parte longiori, & in Rhomboide in partes inæquales. Quæ omnia perspicua erunt, si prius ostenderimus, quatuor hasce figuras, Quadratum, Altera parte longius, Rhombum, & Rhomboidem, esse parallelogramma. Hoc autem demonstrabimus tribus sequentibus theorematibus, quorum primum est.

OMNE quadrilaterum habens latera opposita æqualia, est parallelogrammum.
SINT in quadrilatero A B C D, latera opposita A B, D C, æqualia; Item opposita latera A D, B C. Dico A B C D, esse parallelogrammum; hoc est, lineas A B, D C, esse parallelas; Itemque lineas A D, B C. Ducta enim diametro A C, erunt duo latera A B, B C, trianguli A B C, æqualia duobus lateribus C D, D A, trianguli C D A, utrumque utrique, & basis A C, communis. Igitur erit angulus B, angulo D, æqualis. Rursus quia latera A B, B C, æqualia sunt lateribus C D, D A, utrumque utrique, & anguli, B, D, ostensi æquales; erit angulus B A C, angulo D C A, alterno æqualis, & angulus B C A, alterno angulo D A C. Quare erunt A B, & D C, parallelæ; Item A D, & B C, quod est propositum.
HINC constat, Rhombum, & Rhomboidem esse parallelogramma; quoniam opposita eorum latera sunt inter se æqualia, ut manifestum est ex eorum definitionibus. Pariratione quadratum, parallelogrammum erit, quod latera opposita habeat æqualia. Sunt enim omnia quatuor eius latera inter se æqualia, per eius definitionem. Conuertit autem hoc theorema primam partem propositionis 34. ut patet.
Secundum tbeorema tale est.
OMNE quadrilaterum habens angulos oppositos æquales, est parallelogrammum.
SINT in quadrilatero A B C D, anguli oppositi A, & C, æquales: item oppositi anguli B, & D. Dico A B C D, esse parallelogrammum: hoc est, lineas A B, D C, esse parallelas: Itemque lineas A D, B C. Nam si æqualibus angulis A, & C, addantur æquales anguli B, & D: erunt duo anguli A, & B, duobus angulis D, & C, æquales & idcirco anguli A, & B, dimidium facient quatuor angulorum A, B, C, & D. Cum igitur hi quatuor æquales sint quatuor rectis, ut ad propositionem 32. demonstrauimus, erunt duo A, & B, duobus rectis æquales. Quare A D, B C, parallelæ sunt. Eadem ratione erunt A B, D C, parallelæ. Erunt enim duo quoque anguli A, & D, duobus angulis B, & C, æquales, & c. Quod est propositum. Ex hoc etiam manifestum est, Rhomboidem esse parallelogrammum, cum eius anguli oppositi æquales sint, per definitionem. Similiter quadratum, & altera parte longius. Sunt enim & eorum anguli oppositi æquales, cum sint recti, ex eorum definitionibus.
HOC theorema conuertit secundam partem propositionis 34. ut constat. Tertia autem pars non potest conuerti. Nam & trapezium aliquod bifariam secari potest a diametro, & tamen non est parallelogrammum. Sit enim altera parte longius, vel Rhomboides A B C D, quod parallelogrammum esse ostensum est: in quo, ducta diametro A C, constituatur super A C, triangulo A B C, æquale triangulum A E C, inuerso ordine, ita ut latus C E, sit æquale lateri A B, & A E, ipsi C B, fiatque trapezium A E C D. Quoniam vero triangulum A B C, triangulo A D C, æquale est, quod diameter A C, bifariam secet parallelogrammum D B: Erit & triangulum A E C, triangulo A D C, æquale: Ac proinde trapezium A E C D, bifariam diuidetur a diametro A C.
QVOD si quadrilaterum aliquod diuidatur bifariam ab utraque diametro, illud parallelogrammum erit, ut ostendemus ad propositionem 39. Quod quidem in nullo trapezio fieri potest.
Tertium Tbeorema huiusmodi est.
OMNE quadrilaterum habens omnes angulos rectos, est parallelogrammum.
SINT in quadrilatero A B C D, omnes quatuor anguli recti. Dico ipsum esse parallelogrammum: hoc est, lineas A B, D C, esse parallelas: itemque A D, B C. Quoniam enim due anguli A, & B, æquales sunt duobus rectis, cum sint duo recti: erunt A D, B C, parallelæ. Eodem modo erunt A B, D C, parallelæ: atque adeo A B C D, parallelogrammum. quod est propositum.
HINC rursum constat, Quadratum, & alter aparte longius, esse parallelogramma, cum eorum anguli omnes existant recti, ut liquet ex eorum definitionibus.
HIS in hunc medum demonstratis, Quadratum scilicet, Altera parte longius, Rhombum, & Rhomboidem, esse parallelogramma, facile ostendemus, angulos Quadrati, & Rhombi, bifariam secari a diametro: Angulos vero figuræ, Altera parte longioris, & Rhomboidis, non bifariam, ut paulo ante monuimus. Sit enim Quadratum, vel Rhombus A B C D, in quo diameter A C. Quoniam igitur duo latera B A, A C, trianguli B A C, æqualia sunt duobus lateribus D A, A C, trianguli D A C, utrumque utrique, & basis B C, basi D C, (sunt enim hæ figuræ æquilateræ) erunt anguli B A C, D A C, æquales. Quare angulus B A D, diuiditur bifariam. Eodem modo demonstrabimus, reliquos angulos bifariam secari a diametro.
SIT rursus Altera parte longius, vel Rhomboides A B C D, in quo diameter A C, sitque maius latus A B. Quoniam igitur in triangulo A B C, latus A B, maius est latere B C, erit angulus B C A, maior angulo B A C. Est autem angulus B C A, æqualis angulo D A C, alterno: quod B C, A D, parallelæ sint. (Est enim A B C D, ostensum esse parallelogrammum.) Igitur & angulus D A C, maior erit angulo B A C. Atque propterea angulus B A D, inæqualiter diuiditur a diametro A C. Eadem est ratio aliorum angulorum. Quamobrem apposite Euclides in tertia parte buius propositionis dixit, solum parallelogramma bifariam a diametro secari, non autem & angulos.
EODEM fere pacto ostendemus, duas diametros in Quadrato, & Altera parte longiore æquales esse; At vero in Rhombo, & Rhomboide inæquales, maiorem quidem eam, quæ angulos acutos, minorem vero eam, quæ obtusos angulos dispertit. Sit enim quadratum, vel altera parte longius A B C D, in quo diametri A C, B D, quas dico esse æquales. Cum enim duo latera A B, B C, trianguli A B C, æqualia sint duobus lateribus A B, A D, trianguli B A D: utrumque utrique, & angulus A B C, angulo B A D, quia uterque rectus: erit basis A C, basi B D, æqualis: Ac proinde diametri in quadrate, & figura altera parte longtore æquales sunt.
SIT rursus Rhombus, vel Rhomboides, A B C D, in quo diametri A C, B D; sitque angulus B A D, maior; A B C, minor. Non enim æquales sunt, quia alias uterque esset rectus, cum ambo æquales sint duobus rectis; quod est absurdum, & contra definitiones Rhombi, & Rhomboidis. Dico diametrum B D, maiorem esse diametro A C. Quoniam enim duo latera A B, A D, trianguli B A D, æqualia sunt duobus lateribus A B, B C, trianguli A B C, utrumque utrique, & angulus B A D, angulo A B C, maior existit; erit basis B D, maior base A C. quod est propositum. Ex quo manifestum est, cur in propositione 33. Euclides asseruerit, eas tantum lineas, quæ coniungunt parallelas æquales ad easdem partes, æquales esse, ut ibidem annotuimus. Nam in Rhombo, & Rhomboide rectæ A C, B D, inæquales sunt, licet coniungant parallelas æquales A B, D C: quia non ad easdem partes ipsas coniungunt, ut perspicuum est.
IN omni tamen parallelogrammo diametri se mutuo bifariam diuidunt. Cum enim duo anguli E A D, E D A, trianguli A E D, æquales sint alternis angulis E C B, E B C, trianguli B E C, uterque utrique; & latus A D, æquale lateri B C, opposito in parallelogrammo A B C D, quorum utrumque æqualibus adiacet angulis; Erit & A E, recta rectæ C E, & recta D E, rectæ B E, æqualis. Quare utraque diameter bifariam diuiditur in puncto E.
HVIVS autem, quod modo diximus, conuersum etiam demonstrabimus, nimirum.

OMNE quadrilaterum, in quo diametri se mutuo bifariam diuidunt, parallelogrammum est.
IN quadrilatero enim A B C D, diametri A C, B D, se mutuo bifariam diuidant in E. Dico A B C D, parallelogrammum esse. Cum enim latera A E, E B, trianguli A E B, æqualia sint lateribus C E, E D, trianguli C E D, & anguli contenti ad verticem E, æquales quoque; erunt & bases A B, C D, æquales & angulus A B E, angulo alterno C D E, æqualis. Quare rectæ A B, C D, parallelæ sunt. Eadem ratione parallelæ ostendentur A D, C B. Parallelogrammum ergo est A B C D.
HVC quoque referri potest hoc theorema.

RECTA linea secans diametrum parallelogrammi bifariam quomodocunque, diuidit parallelogrammum bifariam quoque: & recta linea diuidens parallelogrammum bifariam quouis modo, secat quoque diametrum bifariam.
IN parallelogrammo A B C D, diametrum A C, bifariam secet recta E F, in puncto G. Dico parallelogrammum diuidi bifariam. Quoniam enim angulus E A G, æqualis est angulo alterno B C G, & angulus E G A, angulo F G C, Est autem & latus A G, lateri C G, æqualœ, per hypothesin, quæ ambo æqualibus angulis adiacent; erunt & latera E G, F G, æqualia. Quare cum latera A G, G E, æqualia sint lateribus C G, G F, & anguli quoque contenti æquales; erunt triangula A G E, C G F, æqualia. Addita igitur communi quantitate B C G E, erit triangulum A B C, trapezio B C F E, æquale: Sed triangulum A B C, dimidium est parallelogrammi A B C D. Igitur & trapezium dimidium erit eiusdem parallelogrammi, ideoque recta E F, parallelogrammum bifariam secat.
SECET iam E F, parallelogrammum bifariam; Dico & diametrum A C, bifariam secari in G. Si enim diameter A C, non bifariam diuiditur in G, diuidatur bifariam in alio puncto, ut in H, per good ducatur recta E H I. Erit ergo, ut iam demonstrauimus, E I C B, trapezium dimidium parallelogrammi A B C D, atque adeo æquale trapezio E F C B, quod etiam dimidium ponitur eiusdem parallelogrammi, pars toti, quod est absurdum. Diuiditur igitur A C, bifariam in G, & non in alio puncto, quod erat propositum.
HINC facile colligitur, si in latere aliquo parallelogrammi cuiusque punctum signetur, vel etiam intra parallelogrammum, vel extra, quod tamen non sit in diametro, nisi ipsum secet diametrum bifariam; qua ratione ab illo puncto linea duci debeat, quæ parallelogrammum bifariam secet. Si enim diameter ducatur, & a puncto dato per medium punctum diametri recta ducatur, factum erit, quod proponitur. Ut si punctum sit E, in latere A B, ducenda est recta E F, per G, punctum, in quo diameter A C, bifariam diuiditur; & sic de aliis punctis.
DEMONSTRAT quoque bic Peletarius problema non iniucundum. videlicet.

INTER duas lineas rectas infinitas angulum facientes, lineam rectam datæ lineæ æqualem collocare, quæ cum altera illarum faciat angulum cuiuis angulo dato æqualem. Oportet autem hunc angulum datum, & eum, qui lineis datis continetur, minores esse duobus rectis.
D rectæ infinitæ A B, A C, contineant angulum B A C, sitque data recta finita quæcunque D, & angulus datus E, hac lege, ut duo anguli E, & B A C, minores sint duobus rectis. Oportet igitur inter rectas A B, A C, collocare rectam æqualem quidem rectæ D, cum alterutra vero illarum, nimirum cum A C, facientem angulum æqualem angulo dato E. Fiat angulus C A F, æqualis angulo E, & producta F A, ad G, sit A G, æqualis rectæ D; & per G, ducatur G B, parallela ipsi A C, secans A B, in B: Deinde per B, ducatur B C, parallela ipsi A G, secans A C, in C. Dico rectam B C, collocatam inter rectas A B, A C, æqualem esse rectæ D, angulumque B C A, angulo E. Cum enim parallelogrammum sit per constructionem, A C B G, erit recta B C, rectæ G A, æqualis: At G A, æqualis est, per constructionem, rectæ D. Igitur & B C, rectæ D, æqualis erit. Rursus quia angulus B C A, angulo alterno C A F, æqualis est; & eidem angulo C A F, æqualis est, per constructionem, angulus E; erunt anguli E, & B C A, æquales, Quod est propositum. Caterum ex contructione manifestum esse cuilibet potest, cur duo anguli dati minores esse debeant duobus rectis. Nam alias non fieret triangulum A B C, si anguli B A C, & B C A, æquales essent duobus rectis, vel maiores, ut constat ex propositio 17. vel 32.

EX PROCLO
IN omni figura rectilinea latera habente numero paria, si quidem fuerit æquilatera, & æquiangula: erunt duo quælibet latera opposita, parallela inter sese.
LATERA opposita dicuntur illa duo, quæ ex utraque parte latera habent æqualia numero: ut in hexagono A B C D E F, latera opposita erunt A B, E D: quoniam tam ad partes A, & E, duo sunt latera, quam ad partes B, & D. In octogono vero A B C D E F G H, latera opposita erunt A B, F E, quia tam ad partes A, & F, tria sunt latera, quam ad partes B, & E. Et sic in aliis figuris æquilateris parium laterum, ex utraque parte oppositorum laterum erunt tot latera, quot sunt in dimidio numero laterum, minus uno. Ut in quadrangulo erit unum, in hexagono erunt duo, in octogono tria, in decagono quatuor, in figura 12. laterum quinque, & c. Dico igitur qualibet latera opposita esse parallela; A B, nimirum ipsi E D, in hexagono; & A B, ipsi F E, in octogono, & sic de cæteris. Connectantur enim duo extrema oppositorum laterum ad easdem partes linea recta, qualis est in hexagono B D, & in octogono B E. Et quoniam, ut ad 32 propos. demonstrauimus, sex anguli hexagoni æquales sunt octo rectis, erunt tres anguli B, C, D, eiusdem hexagoni æquales quatuor rectis, proptereaquod omnes anguli ponuntur æquales; Sunt autem anguli B C D, C B D, C D B, trianguli B C D, duobus rectis æquales. Reliqui igitur anguli A B D, E D B, duobus rectis æquales erunt; Quare parallela erunt A B, & E D. Rursus quia octo anguli octogoni æquales sunt duodecim rectis, erunt quatuor eius anguli B, C, D, E, sex rectis æquales: Sunt autem quatuor anguli quadrilateri B C D E, æquales quatuor rectis. lgitur duo reliqui anguli A B E, F E B, duobus erunt rectis æquales, atque adeo A B, F E, parallela erunt. Eodem modo demonstrabitur, in omnibus aliis figuris huiusmodi, angulos duos ad lineam rectam extrema oppositorum laterum coniungentem existentes, duobus esse rectis æquales. Nam in decagono aufert ea linea pentagonum, cuius anguli æquales sunt sex rectis: At quinque anguli decagoni æquales sunt octo rectis. Ablatis igitur sex, relinquuntur duo recti. In figura æquilatera, & æquiangula duodecim laterum eadem linea abscindet hexagonum, cuius anguli sunt octo rectis æquales: At sex anguli totius figuræ æquales sunt decem rectis. Demptis igitur octo, remanent duo recti, &c.
QVAMVIS autem omnis figura æquiangula parium laterum habeat latera opposita parallela, ut ostendimus; tamen sola quadrilatera figura latera opposita habens parallela, ab Euclide, & aliis Geometris parallelogrammum dici consueuit, proptereaque in definitionibus, Parallelegrammum diximus esse figuram quadrilateram, &c.
 
Proposition 35. 
THEOR. 25. PROPOS. 35. 
第三十五題 
Parallelograms which are on the same base and in the same parallels are equal to one another. 
PARALLELOGRAMMA super eadem basi, & in eisdem parallelis constituta, inter se sunt æqualia. 
兩平行方形。若同在平行線內。又同底。則兩形必等。 
Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;  I say that ABCD is equal to the parallelogram EBCF. 
INTER duas parallelas A B, C D, super basi C D, existant duo parallelogramma C D E A, C D B F. (Dicuntur autem parallelogramma in eisdem esse parallelis, quando duo latera opposita partes sunt parallelarum, ut in exemplo proposito cernitur)  Dico ipsa parallelogramma inter se esse æqualia, non quoad angulos & latera, sed quoad aream, seu capacitatem. 
解曰。甲乙、丙丁、兩平行線內。有丙丁戊甲、與丙丁乙己、兩平行方形。同丙丁底。題言此兩形等。等者。不謂腰等、角等。謂所函之地等。  後言形等者、多倣此。 
For, since ABCD is a parallelogram, AD is equal to BC. [I. 34]  For the same reason also EF is equal to BC,  so that AD is also equal to EF; [C.N. 1]  and DE is common; therefore the whole AE is equal to the whole DF. [C.N. 2]  But AB is also equal to DC; [I. 34]  therefore the two sides EA, AB are equal to the two sides FD, DC respectively,  and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29]  therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4]  Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]  Let the triangle GBC be added to each;  therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2] 
Cadat enim primo punctum F, inter A, & E. Quoniam igitur in parallelogrammo C D E A, recta A E, æqualis est rectæ C D,  oppositæ & eidem C D, æqualis est F B, in parallelogrammo C D B F, opposita;  Erunt A E, F B, inter se æquales.  Dempta igitur communi F E, remanebit A F, ipsi E B, æqualis:  Est autem & A C, ipsi E D, oppositæ æqualis in parallelogrammo C D E A;    & angulus B E D, angulo F A C, externus interno.  Quare triangulum F A C, triangulo B E D, æquale erit.    Addito igitur communi trapezio C D E F,  fiet totum parallelogrammum C D E A, toti parallelogrammo C D B F, æquale. 
先論曰。設己在甲戊之內。其丙丁戊甲、與丙丁乙己。皆平行方形。丙丁同底。  則甲戊、與丙丁。己乙、與丙丁。各相對之兩邊各等。本篇三四。  而甲戊、與己乙、亦等。公論一。  試於甲戊、己乙、兩線。各減己戊。卽甲己、與戊乙、亦等。公論三。  而甲丙、與戊丁、元等。本篇三四。    乙戊丁外角。與己甲丙內角乂等。本篇廿九。  則乙戊丁、與己甲丙、兩角形必等矣。本篇四。    次於兩角形。每加一丙丁戊己無法四邊形。  則丙丁戊甲、與丙丁乙己、兩平行方形等也。公論二。 
Therefore etc.  Q. E. D. 
  Quod est propsitum.
CADAT secundo punctum F, in punctum E. Dico rursus, parallelogramma C D E A, C D B E, æquala esse. Erunt enim, ut prius, rectæ AE, EB, æquales, nec non & anguli B E D, E A C; atque adeo triangula, E A C, B E D, æqualia. Addito igitur communi triangulo C D E, fient parallelogramma C D E A, C D B E, æqualia.
CADAT tertio punctum F, ultra E, ita ut recta CF, secet rectam D E, in G. Quoniam igitur, ut prius, rectæ A E, F B, sunt æquales; si communis addatur E F; erit tota A F, toti E B, æqualis, nec non & anguli B E D, F A C, æquales erunt; atque adeo triangulum F A C, triangulo B E D, æquale. Ablato ergo communi triangulo E G F, remanebit trapezium A E G C, trapezio FGDB, æquale. Quocirca addito communi triangulo C D G, fiet totum parallelogrammum C D E A, toti parallelogrammo C D B F, æquale. Parallelogramma igitur super eadem basi, & in eisdem parallelis constituta, inter se sunt æqualia. Quod erat demonstrandum.

SCHOLION
CONVERTEMVS facile hanc propositionem, hoc modo.
PARALLELOGRAMMA æqualia super eandem basin, ad easdemque partes censtituta, erunt inter easdem parallelas.
SINT duo parallelogramma æqualia A B C D, C D E F, super eandem basin C D, & ad easdem partes. Dico rectam A B, productam in directum iacere ipsi E F, & propterea ipsa parallelogramma inter easdem esse parallelas. Alias enim A B, producta vel cadet infra E F, vel supra. Cadat primo infra, qualis est A H. Erit igitur parallelogrammum C D G H, æquale parallelogrammo A B C D; Ponitur autem eidem parallelogrammo A B C D, æquale parallelogrammum C D E F. Quare parallelogramma C D E F, C D G H, æqualia erunt, totum & pars, quod est absurdum. Non ergo cadet A B infra E F.
CADAT secundo A B, producta supra E F. Cadet igitur F E, protracta infra AB. Quare, ut prius, erunt parallelogramma A B C D, C D H G, æqualia, totum & pars, quod est absurdum. Idem absurdum consequeretur, si C F, D E, producerentur usque ad A B, protractam. Eademque demonstratio conueniet omnibus casibus, qui occurrere possunt, hoc est, siue punctum E, sit ultra B, siue non, ut perspicuum est. Non ergo cadet A B, supra E F; sed nec infra, ut demonstratum est; ergo producta in directum iacet ipsi E F; ac proinde parallelogramma A B C D, C D E F, in eisdem sunt parallelis.
 
  次論曰。設己、戊、同點。依前甲戊、與戊乙等。乙戊丁、與戊甲丙、兩角形等。本論四。而每加一戊丁丙角形。則丙丁戊甲、與丙丁乙戊、兩平行方形必等。公論二。

後論曰。設己點在戊之外。而丙己、與戊丁、兩線交於庚。依前甲戊、與己乙、兩線等。而每加一戊己線。卽戊乙、與甲己、兩線亦等。公論二。因顯己甲丙、與乙戊丁、兩角形亦等。本篇四。次每減一己戊庚角形。則所存戊庚丙甲、與乙己庚丁、兩無法四邊形、亦等。公論三。次於兩無法形每加一庚丁丙角形則丙丁戊甲、與丙丁乙己、兩平行方形必等。公論二。 

Proposition 36. 
THEOR. 26. PROPOS. 36. 
第三十六題 
Parallelograms which are on equal bases and in the same parallels are equal to one another. 
PARALLELOGRAMMA super æqualibus basibus, & in eisdem parallelis constituta, inter se sunt æqualia. 
兩平行線內。有兩平行方形。若底等。則形亦等。 
Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG;  I say that the parallelogram ABCD is equal to EFGH. 
SINT duo parallelogramma A C E F, G H D B, super æquales bases C E, H D, inter easdem parallelas A B, C D.  Dico ea esse æqualia. 
For let BE, CH be joined.  Then, since BC is equal to FG while FG is equal to EH, BC is also equal to EH. [C.N. 1]  But they are also parallel.  And EB, HC join them;  but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I. 33]  ()  Therefore EBCH is a parallelogram. [I. 34]  And it is equal to ABCD;  for it has the same base BC with it, and is in the same parallels BC, AH with it. [I. 35]  For the same reason also EFGH is equal to the same EBCH; [I. 35]  so that the parallelogram ABCD is also equal to EFGH. [C.N. 1] 
Connectantur enim extrema rectarum C E, G B, ad easdem partes lineis rectis C G, E B.  Quoniam igitur recta C E, æqualis ponitur rectæ H D, & eidem H D, æqualis est G B, in parallelogrammo G H D B, opposita; erunt C E, G B, æquales inter se:  Sunt autem & parallelæ, per hypothesin.  Quare & C G, E B, ipsas coniungentes,    parallelæ erunt, & æquales,  ideoque C E B G, parallelogrammum erit.    Itaque cum parallelogramma A C E F, G C E B, sint inter easdem parallelas, & super eandem basin C E, erit parallelogrammum A C E F, parallelogrammo G C E B, æquale.  Rursus quia parallelogramma G C E B, G H D B, sunt inter easdem parallelas, & super eandem basin G B, erit quoque parallelogrammum G H D B, eidem parallelogrammo G C E B, æquale.  Quare & parallelogramma A C E F, G H D B, inter se æqualia erunt. 
Therefore etc.  Q. E. D. 
Parallelogramma igitur super æqualibus basibus, & in eisdem parallelis constituta, & c.  Quod ostendendum erat.

SCHOLION
CONVERSVM huius theorematis duplex est, ad hunc modum.
PARALLELOGRAMMA æqualia super bases æquales, & ad easdem partes constituta, inter easdem sunt parallelas: Et parallelogramma æqualia inter easdem parallelas, si non eandem habuerint basin, super æquales bases sunt constitura.
SINT primum duo parallelogramma æqualia A B C D, E F G H, super bases æquales B C, F G, & ad easdem partes constituta. Dico ea esse inter easdem parallelas, hoc est, A D, protractam coire in directum cum E H. Nam alias cadet aut infra E H, aut supra. Quo posito sequitur, totum & partem esse æqualia, quemadmodum in conuersa præcedentis propositionis est dictum, & figura facile commonstrat. Intelligendæ sunt autem bases æquales datæ in eadem linea recta B G.
SINT secundo eadem parallelogramma æqualia inter easdem parallelas A H, B G. Dico bases B C, F G, esse æquales. Si enim altera, nempe B C, dicatur maior, abscindatur B I, æqualis rectæ F G, & ducatur I K, parallela ipsi C D. Erit ergo parallelogrammum A B I K, æquale parallelogrammo E F G H; & ideo parallelogrammo A B C D, pars toti, quod est absurdum. Non ergo B C, maior est, quam F G. Eadem ratione neque minor erit. Quare bases B C, F G, æquales sunt.
 
Proposition 37. 
THEOR. 27. PROPOS. 37. 
第三十七題 
Triangles which are on the same base and in the same parallels are equal to one another. 
TRIANGVLA super eadem basi constituta, & in eisdem parallelis, inter se sunt æqualia. 
兩平行線內。有兩三角形。若同底。則兩形必等。 
Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC;  I say that the triangle ABC is equal to the triangle DBC. 
INTER parallelas A B, C D, & super basin C D, sint constituta duo triangula A C D, B C D. (Dicitur autem triangulum inter duas esse parallelas constitutum, quando basis est pars unius, & angulus oppositus alteram attingit.)  Dico ea triangula esse æqualia. 
Let AD be produced in both directions to E, F;  through B let BE be drawn parallel to CA, [I. 31]  and through C let CF be drawn parallel to BD. [I. 31]  Then each of the figures EBCA, DBCF is a parallelogram; and they are equal,  for they are on the same base BC and in the same parallels BC, EF. [I. 35]  Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34]  And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I. 34]  [But the halves of equal things are equal to one another.]  Therefore the triangle ABC is equal to the triangle DBC. 
  Per D, enim ducatur D E, parallela rectæ A C,  & D F, parallela rectæ B C.  Erunt igitur parallelogramma A C D E, B C D F, æqualia.  Sunt enim super eandem basin C D, & intereasdem parallelas.  Sed horum dimidia sunt triangula A C D, B C D; quod A D, B D, diametri bifariam secent parallelogramma A C D E, B C D F.      Igitur & triangula A C D, B C D, æqualia erunt. 
Therefore etc.  Q. E. D. 
Triangula igitur super eadem basi, & c.  Quod erat demonstrandum.

SCHOLION
CONVERSA huius propositionis demonstrabitur ab Euclide propos. 39. Porro ex hac propositione facile cum Proclo demonstrabimus: Triangula, quorum duo latera unius æqualia sint duobus lateribus alterius, utrumque utrique, & angulus unius illis lateribus contentus maior angulo alterius, aliquando esse æqualia, & aliquando inæqualia: Id quod ad propositionem 24. polliciti sumus. Sint enim duo triangula A B C, D E F, & latera A B, A C, æqualia lateribus D E, D F, & angulus A, maior angulo E D F, sintque primum hi duo angulis duobus rectis æquales. Dico triangula esse æqualia. Producatur enim E D, ad H, & F D, ad I; fiatque angulus E D G, æqualis angulo A, versus angulum acutum D F E. Nam semper alter angulorum D E F, D F E, acutus erit, cum ambo minores sint duobus rectis, & recta D G, rectæ D F, seu A C; ducanturque rectæ E G, G F, cadetque semper E G, supra F, ob angulum acutum D F E, ut ad finem scholii propositio 24 ostendimus. Quoniam igitur duo anguli A, & E D F, ponuntur æquales duobus rectis, & angulus E D G, æqualis factus est angulo A; erunt & anguli E D G, E D F, duobus rectis æquales: Sunt autem & anguli E D G, G D H, duobus rectis æquales. Igitur ænguli E D G, E D F, angulis E D G, G D H, æquales erunt. Quare ablato comumni angulo E D G, remanebit angulo E D F, æqualis angulus G D H: Est autem eidem angulo E D F, æqualis angulus H D I. Igitur & anguli G D H, H D I, æquales erunt; atque adeo angulus G D H, dimidium erit totius anguli G D I. Rursus quia latera D F, D G, sunt æqualia in triangulo D F G; erunt anguli D F G, D G F, æquales; qui cum æquales sint externo angulo G D I, erit uterlibet eorum, nempe D G F, dimidium anguli G D I: Ostensum est autem, angulum G D H, dimidium quoque esse eiusdem anguli G D I. Quare anguli G D H, D G F, æquales erunt. Et quia sunt alterni inter E H, F G; erunt E H, F G, parallela. Quamobrem triangula D E G, D E F, æqualia erunt, cum habeant eandem basin D E, sintque inter easdem parallelas D E, F G. Quoniam vero triangulum D E G, æquale est triangulo A B C, propterea quod latera D E, D G, æqualia sunt lateribus A B, A C, & angulo A, æqualis angulus E D G; erit & triangulum A B C, triangulo D E F, æquale, quod est propositum.
SINT deinde anguli A, & E D F, duobus rectis maiores. Dico triangulum A B C, quod maiorem habet angulum, minus esse triangulo D E F. Producatur enim E D, ad H, & F D, ad I; fiatque angulus E D G, æqualis angulo A, & recta D G, rectæ D F, seu A C, æqualis, ducanturque recta E G, G F. Quoniam igitur anguli A, & E D F, ponuntur maiores duobus rectis, erunt & anguli E D G, E D F, duobus rectis maiores: Sunt autem anguli E D G, G D H, æquales duobus rectis. Igitur anguli E D G, E D F, maiores sunt angulis E D G, G D H. Quare ablato communi E D G, remanebit angulus E D F, maior angulo G D H. Quoniam vero angulus E D F, angulo H D I, æqualis est, erit quoque H D I, maior quam G D H; atque adeo G D H, minor, quam dimidium anguli G D I. Rursus quia latera D G, D F, æqualia sunt: erunt anguli D F G, D G F, æquales: qui cum sint æquales externo G D I, erit uteruis eorum, nempe D G F, dimidium anguli G D I: Ostensum est autem, angulum G D H, minorem esse dimidio eiusdem G D I. Quare D G F, maior erit, quam G D H. Abscindatur exangulo D G F, angulus D G K, æqualis angulo alterno G D H. Erit ergo G K, parallela ipsi D E, secabitque G K, rectam, E F. Ducatur ex D, ad K, ubi G K, secat rectam E F, recta D K. Erit igitur triangulum D E G, æquale triangulo D E K. Quoniam autem triangulum D E G, æquale est triangulo A B C, propterea quod latera D E, D G, æqualia sunt lateribus A B, A C, & angulo A, angulus E D G, æqualis; erit & triangulum A B C, triangulo D E K, æquale. Cum igitur D E K, minus sit triangulo D E F; erit quoque A B C, triangulum triangulo D E F, minus. Quod est propositum.
SINT tertio anguli A, & E D F, duobus rectis minores. Dico triangulum A B C, quod maiorem habet angulum, maius esse triangulo D E F. Producatur E D, ad H, & F D, ad I; fiatque angulus E D G, æqualis angulo A, & recta D G, rectæ D E, seu A C; ducanturque rectæ E G, F G. Quoniam igitur anguli A, & E D F, ponuntur minores duobus rectis, erunt quoque anguli E D G, E D F, duobus rectis minores. Sunt autem anguli E D G, G D H, duobus rectis æquales. Igitur E D G, E D F, minores sunt, quam E D G, G D H; demptoque communi E D G, remanebit E D F, minor, quam G D H. Est autem E D F, æqualis ipsi H D I. Quare & H D I, minor erit, quam G D H; atque adeo G D H, maior est dimidio anguli G D I. Quoniam autcm D G F, dimidium est eiusdem anguli G D I, ut iam supra ostensum fuit; erit G D H maior, quam D G F. Fiat igitur angulus D G K, æqualis angulo G D H, ducta recta G K, quæ secabit rectam E F, protractam in K; & ducatur recta D K. Erit ergo, ut prius, G K, parallela ipsi D E; triangulumque D E G, triangulo D E K, æquale: Est autem iterum D E G, æquale ipsi A B C. Igitur & A B C, æquale est ipsi D E K. Quocirca cum D E K, maius sit, quam D E F; erit & A B C, maius, quam D E F. Quod demonstrandum erat.
EX his perspicuum est, cur Euclides in propos. 24. solum collegerit inæqualitatem basium, non autem triangulorum, ut ibidem admonuimus.
 
Proposition 38. 
THEOR. 28. PROPOS. 38. 
第三十八題 
Triangles which are on equal bases and in the same parallels are equal to one another. 
TRIANGVLA super æqualibus basibus constituta, & in eisdem parallelis, inter se sunt æqualia. 
兩平行線內。有兩三角形。若底等。則兩形必等。 
Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD;  I say that the triangle ABC is equal to the triangle DEF. 
INTER parallelas A B, C D, & super æquales bases C E, F D, sint constituta triangula A C E, B F D.  Dico ipsa esse æqualia. 
For let AD be produced in both directions to G, H;  through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE.  Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH;  for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]  Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]  And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]  [But the halves of equal things are equal to one another.]  Therefore the triangle ABC is equal to the triangle DEF. 
  Ducatur enim E G, parallela ipsi A C, & D H, ipsi B F.  Eruntque parallelogramma A C E G, B F D H, æqualia.    Cum igitur horum dimidia sint triangula A C E, B E D;      erunt hæc inter se æqualia. 
Therefore etc.  Q. E. D. 
Triangula ergo super æqualibus basibus, &c.  Quod erat ostendendum.

SCHOLION
CONVERSA huius ostendetur ab Euclide propos. 40.
COLLIGITVR autem ex hac propositione, si a quouis angulo trianguli dati linea recta ducatur diuidens latus oppositum bifariam, triangulum quoque bifariam secari. Ducatur enim in triangulo A B C, ex angulo A, recta A D, diuidens bifariam latus B C, in D. Dico triangulum A B C, bifariam quoque secari. Si enim per A, ducatur parallela ipsi B C, erunt duo triangula A B D, A D C, inter easdem parallelas; & super æquales bases. Quare æqualia erunt.

EX PELETARIO
A puncto quouis dato in uno latere trianguli propositi lineam rectam ducere, quæ bifariam secet triangulum datum.
SIT triangulum A B C, & punctum datum D, in latere B C. Oportet igitur ex D, rectam lineam ducere, quæ bifariam diuidat triangulum. Quod si punctum D, diuidat latus B C, bifartam, recta D A, ducta ad A, diuidet triangulum bifariam, ut in hoc scholio est ostensum: Si vero D, non diuidit B C, bifariam, secetur B C, bifariam in E. Deinde ex D, ad angulum oppositum A ducatur recta D A, & per E, parallela E F, ipsi D A, secans A C, in F. Si igitur ducatur recta D F, erit triangulum diuisum bifariam a linea D F. Nam ducta recta E A, erunt triangula E F A, E F D, æqualia, cum sint super eandem basin E F, & inter easdem parallelas E F, A D. Addito igitur communi C F E, erunt tota triangula A E C, C D F, æqualia: Est autem A E C, dimidium totius A B C, ut iam fuit ostensum. Igitur & C D F, dimidium est eiusdem trianguli A B C. quod erat probandum.
QVOD si punctum D, fuerit in altera medietate E C, eodem modo problema conficiemus: sed tunc triangulum abscindetur ad partes B, trapezium vero ad partes C, ut figura præsens satis indicat. Demonstratio autem eadem est, si in ea mutetur litera B, in C, & C, in B. Hoc tamen problema multo nos uniuersalius proponemus ad finem sexti libri.
 
Proposition 39. 
THEOR. 29. PROPOS. 39. 
第三十九題 
Equal triangles which are on the same base and on the same side are also in the same parallels. 
TRIANGVLA æqualia super eadem basi, & ad easdem partes constituta; & in eisdem sunt parallelis. 
兩三角形。其底同。其形等。必在兩平行線內。 
Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it;  [I say that they are also in the same parallels.] 
SINT duo triangula æqualia A B C, D B C, super eandem basin B C, & ad easdem partes.  Dico ipsa esse inter easdem parallelas constituta, hoc est, rectam ductam A D, parallelam esse ipsi B C. 
And [For] let AD be joined; I say that AD is parallel to BC.  For, if not, let AE be drawn through the point A parallel to the straight line BC, [I. 31] and let EC be joined.  Therefore the triangle ABC is equal to the triangle EBC;  for it is on the same base BC with it and in the same parallels. [I. 37]  But ABC is equal to DBC;  therefore DBC is also equal to EBC, [C.N. 1] the greater to the less: which is impossible.  Therefore AE is not parallel to BC.  Similarly we can prove that neither is any other straight line except AD;  therefore AD is parallel to BC. 
  Si enim non est, ducatur ex A, parallela ipsi B C, quæ vel cadet supra A D, vel infra.  Cadat primum supra, qualis est A E, coeatque cum B D, protracta in E, & ducatur recta E C. Quoniam igitur parallelæ sunt A E, B C; erit triangulo A B C, triangulum E B C, æquale:    Est autem per hypothesin, triangulum quoque D B C, æquale eidem triangulo A B C.  Igitur erunt triangula D B C, E B C, æqualia, pars & totum, quod est absurdum.    Quod si parallela ducta per A, cadat infra A D, qualis est A F; ducta recta F C, erunt eadem ratiocinatione triangula B F C, B D C, æqualia, pars & totum; quod est absurdum.  Erit igitur A D, parallela ipsi B C. 
Therefore etc.  Q. E. D. 
Quare triangula æqualia super eadem basi, &c.  Quod ostendendum erat.

SCHOLION
EX his infert Campanus sequens hoc theorema.

LINEA recta secans duo trianguli latera bifariam, erit reliquo lateri parallela.

SECET linea D E, latera A B, A C, trianguli A B C, bifariam in D, & E. Dico D E, parallelam esse lateri B C. Cum enim triangula A D E, B D E, sint super æquales bases A D, D B, & inter easdem parallelas; si per E, duceretur parallela ipsi A B) erit triangulum B D E, triangulo A D E, æquale: Eadem ratione erit triangulum C E D, eidem triangulo A D E, æquale; Igitur triangula D B E, E C D, æqualia erunt: Habent autem eandem basin D E, & sunt ad easdem partes constituta. Quare inter easdem erunt parallelas, & idcirco D E, B C, parallelæ erunt. Quod est propositum.
ID autem, quod ad finem secundi theorematis in scholio propositio 34. polliciti sumus, facile ex hac propositio demonstrabimus. Videlicet.

OMNE quadrilaterum, quod ab utraque diametro bifariam diuiditur, parallelogrammum est.
NAM quadrilaterum A B C D, diuidatur bifariam ab utraque diametro A C, B D. Dico ipsum esse parallelogrammum. Cum enim triangula A D C, B D C, dimidia sint eiusdem quadrilateri A B C D, ipsa inter se æqualia erunt. Quare cum eandem habeant basin D C, ad easdemque partes sint, ipsa in eisdem parallelis erunt; Atque idcirco rectæ A B, D C, parallelæ sunt. Non aliter ostendemus, parallelas esse A D, B C. Parallelogrammum igitur est A B C D, Quod est propositum.
 
[Proposition 40. 
THEOR. 30. PROPOS. 40. 
第四十題 
Equal triangles which are on equal bases and on the same side are also in the same parallels. 
TRIANGVLA æqualia super æqualibus basibus, & ad easdem partes constituta, & in eisdem sunt parallelis. 
兩三角形。其底等。其形等。必在兩平行線內。 
Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side.  I say that they are also in the same parallels. 
SINT duo triangula æqualia A B C, D E F, super bases æquales B C, E F (quæ in eadem recta linea collocentur) & ad easdem partes constituta.  Dico ea esse in eisdem parallelis, 
For let AD be joined; I say that AD is parallel to BE. 
hoc est, rectam ex A, ad D, ductam parallelam esse rectæ B F. 
For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined.  Therefore the triangle ABC is equal to the triangle FCE;  for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38]  But the triangle ABC is equal to the triangle DCE;  therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible.  Therefore AF is not parallel to BE.  Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE. 
Si enim non est, cadet parallela ipsi B F, per A, ducta, vel supra A D, vel infra. Cadat primum supra, coeatque cum E D, producta in G, & ducatur recta G F.  Quoniam igitur parallelæ sunt A G, B F, erit triangulum E F G, triangulo A B C, æquale:    Ponitur autem & triangulum D E F, eidem triangulo A B C, æquale.  Igitur triangula D E F, G E F, æqualia erunt, pars & totum. Quod est absurdum.    Quod si parallela ducta per A, cadat infra A D, qualis est A H; ducta recta H F, erunt eadem argumentatione triangula H E F, D E F, æqualia, pars & totum, quod est ab?urdum. Est igitur A D, parallela ipsi B F. 
Therefore etc.  Q. E. D.] 
Quare triangula æqualia super æqualibus basibus, &c.  Quod erat demonstrandum.

SCHOLION
EODEM modo demonstrari poterit hoc theorema.
TRIANGVLA æqualia inter easdem parallelas, si non eandem habuerint basin, super æquales bases erunt constituta.
SINT triangula æqualia A B C, D E F, inter parallelas A D, B F, & super bases B C, E F, quas dico esse æquales. Si enim non sunt æquales, sit B C, maior. Abscissa ergo recta C G, æquali ipsi E F, & ducta recta G A; erit triangulum A G C, triangulo D E F, æquale: Ponitur autem & triangulum A B C, eidem triangulo D E F, æquale: Igitur triangula A G C, A B C, æqualia erunt, pars & totum, quod est absurdum. Non ergo inæquales sunt bases B C, E F, sed æquales. Quod est propositum.
 
Proposition 41. 
THEOR. 31. PROPOS. 41. 
第四十一題 
If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle. 
SI parallelogrammum cum triangulo eandem basin habuerit, in eisdemque fuerit parallelis, duplum erit parallelogrammum ipsius trianguli. 
兩平行線內。有一平行方形。一三角形。同底。則方形倍大於三角形。 
For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE;  I say that the parallelogram ABCD is double of the triangle BEC. 
INTER parallelas A B, C D, & super basin C D, constituantur parallelogrammum A C D E, & triangulum B C D.  Dico parallelogrammum esse duplum triangnli B C D. 
For let AC be joined.  Then the triangle ABC is equal to the triangle EBC;  for it is on the same base BC with it and in the same parallels BC, AE. [I. 37]  But the parallelogram ABCD is double of the triangle ABC;  for the diameter AC bisects it; [I. 34]  so that the parallelogram ABCD is also double of the triangle EBC. 
Ducta enim diametro A D, in parallelogrammo,  erunt triangula A C D, B C D, æqualia;    At parallelogrammum A C D E, duplum est trianguli A C D;    Igitur & trianguli B C D, duplum erit idem parallelogrammum A C D E. 
Therefore etc.  Q. E. D. 
Quamobrem, si parallelogrammum cum triangulo, &c.  Quod erat demonstrandum.

SCHOLION
HINC sequitur, si triangulum duplam habuerit basin, fueritque in eisdem parallelis cum parallelogrammo, triangulum parallelogrammo æquale fore. Nam si basis C D, producatur ad F, ut sit D F, æqualis ipsi C D, ducaturque recta F B, erit triangulum B C F, duplum trianguli B C D, quod triangula B C D, B D F, æqualia sint: Est autem & parallelogrammum A C D E, duplam eiusdem trianguli B C D. Igitur æqualia erunt triangulum B C F, & parallelogrammum A C D E.
IDEM hoc theorema Euclidis demonstrari potest eodem modo, si parallelogrammum, & triangulum æquales habuerint bases, & non eandem, fuerintque in eisdem parallelis, ut cernis in parallelogrammo A C D E, & triangulo B F G, quorum bases C D, F G, æquales sunt. Ducta enim diametro A D, in parallelogrammo, erunt triangula A C D, B F G, æqualia. Cum igitur parallelogrammum A C D E, duplum sit trianguli A C D: quod diameter A D, secet parallelogrammum A C D E, bifariam: erit quoque idem trianguli B F G, duplum. Eadem ratione si basis F G, duplicaretur, & recta ad B, duceretur, fieret triangulum parallelogrammo æquale, quoniam triangulum hoc esset duplum etiam trianguli B F G, &c.
CONVERSVM huius theorematis duplex est, hoc modo.

SI trianguli parallelogrammum duplum fuerit, eandemque habuerint basin, vel æquales, & ad easdem partes constituta; Erunt ipsa in eisdem parallelis. Et si parallelogrammum duplum fuerit trianguli, in eisdemque parallelis: erunt bases æquales, si non sit eadem.
SIT parallelogrammum A B C D, duplum trianguli E B C, siue eandem habeant basin, siue æquales. Dico parallelam A D, productam cadere in E, punctum. Nam alias cadet aut supra E, aut infra. Vnde, ut in 39. vel 40. propositio ostendetur pars æqualis toti, ut & figura indicat. Nam erit quoque parallelogrammum A B C D, trianguli B F C, vel B G C, duplum. Quare triangula E B C, F B C, vel triangula E B C, G B C, æqualia erunt, pars & totum. Quod est absurdum.
SIT deinde parallelogrammum A B C D, duplum trianguli E F G, in eisdemque parallelis. Dico bases B C, F G, esse æquales. Nam si altera, nempe B C, sit maior, abscissa æquali C H, & ducta H I, parallela ipsi A B, demonstrabimus parallelogramma A B C D, I H C D, esse æqualia, totum & partem; (quia utrumque duplum est trianguli E F G; illud quidem per hypothesin, hoc vero per 41. propositio) Quod est absurdum. Idem ostendemus si basis F G, maior dicatur. Si enim abscindatur ipsi B C, æqualis B H, ducaturque recta H E, erunt triangula E F H, E F G, æqualia, pars & totum; (Nam utrumque dimidium est parallelogrammi A B C D; Illud quidem per propositio 41. hoc vero per hpothesin.) Quod est absurdum.

EX PROCLO
SI triangulum, & trapezium super eadem basi, & in eisdem fuerint parallelis, maior autem linea parallela trapezii sit basis trianguli; erit trapezium minus duplo trianguli: Si vero minor linea parallela trapezii basis sit trianguli, erit trapezium maius duplo trianguli.
INTER lineas parallelas A E, B C, sint constituta trapezium A B C D, & triangulum E B C, super basim B C, eandem, quæ sit tamen maior, qua altera linea A D, parallela in trapezio dato. Dico trapezium A B C D, minus esse duplo trianguli E B C. Cum enim A D, minor ponatur quam B C, sumatur A F, æqualis ipsi B C, & ducatur recta C E, quæ erit parallela ipsi A B; atque adeo parallelogrammum erit A B C F, quod duplum est trianguli E B C. Quare trapezium A B C D, cum sit pars parallelogrammi, minus erit duplo eiusdem trianguli E B C, quod est propositum.
SINT rursus trapezium, & triangulum, ut prius, sed basis B C, sit minor, quam reliqua lineæ parallela A D, in trapezio dato. Dico trapezium A B C D, maius esse duple trianguli E B C. Cum enim A D, maior sit, quam B C, abscindatur D F, æqualis ipsi B C, & ducatur recta B F, quæ erit parallela ipsi C D; atque adeo parallelogrammum erit B C D F: quod duplum est trianguli E B C. Quare totum trapezium A B C D, quod superat parallelogrammum B C D F, maius erit duplo eiusdem trianguli E B C. quod est propositum.
IDEM concludetur, si trapezium, & triangulum constituta fuerint super æquales bases, ita tamen ut nunc quidem basis trapezii sit maior latere opposito parallelo, nunc vero minor.

TRAPEZIVM habens duo latera opposita parallela, duplum est trianguli, quod basin habet unum latus trapezii coniungens duas parallelas, verticem vero in medio puncto lateris oppositi.
SIT trapezium A B C D, cuius duo latera opposita A B, D C, sint parallela, & super basin B C, constituatur triangulum E B C, verticem E, habens in medio puncto E, lateris A D. Dico tiapezium A B C D, duplum esse trianguli E B C. Producatur enim unum latus trianguli ad verticem, nempe B E, donec coeat cum C D, protracto in F. Et quia parallelæ sunt A B, C F, erunt anguli alterni B A E, F D E, æquales: Sunt autem & anguli A E B, D E F, æquales, quippe qui ad verticem E; & latus A E, trianguli A B E, lateri D E, trianguli D E F, æquale, per hypothesin. Igitur & reliqua latera A B, B E, reliquis lateribus, D F, F E, æqualia erunt, utrumque utrique, & reliqui anguli A B E, D F E, æquales: atque idcirco triangula A B E, D F E, ex corollarium propositio 26. buius liber æqualia erunt. Quare addito communi triangulo C D E, erunt triangulo C E F, æqualia triangula simul A B E, C D E. Est autem & triangulum B C E, eidem triangulo C E F, æquale, quod bases B E, E F, ostensæ sint æquales, & ipsa triangula inter easdem sint parallelas, si per C, duceretur parallela ipsi B F. Igitur triangulum C B E, æquale erit triangulis A B E, C D E; & propterea C B E, triangulum dimidium erit trapezii A B C D, quod est propositum.
 
Proposition 42. 
PROBL. 11. PROPOS. 42. 
第四十二題 
To construct, in a given rectilineal angle, a parallelogram equal to a given triangle. 
DATO triangulo æquale parallelogrammum constituere in dato angulo rectilineo. 
有三角形。求作平行方形、與之等。而方形角、有與所設角等。 
Let ABC be the given triangle, and D the given rectilineal angle;  thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC. 
DATVM triangulum sit A B C, & datus angulus rectilineus D.  Oportet igitur constituere parallelogrammum æquale triangulo A B C, habens angulum æqualem angulo D. 
Let BC be bisected at E, and let AE be joined;  on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23]  through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF.  Then FECG is a parallelogram.  And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC,  for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38]  therefore the triangle ABC is double of the triangle AEC.  But the parallelogram FECG is also double of the triangle AEC,  for it has the same base with it and is in the same parallels with it; [I. 41]  therefore the parallelogram FECG is equal to the triangle ABC.  And it has the angle CEF equal to the given angle D. 
Diuidatur latus unum trianguli, nempe B C, bifariam in E,  & fiat angulus C E F, æqualis angulo D,  Ducatur item per A, recta A F, parallela ipsi B C, quæ secet E F, in F. Rursus per C, vel B, ducatur ipsi E F, parallela C G, occurrens rectæ A F, productæ in G.  Eritque constitutum parallelogrammum C E F G, quod dico esse æquale triangulo A B C.  Ducta enim recta E A; quoniam parallelogrammum C E F G, duplum est trianguli A E C;    & triangulum A B C, duplum eiusdem trianguli A E C, quod triangula A E C, A B E, super æquales bases E C, B E, & in eisdem parallelis, sint æqualia.      Erunt parallelogrammum C E F G, & triangulum A B C, æqualia inter se.  Cum igitur angulus C E F, factus sit æqualis angulo D, constat propositum. 
Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D.  Q. E. F. 
Quocirca dato triangulo æquale parallelogrammum constitui-smus in dato angulo rectilineo.  Quod erat faciendum.

SCHOLION
SVBIVNGIT autem hoc loco Peletarius subsequens problema.

DATO parallelogrammo æquale triangulum contituere, in dato angulo rectilineo.
SIT datum parallelogrammum A B C D, & datus angulus G. Fiat angulus C B E, angulo G, æqualis, secetque recta B E, rectam A D, productam in E. Extendatur quoque B C, ad F, sitque C F, æqualis rectæ B C, & iungatur E F. Dico triangulum B E F, habens angulum E B F, angulo dato G, æqualem, æquale esse parallelogrammo A B C D. Ducta enim recta C E, erit parallelogrammum A B C D, duplum trianguli B C E. Item triangulum B E F, eiusdem trianguli B C E, duplum; quod æqualia sint triangula E B C, E B F. Quare æqualia inter se erunt parallelogrammum A B C D, & triangulum B E F.
 
Proposition 43. 
THEOR. 32. PROPOS. 43. 
第四十三題 
In any parallelogram the complements of the parallelograms about the diameter are equal to one another. 
IN omni parallelogrammo complementa eorum, quæ circa diametrum sunt parallelogrammorum, inter se sunt æqualia. 
凡方形對角線旁、兩餘方形。自相等。 
Let ABCD be a parallelogram, and AC its diameter;  and about AC let EH, FG be parallelograms, and BK, KD the so-called complements;  I say that the complement BK is equal to the complement KD. 
IN parallelogrammo A B C D, sint circa diametrum A C,  parallelogramma A E G H, C F G I, & complementa D F G H, E B I G, ut in 36. defin. diximus.  Dico complementa hæc inter se esse æqualia. 
For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD. [I. 34]  Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK.  For the same reason the triangle KFC is also equal to KGC.  Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC, the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2]  And the whole triangle ABC is also equal to the whole ADC;  therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3] 
Cum enim triangula A B C, C D A, æqualia sint;  Itemque triangula A E G, G H A;        si hæc ab illis demantur, remanebunt trapezia C B E G, C D H G, æqualia: Sunt autem & triangula C G I, C G F, æqualia. Quare si detrahantur ex trapeziis, remanebunt æqualia complementa D F G H, E B I G. 
Therefore etc.  Q. E. D. 
In omni igitur parallelogrammo, complementa, &c.  Quod ostendendum erat.

SCHOLION
EODEM modo hoc theorema demonstratur a Proclo, etiam si duo parallelogramma circa diametrum non coniunguntur in puncto G, sed vel unum ab altero sit semotum, vel ambo se mutuo intersecent. Sit enim prius unum ab altero distans, ita ut complementa sint figuræ quinquangulæ. Vt in parallelogrammo A B C D, circa diametrum A C, consistant parallelogramma A E F G, C H I K. Dico complementa D E F I H, B K I F G, esse æqualia. Cum enim triangula A B C, C D A, æqualia inter se sint: Item triangula A E F, G H I, æqualia triangulis A G F, C K I: erunt reliqua complementa D E F I H, B K I F G, æqualia. Quod est propositum.
SECENT se iam mutuo parallelogramma A E F G, C H I K, circa diametrum consistentia, ita ut communem partem habeant I L F M. Dico adhuc complementa D E L H, B G M K, esse æqualia. Cum enim æqualia sint triangula A B C, C D A: Item triangula A F G, A F E: erunt reliqua quadrilatera B C F G, D C F E, æqualia: Sunt autem rursus æqualia triangula I F M, I F L. Igitur si hæc addantur dictis quadrilateris, erunt figura B C I M G, D C I L E, æquales. Cum igitur & æqualia sint triangula C I K, C I H: erunt reliqua complementa B G M K, D E L H, etiam æqualia. Quod est propositum.
CONVERSVM quoque huius theorematis cum Peletario demonstrabimus, hoc modo.

SI parallelogrammum diuisum fuerit in quatuor parallelogramma, ita ut ex illis duo aduersa sint æqualia: consistent reliqua duo circa diametrum.
DVCTIS duabus rectis E F, G H, quæ sint parallelæ rectis B C, C D, seque secent in I, diuisum sit parallelogrammum A B C D, in quatuor parallelogramma, quorum aduersa duo B E I H, D F I G, sint æqualia. Dico reliqua duo A E I G, C F I H, circa diametrum consistere, hoc est, diametrum a puncto C, ad punctum A, ductam transire per punctum I. Si enim non transit, secet diameter C K A, rectam G H, in K, si fieri potest, & per K, ducatur L M, parallela ipsi B C. Erunt igitur complementa B H K L, D G K M, æqualia: Est autem D G K M, maius quam D G I F. Quare & maius erit B H K L, quam D G I F. Cum ergo D G I F, æquale ponatur ipsi B E I H; erit etiam B H K L, maius quam B E I H, pars quam totum. Quod est absurdum. Non ergo diameter A C, rectam G H, in K, secat, sed per punctum I, transit. Quodest propositum.
 
Proposition 44. 
PROBL. 12. PROPOS. 44. 
第四十四題 
To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle. 
AD datam rectam lineam, dato triangulo æquale parallelogrammum applicare in dato angulo rectilineo. 
一直線上。求作平行方形。與所設三角形等。而方形角、有與所設角等。 
Let AB be the given straight line, C the given triangle and D the given rectilineal angle;  thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C. 
DATA recta linea sit A, datum triangulum B, & datus angulus rectilineus C.  Oportet igitur constituere parallelogrammum æquale triangulo B, angulum habens æqualem angulo C, & unum latus æquale rectæ A. 
Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42];  let it be placed so that BE is in a straight line with AB;  let FG be drawn through to H,  and let AH be drawn through A parallel to either BG or EF. [I. 31]  Let HB be joined.  Then, since the straight line HF falls upon the parallels AH, EF,  the angles AHF, HFE are equal to two right angles. [I. 29]  Therefore the angles BHG, GFE are less than two right angles;  and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]  therefore HB, FE, when produced, will meet.  Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31]  and let HA, GB be produced to the points L, M.  Then HLKF is a parallelogram, HK is its diameter,  and AG, ME are parallelograms, and LB, BF the so-called complements, about HK;  therefore LB is equal to BF. [I. 43]  But BF is equal to the triangle C; therefore LB is also equal to C. [C.N. 1]  And, since the angle GBE is equal to the angle ABM, [I. 15]  while the angle GBE is equal to D, the angle ABM is also equal to the angle D. 
Constituatur triangulo B, æquale parallelogrammum D E F G, habens angulum E F G, angulo C, æqualem,    producaturque G F, ad H, ut F H, sit æqualis rectæ A,  & per H, ducatur H I, parallela ipsi F E, occurrens D E, productæ in I.  Extendatur deinde ex I, per F,              diameter I F, occurrens rectæ D G, productæ in K; & per K, ducatur K L, parallela ipsi G H, secans I H, protractam in L, producaturque E F, ad M.        Dico parallelogrammum L M F H, esse id, quod quæritur.  Habet enim latus F H, æquale datæ rectæ A, & angulum H F M, angulo dato C, æqualem,  cum angulus H F M, æqualis sit angulo E F G, qui factus est æqualis angulo C: 
Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which is equal to D.  Q. E. F. 
Denique parallelogrammum L M F H, æquale est triangulo B, cum æquale sit complemento D E F G, quod factum est æquale triangulo B. Ad datam igitur rectam lineam dato triangulo, &c.  Quod erat faciendum.

SCHOLION
QVOD si quis optet, lineam ipsam A, datam, esse unum latus parallelogrammi, non difficile erit transferre parallelogrammum F M L H, ad rectam A, ex iis, quæ in scholio propositio 31. huius liber docuimus. Si enim in N, extremitate rectæ A, fiat angulus æqualis angulo M F H, & sumatur recta N O, æqualis rectæ F M, compleaturqúe parallelogrammum, ceu in dicto scholio traditum fuit, effectum erit, quod quæritur.
ADDIT hic aliud problema Peletarius, hoc modo.

AD datam rectam lineam, dato parallelogrammo constituere æquale triangulum in dato angulo rectilineo.
SIT data recta A B; datum parallelogrammum C D E F, & datus angulus L. Producatur C D, ad G, ut D G, æqualis sit ipsi C D, & iungatur G E, recta: Eritque triangulum C E G, parallelogrammo C D E F, æquale, ut demonstrauimus scholio propos. 41. Fiat iam super data recta A B, parallelogrammum A B H I, æquale triangulo C E G, hoc est, parallelogrammo C D E F, habens angulum A, angulo L, æqualem; & producatur A I, ad K, ut sit I K, æqualis ipsi A I, iungaturque recta B K. Dico triangulum A B K, constitutum super datam rectam A B, habensque angulum A, æqualem dato angulo L; æquale esse dato parallelogrammo C D E E. Cum enim triangulum A B K, æquale sit parallelogrammo A B H I, ex scholio propos. 41. quod æquale est constructum parallelogrammo C D E F, constat propositum.
 
Proposition 45. 
PROBL. 13. PROPOS. 45. 
第四十五題 
To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure. 
Dato rectilineo æquale parallelogrammum constituere, in dato angulo rectilineo. 
有多邊直線形。求作一平行方形與之等。而方形角、有與所設角等。 
Let ABCD be the given rectilineal figure and E the given rectilineal angle;  thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD. 
DATVM rectilineum sit A B C, & datus angulus D:  oportet igitur construere parallelogrammum æquale rectilineo A B C, quod habeat angulum æqualem angulo D. 
Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42]  let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44]  Then, since the angle E is equal to each of the angles HKF, GHM, the angle HKF is also equal to the angle GHM. [C.N. 1]  Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM.  But the angles FKH, KHG are equal to two right angles; [I. 29]  therefore the angles KHG, GHM are also equal to two right angles.  Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles;  therefore KH is in a straight line with HM. [I. 14]  And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29]  Let the angle HGL be added to each;  therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C.N. 2]  But the angles MHG, HGL are equal to two right angles; [I. 29]  therefore the angles HGF, HGL are also equal to two right angles. [C.N. 1]  Therefore FG is in a straight line with GL. [I. 14]  And, since FK is equal and parallel to HG, [I. 34]  and HG to ML also,  KF is also equal and parallel to ML; [C.N. 1; I. 30]  and the straight lines KM, FL join them (at their extremities);  therefore KM, FL are also equal and parallel. [I. 33]  Therefore KFLM is a parallelogram.  And, since the triangle ABD is equal to the parallelogram FH, and DBC to GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. 
Resoluat rectilineum in triangula A, B, & C. Deinde triangulo A, æquale parallelogrammum constituatur E F G H, habens angulum F, angulo D, æqalem.  Item super rectam G H, parallelogrammum G H I K, æquale triangulo B, habens angulum G, æqualem angulo D. Item super rectam I K, parallelogrammum I K L M, æquale triangulo C, habens angulum K, æqualem anglo D; Et sic deinceps procedatur, si plura fuerint triangula in dato rectilineo, factumque erit, quod iubetur. Nam tria parallelogramma constructa, quæ quidem æqualia sunt rectilineo dato A B C, conficiunt totum unum parallelogrammum, quod sic demonstratur.  Duo anguli E F G, H G K, inter se sunt æquales, cum uterque æqualis sit angulo D.  Addto igitur communi angulo F G H, erunt duo anguli E F G, F G H, qui duobus rectis æquivalent, æquales duobus angulis H G K, F G H,    ideoque hi anguli duobus rectis æquales sunt.    Quare F G, G K, unam rectam lineam efficient.            Eadem ratione ostendemus, E H, H I, unam rectam lineam efficere, propterea quod duo anguli E H G, H I K, æquales inter se sint, (cum sint æquales oppositis angulis æqualibus E F G, H G K.) & duo anguli H I K, I H G, duobus sint rectis æquales, &c.          Cum igitur E I, F K, sint parallelæ;  Itemque E F, I K, quod utraque parallela sit rectæ H G; Parallelogrammum erit E F K I.  Eodemque modo demonstrabitur, parallelogrammum I K L M, adiunctum parallelogrammo E F K I, constituere totum unum parallelogrammum E F L M. 
Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E.  Q. E. F. 
Dato ergo rectilineo A B C, constituimus æquale parallelogrammum E F L M, habens angulum F, æqualem angulo D, dato.  Quod erat efficiendum.

SCHOLION
QVAMVIS in hoc problemate Euclides absolute, & simpliciter docuerit, quanam arte parallelogrammum constituatur æquale rectilineæ figuræ datæ non astringendo nos ad certam aliquam rectam lineam datam, ut in propos. 44. fecerat; Tamen eodem modo, quod iubetur, efficiemus, si recta aliqua lineanobis fuerit assignata. Nam si detur recta linea E F, super ipsam construemus parallelogrammum E F G H, æquale triangulo A. Et eodem modo super G H, constituemus aliud G H I K, æquale triangulo B, &c. Quibus peractis, constitutum erit super datam rectam E F, parallelogrammum E F L M, æquale rectilineo dato, in dato angulo F, qui æqualis est angulo D, proposio.
PARI ratione, propositis quotcunque rectilineis, constituemus illis parallelogrammum æquale, si omnia resoluantur in triangula, quibus æqualia parallelogramma exhibeantur, singulis singula, per propos. 44. veluti factum est in hoc problemate. Nam cum omnia hæc parallelogramma efficiant unum parallelogrammum, uti hic demonstratum fuit, constitutum erit parallelogrammum æquale rectilineis propositis. Vt si quis intelligat duo rectilinea proposita A B, & C; Atque A B, resoluatur in triangula A, & B, singulisque triangulis, A, B, C, singula parallelogramma E G, G I, I L, super rectas E F, H G, I K, iuxta artem huius problematis, æqualia constituantur, ex propos. 42. & 44. erit constructum parallelogrammum totum E F L M, æquale duobus rectilineis A B, & C. Et sic de pluribus.
HVC referri poterit problema utilissimum ex Peletario, quod nos tamen alia ratione, et breuiori demonstrabimus in hunc modum.

DATIS duobus rectilineis inæqualibus, excessum maioris supra minus inquirere.
SINT data rectilinea A, & B, sitque A, maius. Oportet igitur indagare, qua magnitudine rectilineum A, superet rectilineum B. Fiat parallelogrammum C D E F, in quocunque angulo D, æquale maiori rectilineo A. Et super rectum C D, parallelogrammum C D G H, in eodem angulo D, æquale rectilineo minori B. Quoniam igitur parallelogrammum C D E F, superat parallelogrammum C D G H, parallelogrammo E F H G, superabit quoque figura A, figuram B, eodem parallelogrammo E F H G. Quod est propositum.
 
Proposition 46. 
PROBL. 14. PROPOS. 46. 
第四十六題 
On a given straight line to describe a square. 
A DATA recta linea quadratum describere. 
一直線上。求立直角方形。 
Let AB be the given straight line; thus it is required to describe a square on the straight line AB. 
SIT data recta A B, super quam oporteat quadratum describere. 
Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB;  through the point D let DE be drawn parallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31]  Therefore ADEB is a parallelogram;  therefore AB is equal to DE, and AD to BE. [I. 34]  But AB is equal to AD;  therefore the four straight lines BA, AD, DE, EB are equal to one another;  therefore the parallelogram ADEB is equilateral.  I say next that it is also right-angled.  For, since the straight line AD falls upon the parallels AB, DE, the angles BAD, ADE are equal to two right angles. [I. 29]  But the angle BAD is right; therefore the angle ADE is also right.  And in parallelogrammic areas the opposite sides and angles are equal to one another; [I. 34]  therefore each of the opposite angles ABE, BED is also right.  Therefore ADEB is right-angled.  And it was also proved equilateral. 
Ex A, & B, educantur A D, B C, perpendiculares ad A B, sintque ipsi A B, æquales,  & connectatur recta C D.  Dico A B C D, esse quadratum.  Cum enim anguli A, & B, sint recti, erunt A D, B C, parallelæ: Sunt autem & æquales, quod utraque æqualis sit ipsi A B. Igitur & A B, D C, parallelæ sunt & æquales: & ideo parallelogrammum est A B C D, in quo, cum A D, D C, C B, æquales sint ipsi A B,     omnes quatuor lineæ æquales existunt:  Sunt autem & omnes quatuor anguli recti, cum C, & D, æquales sint oppositis rectis A, & B. Quadratum igitur est A B C D, ex definitione;  s  s  s  s  s  s   
Therefore it is a square; and it is described on the straight line AB.  Q. E. F. 
Ac proinde a data recta linea quadratum descripsimus.  Quod faciendum erat.

EX PROCLO
LINEARVM æqualium æqualia sunt quadrata: & quadratorum æqualium æquales sunt lineæ.
SINT primum rectæ A B, C D, æquales. Dico eorum quadrata A B E F, C D G H, æqualia quoque esse. Ductis enim diametris B F, D H, erunt duo latera B A, A F, trianguli B A F, duobus lateribus D C, C H, trianguli D C H, æqualia, utrumque utrique, cum ex definitione quadrati rectæ A F, C H, æquales sint rectis A B, C D. Sunt autem & anguli A, & C, æquales, nempe recti. Igitur triangula B A F, D C H, æqualia erunt. Quæ cum sint dimidia quadratorum, erunt & quadrata tota æqualia. Quod est propositum.
SINT deinde quadrata A B D E, B C F G, æqualis. Dico lineas quoque ipsorum A B, B C, æquales esse. Coniungantur enim quadrata ad angulum B, ut rectæ A B, B C, in directum constituantur. Et quoniam anguli A B G, A B D, sunt recti, erunt & rectæ G B, B D, in directum constitutæ. Ducantur diametri A D, C G, iunganturque rectæ A G, C D. Quoniam igitur quadrata A B D E, B C F G, æqualia sunt, erunt & triangula A B D, B C G, eorum dimidia, æqualia. Addito ergo communi triangulo B C D, fiet totum triangulum A C D, toti triangulo G D C, æquale. Quare triangula A C D, G D C, cum eandem habeant basin C D, ad easdemque sint partes, in eisdem sunt parallelis: ideoque parallelæ sunt A G, C D. Et quoniam, ut in scholio propositio 34. ostendimus, diameter in quadrato secat angulos quadrati bifariam, erunt anguli D A C, G C A, alterni semirecti, ideoque æquales. Quamobrem, & parallelæ sunt A D, C G. Igitur parallelogrammum est A D C G: ac propterea rectæ A D, C G, æquales. Quoniam ergo in triangulis A B D, B C G, latera A D, C G, æqualia sunt, & anguli quibus ea latera adiacent, inter se etiam æquales, cum sint semirecti, ut in scholio propositio 34. ostensum fuit: erunt reliqua latera æqualia, nempe A B, ipsi B C, &c. Quod est propositum.

SCHOLION
POSSENT hæc omnia multo breuius probari per superpositionem quadrati unius super aliud. Nam si lineæ sint æquales, si unæ alteri superponatur, congruent ipsæ inter se. Cum ergo & anguli sint æquales, nempe recti, conuenient quoque ipsi inter se, ideoque totum quadratum toti quadrato congruet. Quod si quadrata sint æqualia, congruent ipsa inter se, propter æqualitatem angulerum. Igitur & lineæ; alias unum quadratum alio maius esset.
 
Proposition 47. 
THEOR. 33. PROPOS. 47. 
第四十七題 
In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle. 
IN rectangulis triangulis, quadratum, quod a latere rectum angulum subtendente describitur, æquale est eis, quæ a lateribus rectum angulum continentibus describuntur, quadratis. 
凡三邊直角形。對直角邊上、所作直角方形。與餘兩邊上、所作兩直角方形幷、等。 
Let ABC be a right-angled triangle having the angle BAC right;  I say that the square on BC is equal to the squares on BA, AC. 
IN triangulo A B C, angulus B A C, sit rectus, describanturque super A B, A C, B C, quadrata A B F G, A C H I, B C D E.  Dico quadratum B C D E, descriptum super latus B C, quod angulo recto opponitur, æquale esse duobus quadratis A B F G, A C H I, quæ super alia duo latera sunt descripta, siue hæc duo latera æqualia sint, siue inæqualia. 
解曰。甲乙丙角形。  於對乙甲丙直角之乙丙邊上、作乙丙丁戊直角方形。本篇四六 題言此形、與甲乙邊上、所作甲乙己庚、及甲丙邊上、所作甲丙辛壬、兩直角方形幷。等。  
For let there be described on BC the square BDEC, and on BA, AC the squares GB, HC; [I. 46]  through A let AL be drawn parallel to either BD or CE,  and let AD, FC be joined.  Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines AC, AG not lying on the same side make the adjacent angles equal to two right angles;  therefore CA is in a straight line with AG. [I. 14]  For the same reason BA is also in a straight line with AH.  And, since the angle DBC is equal to the angle FBA: for each is right:  let the angle ABC be added to each; therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2]  And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively,  and the angle ABD is equal to the angle FBC;  therefore the base AD is equal to the base FC,  and the triangle ABD is equal to the triangle FBC. [I. 4]  Now the parallelogram BL is double of the triangle ABD,  for they have the same base BD and are in the same parallels BD, AL. [I. 41]  And the square GB is double of the triangle FBC,  for they again have the same base FB and are in the same parallels FB, GC. [I. 41]  [But the doubles of equals are equal to one another.]  Therefore the parallelogram BL is also equal to the square GB.  Similarly, if AE, BK be joined, the parallelogram CL can also be proved equal to the square HC;  therefore the whole square BDEC is equal to the two squares GB, HC. [C.N. 2]  And the square BDEC is described on BC, and the squares GB, HC on BA, AC.  Therefore the square on the side BC is equal to the squares on the sides BA, AC. 
See two records above87  
Ducatur enim recta A K, parallela ipsi B E, vel ipsi C D, secans B C, in L,  & iungantur rectæ A D, A E, C F, B H.  Et quia duo anguli B A C, B A G, sunt recti,  erunt rectæ G A, A C, una linea recta;  eodemque modo I A, A B, una recta linea erunt.  Rursus quia anguli A B F, C B E, sunt æquales, cum sint recti,   si addatur communis angulus A B C, fiet totus angulus C B F, toti angulo A B E, æqualis; similiterque totus angulus B C H, toti angulo A C D.  Quoniam igitur duo latera A B, B E, trianguli A B E, æqualia sunt duobus lateribus F B, B C, trianguli F B C, utrumque utrique, ut constat ex definitione quadrati:  Sunt autem & anguli A B E, F B C, contenti hisce lateribus æquales, ut ostendimus;    Erunt triangula A B E, F B C, æqualia.  Est autem quadratum, seu parallelogrammum A B F G, duplum trianguli F B C,  cum sint inter parallelas F B, C G, & super eandem basin B F:   Et parallelogrammum B E K L, duplum trianguli A B E,  quod sint inter parallelas B E, A K, & super eandem basin B E.     Quare æqualia erunt quadratum A B F G, & parallelogrammum B E K L.  Eadem ratione ostendetur, æqualia esse quadratum A C H I, & parallelogrammum C D K L.  Erunt enim rursus triangula A C D, H C B, æqualia, ideoque eorum dupla, parallelogrammum videlicet C D K L, & quadratum A C H I.    Quamobrem totum quadratum B C D E, quod componitur ex duobus parallelogrammis B E K L, C D K L, æquale est duobus quadratis A B F G, A C H I. 
論曰。試從甲作甲癸直線。與乙戊丙丁平行。本篇卅一  分乙丙邊於子。次自甲至丁、至戊、各作直線。  末自乙至辛、自丙至己、各作直線。  其乙甲丙、與乙甲庚、旣皆直角。  卽庚甲、甲丙、是一直線。本篇十四  依顯乙甲、甲壬、亦一直線。  又丙乙戊、與甲乙己、旣皆直角。  而每加一甲乙丙角。卽甲乙戊、與丙乙己、兩角亦等。公論二  依顯甲丙丁、與乙丙辛、兩角亦等。又甲乙戊角形之甲乙、乙戊、兩邊、與丙乙己角形之己乙、乙丙、兩邊等。  甲乙戊、與丙乙己、兩角復等。  則對等角之甲戊、與丙己、兩邊亦等。  而此兩角形、亦等矣。本篇四  夫甲乙己庚直角方形。倍大於同乙己底、  同在平行線內之丙乙己角形。本篇四一  而乙戊癸子直角形。亦倍大於同乙戊底、  同在平行線內之甲乙戊角形。  則甲乙己庚、不與乙戊癸子等乎。公論六  依顯甲丙辛壬直角方形、與丙丁癸子直角形、等。  則乙戊丁丙形一、與甲乙己庚、甲丙辛壬、兩形幷、等矣。 
Therefore etc.  Q. E. D. 
In rectangulis ergo triangulis, quadratum, &c.   Quod demonstrandum erat. 
 


SCHOLION
INVENTIO huius theorematis ad Pythagoram refertur, qui, ut scribit Vitruvius lib. 9. hostias Musis immolauit, quod se in tam præclaro inuento adiuverint. Sunt qui putent, eum immolasse centum boues: si tamen Proclo credendum est, unum tantummodo obtulit. Fortasse autem Pythagoras, ut nonnulli volunt, ex numeris occasionem sumpsit; ut theorema hoc inuestigaret. Cum enim hos tres numeros 3. 4. 5. diligenter esset contemplatus, vidissetque quadratum numerum maioris æqualem esse quadratis numeris reliquorum, composuit triangulum scalenum, cuius maximum latus diuisum erat in 5. partes æquales, minimum in 3. eiusdem magnitudinis, & reliquum in 4. Quo facto, considerauit angulum sub his duobus lateribus contentum, inuenitque eum esse rectum. Idque in quamplurimis aliis numeris, ut in 6. 8. 10. & 9. 12. 15. &c. obseruauit. Quare inquirendum esse iudicauit, num in omni triangulo rectangulo quadratum lateris, quod recto angulo opponitur, reliquorum laterum quadratis æquale esset, quandoquidem omnia triangula, quorum latera habebant magnitudinem secundum dictos numeros, continebant unum angulum rectum: Atque itæ tandem mirabile hoc theorema maxima animi voluptate adinuenit, firmaque ratione demonstrauit. Quod tamen Euclides mirandum in modum amplificauit lib. 6. propos. 31. Ubi demonstrauit, non solum quadratum lateris, quod recto angulo opponitur, æquale esse quadratis reliquorum duorum laterum; Verum etiam figuram quamlibet rectilineam super latus recto angulo oppositum constructam, sine ea sit triangulum, sine quadrangulum, &c. æqualem esse duabus figuris, quæ super reliqua latera describuntur, dummodo priori sint similes, similiterque descriptæ, ut ibidem ostendemus.
CAETERVM quoniam mentionem fecimus trium numerorum, quorum maximi quadratum æquale est quadratis reliquorum, non abs re fuerit, paucis explicare, quonam pacto huiusmodi numeri inueniantur. Habitis igitur his tribus numeris 3. 4. 5. si duplicentur, habebuntur alii tres, 6. 8. 10. si iidem triplicentur, exurgent alii tres 9. 12. 15. & si quadruplicentur, inuenientur hi tres 12. 16. 20. Atque ita reperientur quotcunque alii, si primi illi tres per quemcunque multiplicentur numerum. Traduntur tamen a Proclo duæ regulæ, quibus inueniuntur prædicti numeri, nulla babita ratione illorum trium. Prima ascribatur Pythagoræ, & est huiusmodi. Sumatur pro minimo quicunque numerus impar, ut 5. ex quo ita alios reperies. Ex quadrato numeri accepti, ut hic ex 25. reiice unitatem. Nam reliqui numeri dimidium, videlicet 12. erit alter numerus, cui si addatur unitas, exurget tertius numerus 13. Huius igitur quadratum æquale est quadratis aliorum. Quod si numerus impar acceptus fuiet 3. essent reliqui duo inuenti per hanc regulam 4. & 5. Secunda regula tribuitur Platoni, quæ talis est. Accipiatur numerus quicunque par, nempe 6. Ex huius dimidii quadrato, nimirum 9. detrahe unum, eidemque adde unum, habebisque reliquos dues numeros 8. & 10. primus autem est 6. nimirum numerus par acceptus. Hac regula si accipiatur par 10. reperientur alii due 24. & 26.
COLLIGVNTVR ex celeberrimo hoc Pythagoræ inuento plurimæ scitu non iniucunda tam theoremata, quam problemata, e quibus visum est ea duntaxat in medium proferre, quæ utilitatem magnam rebus Geometricis allatura creduntur, initium hinc sumentes.

SI in quadrato quouis diameter ducatur, quadratum a diametro descriptum duplum erit prædicti quadrati.
IN quadrate A B C D, ducatur diameter A C. Dico quadratum diametri A C, duplum esse quadrati A B C D. Cum enim in triangulo A B C, angulus B, rectus sit, erit quadratum lateris A C, æquale duobus quadratis laterum A B, B C. Cum igitur quadrata linearum A B, B C, æqualia sint, quod lineæ A B, B C, sint æquales, erit quadratum diametri A C, duplum cuiuslibet illorum, ut quadrati lineæ A B, hoc est, quadrati A B C D. Quod est propositum.

QVADRATVM diametri figuræ altera parte longioris æquale est duobus quadratis laterum inæqualium.
IN altera parte longiori A B C D, ducatur diameter A C; & quia in triangulo A B C, angulus B, est rectus, erit quadratum lateris A C, æquale duobus quadratis laterum inæqualium A B, B C. Quod est propositum.

SI fuerint duo triangula rectangula, quorum latera rectis angulis opposita sint æqualia, erunt duo quadrata reliquorum duorum laterum unius trianguli æqualia duobus quadratis reliquorum duorum laterum alterius.
TRIANGVLORVM A B C, D E F, anguli A, & D, sint recti, lateraque opposita B C, E F, æqualia. Dico duo quadrata laterum A B, A C, simul sumpta æqualia esse duobus quadratis laterum D E, D F, simul sumptis. Nam quadratæ linearum B C, E F, æqualia inter se sunt, cum & ipsæ lineæ inter se ponantur æquales. Quadrato autem lineæ B C, æqualia sunt quadrata linearum A B, A C. Et quadrato lineæ E F, æqualia sunt quadrata linearum A B, A C; Et quadrato lineæ E F, æqualia sunt quadrata linearum D E, D F: Perspicuum ergo est, quod ponitur.

DVOBVS quadratis inæqualibus propositis, inuenite alia duo quadrata, quæ & æqualia sint inter se, & simul sumpta æqualia duobus inæqualibus propositis simul sumptis.
SINT A, & B, latera duorum quadratorum in æqualium; Fiat angulus rectus D C E, sitque D C, recta æqualis rectæ B, & recta C E, rectæ A. Ducta deinde recta D E, coniungente duo puncta D, E, constituantur super ipsam duo anguli semirecti D E F, E D F, coeantque rectæ D F, E F, in F. Quoniam igitur in triangulo F D E, ænguli F D E, F E D, æquales sunt; erunt & latera D F, E F, æqualia, ideoque & quadrata eorundem laterum æqualia. Dico iam, eadem quadrata linearum D F, E F, æqualia esse quadratis linearum A, & B, hoc est, quadratis linearum C E, & C D. Nam cum in triangulo D E F, anguli F D E, F E D, faciant unum rectum; erit reliquus angulus F, rectus. Quamobrem erunt quadrata linearum D F, E F, æqualia quadrato lineæ D E; Sed eidem quadrato lineæ D E, æqualia sunt quoque quadrata linearum C D, C E. Igitur quadrata linearum D F, E F, æqualia sunt quadratis linearum D C, E C. Quod est propositum.

PROPOSITIS duabus lineis inæqualibus, inuenire id, quo plus potest maior, quam minor.
POTENTIA lineæ rectæ dicitur eius quadratum. Tantum enim quæuis recta linea posse dicitur, quantum est eius quadratum. Sint ergo duæ lineæ inæquales A, & B, oporteatque cognoscere, quanto maius sit quadratum maioris lineæ A, quam minoris B. Ex quauis linearecta C D; sumatur C E, æqualis rectæ A, & E F, æqualis rectæ B. Deinde centro E, & interteruallo E C, semicirculus deseribatur C G D; & ex F, ducatur F G, perpendicularis ad C D. Dico quadratum rectæ A, hoc est rectæ C E, sibi æqualis, maius esse, quam quadratum rectæ B, hoc est, recta E F, sibi æqualis, quadrato rectæ F G. Ducta enim recta E G, erit eius quadratum æquale quadratis rectarum E F, F G, hoc est, quadratum rectæ E C, illi æquale, superabit quadratum rectæ E F, quadrato rectæ F G. Quod est propositum.

PROPOSITIS quotcunque quadratis, siue æqualibus, siue inæqualibus, inuenire quadratum omnibus illis æquale.
SINT latera quinque quadratorum A, B, C, D, E. oporteatque inuenire quadratum æquale omnibus illis quinque. Fiat angulus rectus F G H, sitque recta F G, æqualis rectæ A, & recta G H, rectæ B; Ducta deinde recta H F, fiat angulus rectus F H I, sitque H I, æqualis rectæ C. Ducta rursus recta I F, fiat angulus rectus F I K, sitque I K, æqualis rectæ D. Ducta denique recta K F, fiat angulus rectus F K L, sitque K L, æqualis rectæ E, ducaturque recta F L. Dico quidratum rectæ F L, æquale esse quinque quadratis propositis. Quadratum enim rectæ F H, æquale est quadratis rectarum F G, G H, hoc est, quadratis rectarum A, & B. Rursus quadratum rectæ F I, æquale est quadratis rectarum F H, H I, & idcirco quadratis rectarum A, B, & C. Item quadratum rectæ F K, æquale est quadratis rectarum F I, I K, idcoque quadratis rectarum A, B, C, & D Denique quadratum rectæ F L, æquale est quadratis rectarum F K, K L, ac propterea quadratis rectarum A, B, C, D, & E. Quod est propositum.

PROPOSITIS duobus quadratis quibuscunque, alteri illorum adiungere figuram, quæ reliquo quadrato sit æqualis, ita ut tota figura composita sit etiam quadrata.
SINT duo quadrata A B C D, E F G H, propositumque sit quadrato A B C D, apponere figuram, quæ sit æqualis quadrato E F G H, &c. Sumatur recta B I, æqualis rectæ F G, lateri quadrati E F G H, Ducta autem recta A I, & producta recta B A, ad partes A, accipiatur B K, æqualis rectæ A I, perficiaturque quadratum B K L M. Dico figuram A D C M L K, quadrato A B C D, adiunctam, æqualem esse quadrato E F G H. Quoniam enim quadratum rectæ A I, hoc est, quadratum B K L M, æquale est quadratis rectarum A B, B I, hoc est, quadratis A B C D, E F G H; Si auferatur commune quadratum A B C D, remanebit figura A D C M L K æqualis quadrato E F G H. Quod est propositum.

COGNITIS duobus lateribus quibuscunque trianguli rectanguli, in cognitionem reliquilateris peruenire.
SIT angulus A, rectus in triangulo A B C, sintque primum cognita latera A B, A C, circa angulum rectum, quorum A B, ponatur 6. palmorum, & A C, 8. Quoniam igitur quadrata rectarum A B, A C, nempe quadrati palmi 36. & 64. æqualia sunt quadrato rectæ B C; Si illa coniungantur simul, efficietur hoc quadratorum palmorum 100. Latus ergo B C, continebit 10. palmos. Tantum enim est latus, seu radix quadrata 100. palmorum, ut perspicuum est apud Arithmeticos. Sint deinde cognita latera A B, B C, sitque A B, 6. palmorum, & B C, 10. Quoniam igitur quadrata rectarum A B, A C, æqualia sunt quadrato rectæ B C; Si quadratum rectæ A B, quod continet palmos 36. detrabatur ex quadrato rectæ B C, quod est palmorum 100. remanebit quadratum rectæ A C, 64. palmorum. Latus ergo A C, continebit 8. palmos. Tan- ta enim est radix quadrata, seu radix 64. palmorum. Quod est propositum. Cæterum non semper hac arte inuenientur numeri rationales, quia non omnis numerus habet latus, radicemve quadratam, ut notum est apud Arithmeticos. Unde latus inuentum sæpe numero exprimi nequit, nisi per radicem surdam, quam vocant: Sed de his alias.

THEOREMATE porro hoc Pythagoreo multo uniuersalius est illud, quod a Pappo demonstratur in omni triangulo, siue illud rectangulum sit, siue non, & de quibuscunque parallelogrammis super latera trianguli constructis tam rectangulis, quam non rectangulis, etiamsi non sint inter se æquiangula. Quod nos in formam theorematis redigentes, clarius hoc modo proposuimus, & meo iudicio generalius adbuc, quam Pappus.
IN omni triangulo, parallelogramma quæcunque super duobus lateribus descripta, æqualia sunt parallelogrammo super reliquo latere constituto, cuius alterum latus æquale sit, & parallelum rectæ ductæ ab angulo, quem duo illa latera comprehendunt, ad punctum, in quo conueniunt latera parallelogrammorum lateribus trianguli opposita, si ad partes anguli illius producantur.
SIT triangulum quodcunque A B C, constituanturque super latera A B, A C, parallelogramma quæcunque A B D E, A C F G, quorum latera D E, F G, quæ lateribus A B, A C, assumptis in triangulo opponuntur, producta ad partes anguli A, dictis lateribus A B, A C, comprehensi, conueniant in H, ducaturque recta A H. Dico parallelogramma A D, A F, æqualia esse parallelogrammo super latus B C, descripto, cuius alterum latus æquale sit, & parallelum rectæ A H. Producta enim H A, secet B C, in L, & per B, C, agantur B I, C K, parallelæ ipsi A H, iungaturque rectæ I K. Quoniam igitur parallelogramma sunt B I, H A, C K H A; erit utraque B I, C K, ipsi A H, æqualis; atque adeo & inter se æquales erunt B I, C K; quæ cum sint etiam parallelæ, quod eidem A H, parallelæ sint; erunt quoque B C, I K, parallelæ, & æquales. Quare parallelogrammum est B C K I, super latus B C, habens alterum latus B I, rectæ A H, æquale, & parallelum: Cuiquidem æqualia ostendenda sunt parallelogramma A D, A F. Quia ergo æqualia sunt parallelogramma A D, A B I H, quod eandem habeant basin A B, in eisdemque sint parallelis A B, H D; Est autem A B I H, parallelogrammo I L, æquale, quod illud cum hoc etiam eandem babeat basin B I, in eisdemque sit parallelis B I, L H: Erit quoque A D, eidem I L, æquale. Non aliter ostendemus, A F, ipsi K L, esse æquale: Quare parallelogramma A D, A F, parallelogrammo B K, æqualia sunt. Quod est propositum.
PAPPVS construit figuram aliter. Nam sumit rectas A C, A E, & A B, A G, in directum positas, ita ut parallelogramma A D, A F, sint æquiangula, habentia angulos A E D, A G F angulo B A C, internos externo æquales, ceu hæc eius figura indicat. Sed nos uniuersalius rem proposuimus, ut manifestum est.
 
一增。凡直角方形之對角線上。作直角方形。倍大於元形。如甲乙丙丁直角方形之甲丙線上。作直角方形。倍大於甲乙丙丁形。

二增題。設不等兩直角方形。如一以甲為邊。一以乙為邊。求別作兩直角方形。自相等。而幷之、又與元設兩形幷、等。
法曰。先作丙戊線、與甲等。次作戊丙丁直角、而丙丁線、與乙等。次作戊丁線相聯末於丙丁戊角、丙戊丁角、各作一角。皆半於直角。己戊己丁、兩腰遇於己。公論十一而等。本篇六卽己戊、己丁、兩線上所作兩直角方形自相等。而幷之、又與丙戊、丙丁、上所作兩直角方形幷、等。
論曰。己丁戊、己戊丁、兩角。旣皆半於直角。則丁己戊為直角。本篇卅二而對直角之丁戊線上、所作直角方形。與兩腰線上、所作兩直角方形幷、等矣。本題己戊、與己丁、旣等。則其上所作兩直角方形、自相等矣。又丁戊線上、所作直角方形。與丙丁、丙戊、線上所作兩直角方形幷、旣等。則己戊、己丁、上兩直角方形幷。與丙戊、丙丁、上兩直角方形幷、亦等。

三增題。多直角方形。求幷作一直角方形。與之等。
法曰。如五直角方形。以甲、乙、丙、丁、戊、為邊。任等不等。求作一直角方形、與五形幷、等。先作己庚辛直角。而己庚線、與甲等。庚辛線、與乙等。次作己辛線。旋作己辛壬直角。而辛壬與丙等。次作己壬線。旋作己壬癸直角。而壬癸與丁等。次作己癸線。旋作己癸子直角。而癸子與戊等。末作己子線。題言己子線上、所作直角方形、卽所求。
論曰。己辛上。作直角方形。與甲、乙、兩形幷等。本題己壬上作直角方形。與己辛、及丙、兩形幷、等。餘倣此推顯。可至無窮。
四增。三邊直角形。以兩邊求第三邊長短之數。
法曰。甲乙丙角形甲為直角。先得甲乙、甲丙、兩邊長短之數。如甲乙六。甲丙八。 求乙丙邊長短之數。其甲乙、甲丙、上所作兩直角方形幷。旣與乙丙上所作直角方形等。本題則甲乙之冪、自乘之數曰冪得三十六。甲丙之冪、得六十四。幷之得百。而乙丙之冪亦百。百開方得十。卽乙丙數十也。又設先得甲乙乙丙。如甲乙六。乙丙十。而求甲丙之數。其甲乙、甲丙、上兩直角方形幷。旣與乙丙上直角方形等。則甲乙之冪、得三十六。乙丙之冪、得百。百減三十六。得甲丙之冪六十四。六十四開方得八。卽甲丙八也。求甲乙倣此。
此以開方盡實者為例。其不盡實者。自具算家分法。 

Proposition 48. 
THEOR. 34. PROPOS. 48. 
第四十八題 
If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right. 
SI quadratum, quod ab uno laterum trianguli describitur, æquale sit eis, quæ a reliquis trianguli lateribus describuntur, quadratis: Angulus comprehensus sub reliquis duobus trianguli lateribus, rectus est. 
凡三角形之一邊上、所作直角方形。與餘邊所作兩直角方形幷、等。則對一邊之角、必直角。 
For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC;  I say that the angle BAC is right. 
DETVR triangulum A B C, sitque quadratum lateris A C, æquale quadratis reliquorum laterum B A, B C.  Dico angulum A B C, esse rectum. 
For let AD be drawn from the point A at right angles to the straight line AC,  let AD be made equal to BA,  and let DC be joined.  Since DA is equal to AB, the square on DA is also equal to the square on AB.  Let the square on AC be added to each;  therefore the squares on DA, AC are equal to the squares on BA, AC.  But the square on DC is equal to the squares on DA, AC,  for the angle DAC is right; [I. 47]  and the square on BC is equal to the squares on BA, AC, for this is the hypothesis;  therefore the square on DC is equal to the square on BC,  so that the side DC is also equal to BC.  And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC;  and the base DC is equal to the base BC;  therefore the angle DAC is equal to the angle BAC. [I. 8]  But the angle DAC is right; therefore the angle BAC is also right. 
Ducatur namque B D, perpendicularis ad B A,  & æqualis rectæ B C,  connectaturque recta A D.  Quoniam igitur in triangulo A B D, angulus A B D, rectus est; erit quadratum rectæ A D, æquale quadratis rectarum B A, B D: Est autem quadratum rectæ B D, quadrato rectæ B C, æquale, ob linearum æqualitatem.      Quare quadratum rectæ A D, quadratis rectarum B A, B C, æquale erit.    Cum ergo quadratum rectæ A C, eisdem quadratis rectarum B A, B C, æquale ponatur;  erunt quadrata rectarum A D, A C, inter se æqualia,   ac propterea & rectæ ipsæ A D, A C, æquales.  Quoniam igitur latera B A, B D, trianguli A B D, æqualia sunt lateribus B A, B C, trianguli A B C;  & basis A D, oftensa est æqualis basi A C;  erunt anguli A B D, A B C, æquales:  Est autem angulus A B D, ex constructione rectus. Igitur & angulus A B C, rectus erit. 
Therefore etc.  Q. E. D. 
Si igitur quadratum, quod ab uno laterum trianguli describitur, &c.  Quod demonstrandum erat.

SCHOLION
CONVERSVM est autem theorema hoc præcedentis theorematis Pythagorici, ut perspicuum est.


FINIS ELEMENTI PRIMI. 
BOOK II. 
EVCLIDIS ELEMENTVM SECVNDVM 
幾何原本
利瑪竇口譯
徐光啟筆受幾何原本第二卷之首 
DEFINITIONS. 
DEFINITIONES. 
界說二則 
1 Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle. 
OMNE parallelogrammum rectangulum contineri dicitur sub rectis duabus lineis, quæ rectum comprehendunt angulum. 
第一界
凡直角形之兩邊、函一直角者。為直角形之矩線。
如甲乙、偕乙丙。函甲乙丙直角。得此兩邊。卽知直角形大小之度。今別作戊線、己線。與甲乙、乙丙、各等。亦卽知甲乙丙丁直角形大小之度。則戊、偕己、兩線。為直角形之矩線。
此例與算法通。如上圖。一邊得三。一邊得四。相乘得十二。則三、偕四、兩邊、為十二之矩數。凡直角諸形之內四角、皆直。故不必更言四邊、及平行線。止名為直角形。省文也。
凡直角諸形。不必全舉四角。止舉對角二字。卽指全形。如甲乙丙丁直角形。止舉甲丙、或乙丁。亦省文也。 
2 And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon. 
IN omni parallelogrammo spatio, vnumquodlibet eorum, quæ circa diametrum illius sunt, parallelogrammorum, cum duobus complementis, Gnomon vocetur. 
第二界
諸方形、有對角線者。其兩餘方形。任偕一角線方形。為罄折形。(p. 八四)
甲乙丙丁、方形。任直、斜角。作甲丙對角線。從庚點作戊己、辛壬、兩線。與方形邊平行。而分本形為四方形。其辛己、庚乙、兩形為餘方形。辛戊、己壬、兩形為角線方形。一卷界 \\ 說三六兩餘方形。任偕一角線方形。為罄折形。如辛己、庚乙、兩餘方形。偕己壬角線方形。同在癸子丑圜界內者。是癸子丑罄折形也。用辛戊角線方形、倣此。 
Proposition 1. 
THEOR. 1. PROPOS. 1. 
幾何原本第二卷本篇論線 計十四題
第一題 
If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments. 
SI fuerint duæ rectæ lineæ, seceturque ipsarum altera in quotcunque segmenta: Rectangulum comprehensum sub illis duabus rectis lineis, æquale est eis, quæ sub insecta, & quolibet segmentorum comprehenduntur, rectangulis. 
兩直線。任以一線、任分為若干分。其兩元線矩內直角形。與不分線、偕諸分線、矩內諸直角形幷、等。 
Let A, BC be two straight lines, and let BC be cut at random at the points D, E;  I say that the rectangle contained by A, BC is equal to the rectangle contained by A, BD, that contained by A, DE and that contained by A, EC. 
   
   
For let BF be drawn from B at right angles to BC; [I. 11]  let BG be made equal to A, [I. 3]  through G let GH be drawn parallel to BC, [I. 31]  and through D, E, C let DK, EL, CH be drawn parallel to BG. 
       
       
Then BH is equal to BK, DL, EH.  Now BH is the rectangle A, BC,  for it is contained by GB, BC, and BG is equal to A;  BK is the rectangle A, BD,  for it is contained by GB, BD, and BG is equal to A;  and DL is the rectangle A, DE,  for DK, that is BG [I. 34], is equal to A.  Similarly also EH is the rectangle A, EC.  Therefore the rectangle A, BC is equal to the rectangle A, BD, the rectangle A, DE and the rectangle A, EC. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
Proposition 2. 
THEOR. 2. PROPOS. 2. 
第二題 
If a straight line be cut at random, the rectangle contained by the whole and both of the segments is equal to the square on the whole. 
SI recta linea secta sit vtcunque: Rectangula, quæ sub tota, & quolibet segmentorum comprehenduntur, æqualia sunt ei, quod à tota sit, quadrato. 
一直線。任兩分之。其元線上直角方形。與元線偕兩分線、兩矩內直角形幷、等。 
For let the straight line AB be cut at random at the point C;  I say that the rectangle contained by AB, BC together with the rectangle contained by BA, AC is equal to the square on AB. 
   
   
For let the square ADEB be described on AB [I. 46],  and let CF be drawn through C parallel to either AD or BE. [I. 31] 
   
   
Then AE is equal to AF, CE.  Now AE is the square on AB;  AF is the rectangle contained by BA, AC,  for it is contained by DA, AC, and AD is equal to AB;  and CE is the rectangle AB, BC, for BE is equal to AB.  Therefore the rectangle BA, AC together with the rectangle AB, BC is equal to the square on AB. 
           
           
Therefore etc.  Q. E. D. 
   
   
Proposition 3. 
THEOR. 3. PROPOS. 3. 
第三題 
If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment. 
SI recta linea secta sit vtcunque: Rectangulum sub tota, & vno segmentorum comprehensum, æquale est & illi, quod sub segmentis comprehenditur, rectangulo, & illi, quod a prædicto segmento describitur, quadrato. 
一直線。任兩分之。其元線、任偕一分線、矩內直角形與分餘線、偕一分線、矩內直角形。及一分線上直角方形幷等。 
For let the straight line AB be cut at random at C;  I say that the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC. 
   
   
For let the square CDEB be described on CB; [I. 46]  let ED be drawn through to F, and through A let AF be drawn parallel to either CD or BE. [I. 31]  Then AE is equal to AD, CE.  Now AE is the rectangle contained by AB, BC,  for it is contained by AB, BE, and BE is equal to BC;  AD is the rectangle AC, CB, for DC is equal to CB;  and DB is the square on CB.  Therefore the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC. 
               
               
Therefore etc.  Q. E. D. 
   
   
Proposition 4. 
THEOR. 4. PROPOS. 4. 
第四題 
If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments. 
SI recta linea secta sit vtcunque: Quadratum, quod a tota describitur, æquale est & illis, quæ a segmentis describuntur, quadratis, & ei, quod bis sub segmentis comprehenditur, rectangulo. 
一直線。任兩分之。其元線上直角方形。與各分上兩直角方形、及兩分互偕、矩線內兩直角形幷、等。 
For let the straight line AB be cut at random at C;  I say that the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB. 
   
   
For let the square ADEB be described on AB, [I. 46]  let BD be joined; through C let CF be drawn parallel to either AD or EB, and through G let HK be drawn parallel to either AB or DE. [I. 31]  Then, since CF is parallel to AD, and BD has fallen on them, the exterior angle CGB is equal to the interior and opposite angle ADB. [I. 29]  But the angle ADB is equal to the angle ABD,  since the side BA is also equal to AD; [I. 5]  therefore the angle CGB is also equal to the angle GBC,  so that the side BC is also equal to the side CG. [I. 6]  But CB is equal to GK, and CG to KB; [I. 34]  therefore GK is also equal to KB;  therefore CGKB is equilateral.  I say next that it is also right-angled.  For, since CG is parallel to BK,  the angles KBC, GCB are equal to two right angles. [I. 29]  But the angle KBC is right;  therefore the angle BCG is also right,  so that the opposite angles CGK, GKB are also right. [I. 34]  Therefore CGKB is right-angled; and it was also proved equilateral;  therefore it is a square; and it is described on CB.  For the same reason HF is also a square;  and it is described on HG, that is AC. [I. 34]  Therefore the squares HF, KC are the squares on AC, CB.  Now, since AG is equal to GE, and AG is the rectangle AC, CB, for GC is equal to CB,  therefore GE is also equal to the rectangle AC, CB.  Therefore AG, GE are equal to twice the rectangle AC, CB.  But the squares HF, CK are also the squares on AC, CB;  therefore the four areas HF, CK, AG, GE are equal to the squares on AC, CB and twice the rectangle contained by AC, CB.  But HF, CK, AG, GE are the whole ADEB,  which is the square on AB.  Therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB. 
                                                         
                                                         
Therefore etc.  Q. E. D. 
   
   
 
 
 
Proposition 5. 
THEOR. 5. PROPOS. 5. 
第五題 
If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half. 
SI recta linea secetur in æqualia, & non æqualia: Rectangulum sub inæqualibus segmentis totius comprehensum, vna cum quadrato, quod ab intermedia sectionum, æquale est ei, quod a dimidia describitur, quadrato. 
直線兩平分之。又任兩分之。其任兩分線、矩內直角形、及分內線上直角方形、幷。與平分半線上直角方形等。 
For let a straight line AB be cut into equal segments at C and into unequal segments at D;  I say that the rectangle contained by AD, DB together with the square on CD is equal to the square on CB. 
   
   
For let the square CEFB be described on CB, [I. 46]  and let BE be joined;  through D let DG be drawn parallel to either CE or BF,  through H again let KM be drawn parallel to either AB or EF,  and again through A let AK be drawn parallel to either CL or BM. [I. 31]  Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each;  therefore the whole CM is equal to the whole DF.  But CM is equal to AL,  since AC is also equal to CB; [I. 36]  therefore AL is also equal to DF.  Let CH be added to each;  therefore the whole AH is equal to the gnomon NOP.  But AH is the rectangle AD, DB,  for DH is equal to DB,  therefore the gnomon NOP is also equal to the rectangle AD, DB.  Let LG, which is equal to the square on CD, be added to each;  therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD.  But the gnomon NOP and LG are the whole square CEFB, which is described on CB;  therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB. 
                                     
                                     
Therefore etc.  Q. E. D. 
   
   
Proposition 6. 
THEOR. 6. PROPOS. 6. 
第六題 
If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line. 
SI recta linea bifariam secetur, & illi recta quædam linea in rectum adiiciatur: Rectangulum comprehensum sub tota cum adiecta, & adiecta, vna cum quadrato a dimidia, æquale est quadrato a linea, quæ tum ex dimidia, tum ex adiecta componitur, tanquam ab vna, descripto. 
一直線。兩平分之。又任引增一直線。共為一全線。其全線、偕引增線、矩內直角形。及半元線上直角方形、幷。與半元線偕引增線、上直角方形等。 
For let a straight line AB be bisected at the point C, and let a straight line BD be added to it in a straight line;  I say that the rectangle contained by AD, DB together with the square on CB is equal to the square on CD. 
   
   
For let the square CEFD be described on CD, [I. 46] and let DE be joined;  through the point B let BG be drawn parallel to either EC or DF, through the point H let KM be drawn parallel to either AB or EF, and further through A let AK be drawn parallel to either CL or DM. [I. 31] 
   
   
Then, since AC is equal to CB, AL is also equal to CH. [I. 36]  But CH is equal to HF. [I. 43]  Therefore AL is also equal to HF.  Let CM be added to each;  therefore the whole AM is equal to the gnomon NOP.  But AM is the rectangle AD, DB,  for DM is equal to DB;  therefore the gnomon NOP is also equal to the rectangle AD, DB.  Let LG, which is equal to the square on BC, be added to each;  therefore the rectangle contained by AD, DB together with the square on CB is equal to the gnomon NOP and LG.  But the gnomon NOP and LG are the whole square CEFD, which is described on CD;  therefore the rectangle contained by AD, DB together with the square on CB is equal to the square on CD. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
Proposition 7. 
THEOR. 7. PROPOS. 7. 
第七題 
If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment. 
SI recta linea secetur vtcunque; Quod a tota, quodque ab vno segmentorum, vtraque simul quadrata, æqualia sunt & illi, quod bis sub tota, & dicto segmento comprehenditur, rectangulo, & illi, quod a reliquo segmento sit, quadrato. 
一直線。任兩分之。其元線上、及任用一分線上、兩直角方形、幷。與元線偕一分線、矩內直角形二、及分餘線上直角方形幷、等。 
For let a straight line AB be cut at random at the point C;  I say that the squares on AB, BC are equal to twice the rectangle contained by AB, BC and the square on CA. 
   
   
For let the square ADEB be described on AB, [I. 46]  and let the figure be drawn. 
   
   
Then, since AG is equal to GE, [I. 43] let CF be added to each;  therefore the whole AF is equal to the whole CE.  Therefore AF, CE are double of AF.  But AF, CE are the gnomon KLM and the square CF;  therefore the gnomon KLM and the square CF are double of AF.  But twice the rectangle AB, BC is also double of AF;  for BF is equal to BC;  therefore the gnomon KLM and the square CF are equal to twice the rectangle AB, BC.  Let DG, which is the square on AC, be added to each;  therefore the gnomon KLM and the squares BG, GD are equal to twice the rectangle contained by AB, BC and the square on AC.  But the gnomon KLM and the squares BG, GD are the whole ADEB and CF,  which are squares described on AB, BC;  therefore the squares on AB, BC are equal to twice the rectangle contained by AB, BC together with the square on AC. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
Proposition 8. 
THEOR. 8. PROPOS. 8. 
第八題 
If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line. 
SI recta linea secetur vtcunque: Rectangulum qua- ter comprehensum sub tota, & vno segmentorum, cum eo, quod à reliquo segmento fit, quadrato, æquale est ei, quod a tota, & dicto segmento, tanquam ab vna linea describitur, quadrato. 
一直線。任兩分之。其元線偕初分線、矩內直角形四、及分餘線上直角方形、幷。與元線偕初分線上直角方形、等。 
For let a straight line AB be cut at random at the point C;  I say that four times the rectangle contained by AB, BC together with the square on AC is equal to the square described on AB, BC as on one straight line. 
   
   
For let [the straight line] BD be produced in a straight line [with AB], and let BD be made equal to CB;  let the square AEFD be described on AD, and let the figure be drawn double. 
   
   
Then, since CB is equal to BD, while CB is equal to GK, and BD to KN, therefore GK is also equal to KN.  For the same reason QR is also equal to RP.  And, since BC is equal to BD, and GK to KN, therefore CK is also equal to KD, and GR to RN. [I. 36]  But CK is equal to RN, for they are complements of the parallelogram CP; [I. 43]  therefore KD is also equal to GR;  therefore the four areas DK, CK, GR, RN are equal to one another.  Therefore the four are quadruple of CK.  Again, since CB is equal to BD, while BD is equal to BK, that is CG, and CB is equal to GK, that is GQ, therefore CG is also equal to GQ.  And, since CG is equal to GQ, and QR to RP,  AG is also equal to MQ, and QL to RF. [I. 36]  But MQ is equal to QL, for they are complements of the parallelogram ML; [I. 43]  therefore AG is also equal to RF;  therefore the four areas AG, MQ, QL, RF are equal to one another.  Therefore the four are quadruple of AG.  But the four areas CK, KD, GR, RN were proved to be quadruple of CK;  therefore the eight areas, which contain the gnomon STU, are quadruple of AK.  Now, since AK is the rectangle AB, BD, for BK is equal to BD,  therefore four times the rectangle AB, BD is quadruple of AK.  But the gnomon STU was also proved to be quadruple of AK;  therefore four times the rectangle AB, BD is equal to the gnomon STU.  Let OH, which is equal to the square on AC, be added to each;  therefore four times the rectangle AB, BD together with the square on AC is equal to the gnomon STU and OH.  But the gnomon STU and OH are the whole square AEFD, which is described on AD,  therefore four times the rectangle AB, BD together with the square on AC is equal to the square on AD.  But BD is equal to BC;  therefore four times the rectangle contained by AB, BC together with the square on AC is equal to the square on AD,  that is to the square described on AB and BC as on one straight line. 
                                                     
                                                     
Therefore etc.  Q. E. D. 
   
   
Proposition 9. 
THEOR. 9. PROPOS. 9. 
第九題 
If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section. 
SI recta linea secetur in æqualia, & non æqualia: Quadrata, quæ ab inæqualibus totius segmentis fiunt, simul duplicia sunt & eius, quod à dimidia, & eius, quod ab intermedia sectionum fit, quadrati. 
一直線。兩平分之。又任兩分之。任分線上、兩直角方形幷。倍大於平分半線上、及分內線上、兩直角方形幷。 
For let a straight line AB be cut into equal segments at C, and into unequal segments at D;  I say that the squares on AD, DB are double of the squares on AC, CD. 
   
   
For let CE be drawn from C at right angles to AB,  and let it be made equal to either AC or CB;  let EA, EB be joined,  let DF be drawn through D parallel to EC,  and FG through F parallel to AB, and let AF be joined.  Then, since AC is equal to CE,  the angle EAC is also equal to the angle AEC.  And, since the angle at C is right,  the remaining angles EAC, AEC are equal to one right angle. [I. 32]  And they are equal;  therefore each of the angles CEA, CAE is half a right angle.  For the same reason each of the angles CEB, EBC is also half a right angle;  therefore the whole angle AEB is right.  And, since the angle GEF is half a right angle, and the angle EGF is right, for it is equal to the interior and opposite angle ECB, [I. 29]  the remaining angle EFG is half a right angle; [I. 32]  therefore the angle GEF is equal to the angle EFG,  so that the side EG is also equal to GF. [I. 6]  Again, since the angle at B is half a right angle, and the angle FDB is right,  for it is again equal to the interior and opposite angle ECB, [I. 29]  the remaining angle BFD is half a right angle; [I. 32]  therefore the angle at B is equal to the angle DFB,  so that the side FD is also equal to the side DB. [I. 6]  Now, since AC is equal to CE, the square on AC is also equal to the square on CE;  therefore the squares on AC, CE are double of the square on AC.  But the square on EA is equal to the squares on AC, CE, for the angle ACE is right; [I. 47]  therefore the square on EA is double of the square on AC.  Again, since EG is equal to GF, the square on EG is also equal to the square on GF;  therefore the squares on EG, GF are double of the square on GF.  But the square on EF is equal to the squares on EG, GF;  therefore the square on EF is double of the square on GF.  But GF is equal to CD; [I. 34]  therefore the square on EF is double of the square on CD.  But the square on EA is also double of the square on AC;  therefore the squares on AE, EF are double of the squares on AC, CD.  And the square on AF is equal to the squares on AE, EF, for the angle AEF is right; [I. 47]  therefore the square on AF is double of the squares on AC, CD.  But the squares on AD, DF are equal to the square on AF, for the angle at D is right; [I. 47]  therefore the squares on AD, DF are double of the squares on AC, CD.  And DF is equal to DB;  therefore the squares on AD, DB are double of the squares on AC, CD. 
                                                                               
                                                                               
Therefore etc.  Q. E. D. 
   
   
Proposition 10. 
THEOR. 10. PROPOS. 10. 
第十題 
If a straight line be bisected, and a straight line be added to it in a straight line, the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line. 
SI recta linea secetur bifariam, adijciatur autem ei in rectum quæpiam recta linea: Quod a tota cum adiuncta, & quod ab adiuncta, vtraque simul quadrata, duplicia sunt & eius, quod a dimidia, & eius, quod a composita ex dimidia & adiuncta, tanquam ab vna, descriptum sit, quadrati. 
一直線。兩平分之。又任引增一線。共為一全線。其全線上、及引增線上、兩直角方形、幷。倍大於平分半線(p. 一○三)幾何原本 卷二上、及分餘半線偕引增線上、兩直角方形、幷。 
For let a straight line AB be bisected at C, and let a straight line BD be added to it in a straight line;  I say that the squares on AD, DB are double of the squares on AC, CD. 
   
   
For let CE be drawn from the point C at right angles to AB [I. 11],  and let it be made equal to either AC or CB [I. 3];  let EA, EB be joined;  through E let EF be drawn parallel to AD, and through D let FD be drawn parallel to CE. [I. 31]  Then, since a straight line EF falls on the parallel straight lines EC, FD,  the angles CEF, EFD are equal to two right angles; [I. 29]  therefore the angles FEB, EFD are less than two right angles.  But straight lines produced from angles less than two right angles meet; [I. Post. 5]  therefore EB, FD, if produced in the direction B, D, will meet.  Let them be produced and meet at G, and let AG be joined.  Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC; [I. 5]  and the angle at C is right;  therefore each of the angles EAC, AEC is half a right angle. [I. 32]  For the same reason each of the angles CEB, EBC is also half a right angle;  therefore the angle AEB is right.  And, since the angle EBC is half a right angle, the angle DBG is also half a right angle. [I. 15]  But the angle BDG is also right,  for it is equal to the angle DCE, they being alternate; [I. 29]  therefore the remaining angle DGB is half a right angle; [I. 32]  therefore the angle DGB is equal to the angle DBG,  so that the side BD is also equal to the side GD. [I. 6]  Again, since the angle EGF is half a right angle, and the angle at F is right,  for it is equal to the opposite angle, the angle at C, [I. 34]  the remaining angle FEG is half a right angle; [I. 32]  therefore the angle EGF is equal to the angle FEG, so that the side GF is also equal to the side EF. [I. 6]  Now, since the square on EC is equal to the square on CA,  the squares on EC, CA are double of the square on CA.  But the square on EA is equal to the squares on EC, CA; [I. 47]  therefore the square on EA is double of the square on AC. [C. N. 1]  Again, since FG is equal to EF, the square on FG is also equal to the square on FE;  therefore the squares on GF, FE are double of the square on EF.  But the square on EG is equal to the squares on GF, FE; [I. 47]  therefore the square on EG is double of the square on EF.  And EF is equal to CD; [I. 34]  therefore the square on EG is double of the square on CD.  But the square on EA was also proved double of the square on AC;  therefore the squares on AE, EG are double of the squares on AC, CD.  And the square on AG is equal to the squares on AE, EG; [I. 47]  therefore the square on AG is double of the squares on AC, CD.  But the squares on AD, DG are equal to the square on AG; [I. 47]  therefore the squares on AD, DG are double of the squares on AC, CD.  And DG is equal to DB;  therefore the squares on AD, DB are double of the squares on AC, CD. 
                                                                                     
                                                                                     
Therefore etc.  Q. E. D. 
   
   
Proposition 11. 
PROBL. 1. PROPOS. 11. 
第十一題 
To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment. 
DATAM rectam lineam secare, vt comprehensum sub tota, & altero segmentorum rectangulum, æquale sit ei, quod a reliquo segmento fit, quadrato. 
一直線。求兩分之。而元線偕初分線矩內直角形。與分餘線上直角方形、等。 
Let AB be the given straight line;  thus it is required to cut AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment. 
   
   
For let the square ABDC be described on AB; [I. 46]  let AC be bisected at the point E,  and let BE be joined;  let CA be drawn through to F,  and let EF be made equal to BE;  let the square FH be described on AF,  and let GH be drawn through to K.  I say that AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on AH. 
               
               
For, since the straight line AC has been bisected at E, and FA is added to it,  the rectangle contained by CF, FA together with the square on AE is equal to the square on EF. [II. 6]  But EF is equal to EB;  therefore the rectangle CF, FA together with the square on AE is equal to the square on EB.  But the squares on BA, AE are equal to the square on EB, for the angle at A is right; [I. 47]  therefore the rectangle CF, FA together with the square on AE is equal to the squares on BA, AE.  Let the square on AE be subtracted from each;  therefore the rectangle CF, FA which remains is equal to the square on AB.  Now the rectangle CF, FA is FK, for AF is equal to FG;  and the square on AB is AD;  therefore FK is equal to AD.  Let AK be subtracted from each;  therefore FH which remains is equal to HD.  And HD is the rectangle AB, BH,  for AB is equal to BD;  and FH is the square on AH;  therefore the rectangle contained by AB, BH is equal to the square on HA. 
                                 
                                 
Τherefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA.  Q. E. F. 
   
   
Proposition 12. 
THEOR. 11. PROPOS. 12. 
第十二題 
In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. 
IN amblygoniis triangulis, quadratum, quod fit a latere angulum obtusum subtendente, maius est quadratis, quæ fiunt a lateribus obtusum angulum comprehendentibus, rectangulo bis comprehenso & ab vno laterum, quæ sunt circa obtusum angulum, in quod, cum protractum fuerit, cadit perpendicularis, & ab assumpta exterius linea sub perpendiculari prope angulum obtusum. 
三邊鈍角形之對鈍角邊上直角方形。大於餘邊上兩直角方形幷之較。為鈍角旁任用一邊、偕其引增線之與對角所下垂線相遇者、矩內直角形。二。 
Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced;  I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD. 
   
   
For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4]  Let the square on DB be added to each;  therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD.  But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47]  and the square on AB is equal to the squares on AD, DB; [I. 47]  therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD;  so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD. 
             
             
Therefore etc.  Q. E. D. 
   
   
Proposition 13. 
THEOR. 12. PROPOS. 13. 
第十三題 
In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle. 
IN oxygoniis triangulis, quadratum a latere angulum acutum subtendente minus est quadratis, quae fiunt à lateribus acutum angulum comprehendentibus, rectangulo bis comprehenso, & ab vno laterum, quæ sunt circa acutum angulum, in quod perpendicularis cadit, & ab assumpta interius linea sub perpendiculari prope acutum augulum. 
三邊銳角形之對銳角邊上直角方形。小於餘邊上兩直角方形幷、之較。為銳角旁任用一邊、偕其對角所下垂線旁之近銳角分線、矩內直角形、二。 
Let ABC be an acute-angled triangle having the angle at B acute,  and let AD be drawn from the point A perpendicular to BC;  I say that the square on AC is less than the squares on CB, BA by twice the rectangle contained by CB, BD. 
     
     
For, since the straight line CB has been cut at random at D,  the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7]  Let the square on DA be added to each;  therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC.  But the square on AB is equal to the squares on BD, DA,  for the angle at D is right; [I. 47]  and the square on AC is equal to the squares on AD, DC;  therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD,  so that the square on AC alone is less than the squares on CB, BA by twice the rectangle contained by CB, BD. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
Proposition 14. 
PROBL. 2. PROPOS. 14. 
第十四題 
To construct a square equal to a given rectilineal figure. 
DATO rectilineo, æquale quadratum constituere. 
有直線形。求作直角方形。與之等。 
Let A be the given rectilineal figure;  thus it is required to construct a square equal to the rectilineal figure A. 
   
   
For let there be constructed the rectangular parallelogram BD equal to the rectilineal figure A. [I. 45]  Then, if BE is equal to ED, that which was enjoined will have been done;  for a square BD has been constructed equal to the rectilineal figure A.  But, if not, one of the straight lines BE, ED is greater.  Let BE be greater, and let it be produced to F;  let EF be made equal to ED, and let BF be bisected at G.  With centre G and distance one of the straight lines GB, GF let the semicircle BHF be described;  let DE be produced to H, and let GH be joined. 
               
               
Then, since the straight line BF has been cut into equal segments at G, and into unequal segments at E, the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5]  But GF is equal to GH;  therefore the rectangle BE, EF together with the square on GE is equal to the square on GH.  But the squares on HE, EG are equal to the square on GH; [I. 47]  therefore the rectangle BE, EF together with the square on GE is equal to the squares on HE, EG.  Let the square on GE be subtracted from each;  therefore the rectangle contained by BE, EF which remains is equal to the square on EH.  But the rectangle BE, EF is BD,  for EF is equal to ED;  therefore the parallelogram BD is equal to the square on HE.  And BD is equal to the rectilineal figure A.  Therefore the rectilineal figure A is also equal to the square which can be described on EH. 
                       
                       
Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilineal figure A.  Q. E. F. 
   
   
BOOK III. 
EVCLIDIS ELEMENTVM III. 
幾何原本第三卷之首 
DEFINITIONS. 
DEFINITIONES. 
界說十則 
1. Equal circles are those the diameters of which are equal, or the radii of which are equal. 
I. AEQUALES circuli sunt, quorum diametri sunt aequales; vel quorum, quae ex centris, rectae lineae sunt aequales.
QUONIAM Euclides hoc 3. liber varias circuli proprietates demonstrat, idcirco explicat prius terminos quosdam, quorum frequens futurus est usus in hoc lib. Primo itaque docet, eos circulos esse aequales, quorum diametri, vel semidiametri aequales sunt. Cum enim circulus describatur ex circumvolutione semidiametri circa alterum extremum fixum, et immobile, ceu in 1. lib. diximus, perspicuum est, eos circulos esse aequales, quorum semidiametri, seu rectae ex centris ductae, sunt aequales; vel etiam quorum totae diametri aequales sunt. Ut si diametri AB, BC, vel rectae DF, EG, e centris D, et E, ductae sint aequales, aequales erunt circuli AFB, et BGC. Sic etiam si circuli sint aequales, erunt diametri, vel rectae e centris ductae, aequales. Ex his liquet, circulos, quorum diametri, vel rectae ductae ex centris sunt inaequales, inaequales esse; atque adeo illum, cuius diameter, vel semidiameter maior, maiorem, etc. 
第一界
凡圜之徑線等。或從心至圜界線等。為等圜。
三卷將論圜、之情。故先為圜界說。此解圜之等者。如上圖、甲乙、乙丙、兩徑等。或丁己、戊庚、從心至圜界等。卽甲己乙、乙庚丙、兩圜等、若下圖、甲乙、乙丙、兩徑不等。或丁己、乙庚、從心至圜界不等。則兩圜亦不等矣。 
2. A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle. 
II. RECTA linea circulum tangere dicitur, quae cum circulum tangat, si producatur, circulum non secat.
UT recta AB, si ita circulum BFD, tangat in B, ut producta ad C, nulla ratione circulum secet, sed tota iaceat extra ipsum, dicetur tangere circulum. At vero recta EF, quia ita eundem circulum tangit in F, ut producta ad G, secet circulum, cadatque intra ipsum; non dicetur circulum tangere, sed secare. 
第二界
凡直線、切圜界過之而不與界交。為切線。
甲乙線、切乙己丁圜之界。乙又引長之至丙、而不與界交。其甲丙線、全在圜外。為切線。若戊己線、先切圜界。而引之至庚。入圜內。則交線也。 
3. Circles are said to touch one another which, meeting one another, do not cut one another. 
III. CIRCVLI sese mutuo tangere dicuntur, qui sese mutuo tangentes, sese mutuo non secant.
EODEM modo duo circuli AC, BC, mutuo dicuntur tangere in C, quia ita sese contingant in C, ut neuter alterum secet. Est autem hic contactus circulorum duplex. Aut enim exterius sese circuli tangunt, ut quando unus extra alterum est positus; aut interius; quando unus intra alterum constituitur. Quod si duo circuli ita se mutuo tangant, ut unus alterum quoque secet, dicentur circuli illi se mutuo secare, et non tangere. 
第三界
凡兩圜相切、而不相交。為切圜。
甲、乙、兩圜。不相交、而相切於丙。或切於外。如第一圖。或切於內。如第三圖。其第二、第四圖、則交圜也。 
4. In a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal. 
IIII. IN circulo aequaliter distare à centro rectae lineae dicuntur, cum perpendiculares, quae à centro in ipsas ducuntur, sunt aequales. 
第四界
凡圜內直線。從心下垂線。其垂線大小之度。卽直線距心遠近之度。 
5. And that straight line is said to be at a greater distance on which the greater perpendicular falls. 
Longius autem abesse illa dicitur, in quam maior perpendicularis cadit.
QUONIAM inter omnes lineas rectas, quae ab aliquo puncto ad quamlibet lineam rectam ducuntur, brevissima est perpendicularis, et semper eadem; aliae vero infinitis modis variari possunt; recte distantia illius puncti a linea accipitur penes lineam perpendicularem. Ut distantia puncti A, a recta BC, dicitur esse perpendicularis AD, non autem AE, vel AF, vel alia quaevis quae non perpendicularis est; quia AD, omnibus est brevior,88 quod angulus AED, vel AFD, minor sit angulo recto ADE. Immo non solum AE, AF, maiores sunt, quam AD, sed etiam ipsae inter se inaequales sunt. Est enim AF, maior, quam AE,89 cum angulus AEF, sit obtusus, et AFE, acutus, et sic de aliis lineis non perpendicularibus. Hinc factum est, ut Euclides aequalem distantiam rectarum in circulo ab ipsius centro definierit per aequales perpendiculares, et inaequalem distantiam per inaequales. Ut duae rectae AB, CD, in circulo ABCD, aequaliter dicentur distaere a centro E, si perpendiculares EF, EG, aequales fuerint. At linea CD, longius abesse dicetur a centro E, quam linea HI, si perpendicularis EG, maior fuerit perpendiculari EK. 
凡一點、至一直線上。惟垂線至近。其他卽遠。垂線一而己。遠者無數也。故欲知點與線相去遠近。必用垂線為度。試如前圖、甲點與乙丙線、相去遠近。必以甲丁垂線為度。為甲丁一線。獨去直線至近。他若甲戊、甲己、諸線。愈大愈遠。乃至無數。故如後圖、設甲乙丙丁圜內之甲乙、丙丁、兩線。其去戊心遠近等。為己戊、庚戊、兩垂線等故。若辛壬線。去戊心近矣。為戊癸垂線小故。 
6. A segment of a circle is the figure contained by a straight line and a circumference of a circle. 
V. SEGMENTVM circuli est figura, quae sub recta linea, et circuli peripheria comprehenditur.
UT si ducatur in ctrculo ABCD, recta BD, utcunque, dicetur tam figura ABD, contenta circumferentia BAD, et recta BD; quam figura BCD, comprehensa recta BD, et circumferentia BCD, circuli segmentum. Ex his colligitur triplex circuli segmentu, semicirculus, quando recta BD, per centrum E, incedit; segmentum semicirculo maius, quandorecta BD, non transit per centrum, in ipso tamen centrum existit, quale est segmentum BAD; Et segmentum semicirculo minus, extra quod centrum circuli constituitur, cuiusmods est segmentum BCD. Vocatur a plerisque Geometris recta BD, chorda, et circumferentia BAD, vel BCD, arcus. 
第五界
凡直線割圜之形。為圜分。
甲乙丙丁圜之乙丁直線。任割圜之一分。如甲乙丁、及乙丙丁、兩形。皆為圜分。凡分有三形。其過心者、為半圜分。函心者、為圜大分。不函心者、為圜小分。又割圜之直線、為絃。所割圜界之一分、為弧。 
7. An angle of a segment is that contained by a straight line and a circumference of a circle. 
VI. SEGMENTI autem angulus est, qui sub recta linea, et circuli peripheria comprehenditur.
DEFINIT iam Euclides tria genera angulorum, qui in circulis consideranmur. Primo loco angulum segmenti, dicens angulum mixtum BAC, vel BCA, contentum subrecta linea AC, et circumferentia ABC, appellari angulum segmenti. Quod si segmentis circuli fuerit semicirculus, dicetur angulus semicirculi. Si vero segmentum maius semicirculo extiterit, vocabitur angulus segmenti maioris: Si denique segmentum minus fuerit semicirculo, angulus segmenti minoris nuncupabitur. 
第六界
凡圜界偕直線內角。為圜分角。
以下三界。論圜角三種。本界所言。雜圜也。其在半圜分內、為半圜角。在大分內、為大分角。在小分內、為小分角。 
8. An angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the extremities of the straight line which is the base of the segment, is contained by the straight lines so joined. 
VII. IN segmento autem angulus est, cum in segmenti peripheria sumptum fuerit quodpiam punctum, et ab illo in termines rectae eius lineae, quae segmenti basis est, adiunctae fuerint rectae lineae: Is, inquam, angulus ab adiunctis illis lineis comprehensus.
SIT segmentum circuli quodcunque ABC, cuius basis recta AC: Ex suscepto quolibet puncto B, in circumferentia, ducantur ad puncta A, et C, extrema basis, rectae linea BA, BC. Angulus igitur rectilinsus ABC, dicitur existere in segmento ABC. 
第七界
凡圜界。任於一點、出兩直線。作一角。為負圜分角。
甲乙丙圜分。甲丙為底。於乙點出兩直線。作甲乙丙角形。其甲乙丙角。為負甲乙丙圜分角。 
9. And, when the straight lines containing the angle cut off a circumference, the angle is said to stand upon that circumference. 
VIII. CVM vero comprehendentes angulum rectae lineae aliquam assumunt peripheriam, illi angulus insistere dicitur.
EX puncto A, quolibet suscepto in circumferentia circuli ABCD, ducantur rectae duae lineae AB, AD, ad duo extrema B, et D, circumferentiae BCD, cuiusque, quam quidem duae rectae AB, AD, assumunt. Angulus itaque rectilineus BAD, insistere dicitur circumferentiae BCD. Perspicuum autem est, hunc angulum a praecedenti non differre, nisi voce tenus. Idem enim angulus rectilineus BAD, iuxta praecedentem quidem definitionem dicitur esse in segmento BAD, si recta BD, basis duceretur; ex hac vero insistere circumferentiae BCD. Non tamen confundendus est angulus in segmento aliquo, cum angulo, qui circumferentiae insistit, quamvis unus et idem sit; ad diversa siquidem referuntur. Angulus enim in segmento, segmentum, in quo existit, angulus autem insistens circumferentiae, circumferentiam, que basis est ipsius anguli, respicit. Unde si sumatur segmentum alsquod circuli BCD, in circulo ABCD, non erit idem angulus in hoc segmento existens, et eius circumferentiae insistens. Angulus enim in eo existens, erit BCD; at eius circumferentiae CBD, insistens, erit BAD, qui multum ab eo differt. Qua in re mirum in modum ballucinati sunt Orontius, Peletarius, et alii interpretes nonnulli. Quod autem angulus in segmento, et angulus circumferentiae cuipiam insistens, ad diversos arcus referantur, luce clarius patebit ex ultima propos. lib. 6. quae solum convenire potest circumferentiis circulorum, quibus anguli insistunt, non autem, in quibus existunt, ut eo in loco ostendemus. Idem quoque facile constat ex verbo graeco βεβηκέναι, quod ascendisse significat. Ascendit enim angulus DAB, supra circumferentiam BCD.
PRÆTER tres dictos angulos consideratur etiam a Geometris angulus contingentiae, qui continetur linea recta tangente circulum, et circumferentia circuli; vel certe duabus circumferentiis se mutuo tangentibus, sive hoc exterius fiat, sive interius. Exemplum. Si recta AB, tangat circulum CDE, in C; angulus mixtus ACD, vel CBE, dicetur angulus contingentiae, sive contactus: Rursus, si circulus CED, tangat circulum EFG, exterius in E; Item circulus HFI, circulum EFG, interius in F; appellabitur tam angulus curvilineus CEF, quam EFH, vel GFI, angulus contactus, seu contingentiae. Sunt itaque, ut vides, tres anguli contingentiae, unus quidem mixtus, reliqui vero duo curvilinei.
 
第八界
若兩直線之角。乘圜之一分。為乘圜分角。
甲乙丙丁圜內。於甲點出甲乙、甲丁、兩線。其乙甲丁角。為乘乙丙丁圜分角。圜角三種之外。又有一種。為切邊角。或直線切圜。或兩圜相切。其兩圜相切者。又或內或外如下圖甲乙線。切丙丁戊圜於丙。卽甲丙丁、乙丙戊、兩角為切邊角。又丙丁戊、己戊庚、兩圜。外相切於戊。及己戊庚、己辛壬、兩圜。內相切於己。卽丙戊己、戊己辛、壬己庚、三角。俱為切邊角。 
10. A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them. 
IX. SECTOR autem circuli est, cum ad ipsius circuli centrum constitutus fuerit angulus, comprehensa nimirum figura et a rectis lineis angulum continentibus, et a peripheria ab illis assumpta.
SI in circulo ABCD, cuius centrum E, rectae AE, CE, constituant angulum AEC, ad centrum E; nominabitur figura AECD, contenta rectis AE, EC, et circumferentia ADC, quam praedictae lineae assumunt, sector circuli. Ex hoc autem perspicue etiam colligitur, angulum, quidefinitione 8. explicatur, referri ad circumferentiam, quae ipsius basis est, non autem ad eam, in qua existit, ut multi interpretes existimarunt. Nam sicut in hac definitione Euclides intelligit circumferentiam ADC, quae basis est anguli ad centrum constituti, quando mentionem facit peripheriae a rectis AE, CE, assumptae: Ita quoque in illa intellexisse eum necesse est nomine peripheriae, quam rectae lineae assumunt, eam, quae basis est anguli ad circumferentiam constituti; quandoquidem in utraque definitione usus est eodem verbo graeco ἀπολαμβάνω. 
第九界
凡從圜心、以兩直線作角。偕圜界、作三角形。為分圜形。甲乙丙丁圜從戊心出戊甲、戊丙兩線、偕甲丁丙圜界、作角形。為分圜形。 
11. Similar segments of circles are those which admit equal angles, or in which the angles are equal to one another. 
X. SIMILIA circuli segmenta sunt, quae angulos capiunt aequales: Aut in quibus anguli inter se sunt aequales.
SEGMENTA, seu ABDF, DCAE, quae capiunt hos duos angulos ABF, DCE, aequales: vel, quod idem est, in quibus idem anguli aequales existunt, iuxta 7. definitionem, similes dicuntur.
EODEM modo segmenta diversorum circulorum tam aequalium, quam inaequalium, a Geometris dicuntur similia, quae vel suscipiunt aequales angulos; vet in quibus aequales anguli existunt. Ut si in circulis ABC, DEF, GHI, fuerint aequales, dicentur segmenta, seu circumferentiae ABC, DEF, GHI, quae dictos angulos suscipiunt, vel in quibus praedicti anguli existunt, similes. Consistit autem haec segmentorum, circumferentiarumve similitudo in eo, quod qualis pars est una circumferentia totius suae circumferentiae, talis quoque sit alter a circumferentia, quae dicitur huic similis, totius suae ctrcumferentiae ita ut qualis, et quanta pars est circumferentiae ABC, totius circumferentiae ABCA, talis et tanta quoque pars sit circumferentia DEF, totius circumferentiae DEFD; Item talis, et tanta circumferentia GHI, totius circumferentiae GHIG. Vel potius segmentorum similitudo in hoc consistit, quod segmenta, seu circumforentiae similes, ad totas circumferentias suas eandem habeant proportionem. Quod autem segmenta, quae vel aequales suscipiunt angulos, vel in quibus existunt aequales anguli, sint huiusmodi, demonstrabimus propositione ultima liber 6. Nunc satis sit, talia segmenta circulorum, vel etiam arcus, circumferentiasve, appellari similes.
 
第十界
凡圜內兩負圜分角、相等。卽所負之圜分相似。
甲乙丙丁圜內。有甲乙己、與丁丙戊、兩負圜分角等。則所負甲乙丁己、與丁丙甲戊、兩圜分相似。又有兩圜或等、或不等。其負圜分角等。卽圜分俱相似。如下三圖、三圜之甲乙丙、丁戊己、庚辛壬、三負圜分角等、卽所負甲乙丙、丁戊己、庚辛壬、三圜分相似相似者。如云同為幾分圜之幾也。 
BOOK III. PROPOSITIONS.
PROPOSITION 1. 
PROBL. 1. PROPOS. 1. 
幾何原本第三卷本篇論圜 計三十七題
第一題 
To find the centre of a given circle. 
DATI circuli centrum reperire. 
有圜。求尋其心。 
Let ABC be the given circle;  thus it is required to find the centre of the circle ABC. 
SIT circulus datus ABCD,  cuius centrum oportet invenire. 
法曰、甲乙丙丁圜。  求尋其心。 
Let a straight line AB be drawn through it at random, and let it be bisected at the point D;  from D let DC be drawn at right angles to AB and let it be drawn through to E;  let CE be bisected at F;  I say that F is the centre of the circle ABC. 
Ducatur in eo linea utcunque AC, quae bifariam dividatur in E,  et per E, ad AC, perpendicularis agatur BD.  Hac igitur bifariam secta in F;  dico F, esse centrum circuli propositi. 
先於圜之兩界、任作一甲丙直線。次兩平分之於戊。一卷十  次於戊上、作乙丁垂線。  兩平分之於己。  卽己為圜心。 
For suppose it is not, but, if possible, let G be the centre, and let GA, GD, GB be joined.  Then, since AD is equal to DB, and DG is common, the two sides AD, DG are equal to the two sides BD, DG respectively;  and the base GA is equal to the base GB, for they are radii;  therefore the angle ADG is equal to the angle GDB. [I. 8]  But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]  therefore the angle GDB is right.  But the angle FDB is also right;  therefore the angle FDB is equal to the angle GDB, the greater to the less: which is impossible.  Therefore G is not the centre of the circle ABC. 
In ipsa enim recta BD, aliud punctum, praeter F, non erit centrum, cum omne aliud punctum ipsam dividat inaequaliter, quandoquidem in F, divisa fuit aequaliter. si igitur F, non est centrum, sit punctum G, extra rectam BD, centrum, a quo ducantur lineae GA, GE, GC.  Quoniam ergo latera AE, EG, trianguli AEG, aequalia sunt lateribus CE, EG, trianguli CEG;  et basis AG, basi CG; (a centro enim ducuntur)  erunt anguli AEG, CEG, aequales,90   ideoque recti:    Erat autem et angulus AEF, rectus ex constructione.  Igitur recti AEF, AEG, aequales sunt, pars et totum. quod est absurdum.  Non est ergo punctum G, centrum; 
論曰。如云不然。令言心何在。彼不得言在己之上、下。何者。乙丁線。旣平分於己。離平分。不能為心故。必 言心在乙丁線外、為庚。卽令自庚至丙、至戊、至甲。各作直線。  則甲庚戊角形之甲戊。旣與丙庚戊角形之丙戊兩邊等。戊庚同邊。  而庚甲、庚丙、兩線。俱從心至界。宜亦等。  卽對等邊之庚戊甲。庚戊丙、兩角。宜亦等。一卷八  而為兩直角矣。一卷界說十  夫乙戊甲、旣直角。  而庚戊甲1 、又為直角。  可不可也。   
Similarly we can prove that neither is any other point except F. 
eademque est ratio de omni alio. 
Therefore the point F is the centre of the circle ABC. 
Quare F, centrum erit. Itaque dati circuli centrum reperimus. Quod etat faciendum. 
 
PORISM
From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line. 
Q. E. F. 
CORROLARIUM
HINC manifestum est, si in circulo recta aliqua linea aliquam rectam lineam bifariam, et ad angulos rectos secet, in secante esse centrum circuli. 
Nam ex eo, quod BD, recta rectam AC, bifariam secat in E, et ad angulos rectos, ostensum fuit, punctum eius medium F, necessario esse circuli centrum.91  
系。因此推顯、圜內有直線。分他線為兩平分、而作直角。卽圜心在其內。   
PROPOSITION 2. 
THEOR.1. PROPOS. 2. 
第二題 
If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle. 
SI in circuli peripheria duo quælibet puncta accepta fuerint; Recta linea, quæ ad ipsa puncta adiungitur, intra circulum cadet. 
圜界、任取二點。以直線相聯。則直線全在圜內。 
Let ABC be a circle, and let two points A, B be taken at random on its circumference;  I say that the straight line joined from A to B will fall within the circle. 
IN circulo ABC, sumantur quaelibet duo puncta A, et C, in eius circumferentia.  Dico rectam ex A, in C, ductam cadere intra circulum, ita ut ipsum secet. 
解曰。甲乙丙圜界上。任取甲、丙、二點。  作直線相聯。題言甲丙線、全在圜內。 
For suppose it does not, but, if possible, let it fall outside, as AEB;  let the centre of the circle ABC be taken [III. 1], and let it be D;  let DA, DB be joined, and let DFE be drawn through. 
Si enim non cadit intra, cadat extra, qualis est linea ADC, recta.  Invento igitur centro E,92   ducantur ab eo ad puncta assumpta A, et C, nec non ad quodvis punctum D, in recta ADC, lineae rectae EA, EC, ED, fecetque ED, circumferentiam in B. 
論曰。如云在外。若甲丁丙線。  令尋取甲乙丙圜之戊心。本篇一  次作戊甲、戊丙、兩直線。次於甲丁丙線上作戊乙丁線 
Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE. [I. 5]  And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16]  But the angle DAE is equal to the angle DBE;  therefore the angle DEB is greater than the angle DBE.  And the greater angle is subtended by the greater side; [I. 19]  therefore DB is greater than DE.  But DB is equal to DF;  therefore DF is greater than DE, the less than the greater: which is impossible.  Therefore the straight line joined from A to B will not fall outside the circle.  Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within. 
Quoniam ergo duo latera EA, EC, trianguli, cuius basis ponitur recta ADC, aequalìa sunt, (e centro enim ducuntur93 ) erunt anguli EAD, ECD, aequales: 
Est autem angulus EDA, angulo ECD, maior, externus interno opposito,94 cum latus CD, in triangulo ECD, sit productum ad A. Igitur et angulo EAD, maior erit idem angulus EDA.  See two records above      Quare recta EA, maiori angulo opposita,95   hoc est, recta EB, sibi aequalis,   maior erit, quam recta ED, pars quam totum. Quod est absurdum.  Non igitur recta ex A, in C, ducta extra circulum cadet, sed intra.  Eodem enim modo demonstrabitur, rectam ductam ex A, in C, non posse cadere super arcum ABC, ita ut eadem sit, quae circumferentia ABC. Esset enim recta EA, maior, quam recta EB. Quod etiam ex definitione rectae lineae patet, cum ABC, arcus sit linea curua, non autem recta. 
而與圜界遇於乙。卽戊甲丁丙。當為三角形。以甲丁丙為底。戊甲戊丙兩腰等。其戊甲丙。戊丙甲、兩角宜等。一卷五  而戊丁甲。為戊丙丁之外角。宜大於戊丙丁角。卽亦宜大於戊甲丁角。一卷十六 則對戊丁甲大角之        戊甲線。宜大於戊丁線矣。一卷十九  夫戊甲、與戊乙。本同圜之半徑。等。據如所論。  則戊乙亦大於戊丁。不可通也。  若云不在圜外、  而在圜界。依前論。令戊甲大於戊乙。亦不可通也。 
Therefore etc.  Q. E. D. 
Itaque si in circuli peripheria duo quaelibet puncta, etc. 
Quod erat ostendendum.

SCHOLION.
IDEM hoc theorema demonstrari poterit affirmative, hoc modo, Recta AB, coniungat duo puncta A, et B, in circumferentia circuli AB, cuius centrum C. Dico rectam AB, intra circulum cadere, ita ut omnia eius puncta media intra circulum existant. Assumatur enim quodcunque eius punctum intermedium D, et ex centro educantur recta CA, CB, CD. Quoniam igitur duo latera CA, CB, trianguli CAB, aequalia sunt, erunt anguli CAB, CBA, aequales.96 Est autem angulus CDA, angulo CBA, maior, externus interno;97 Igitur idem angulus CDA, angulo CAD, maior erit. Quare cum CA, sit ducta a centro ad circumferentiam,98 usque, non perveniet recta CD, ad circumferentiam, ideoque punctum D, intra circulum cadet: Idem ostendetur de quolibet alio puncto assumpto. Tota igitur recta AB, intra circulum cadit. Quod est propositum.


COROLLARIUM.
HINC est manifestum, lineam rectam, quae circulum tangit, ita ut cum non secet, in uno tantum puncto ipsum tangere. Si enim in duobus punctis eum tangeret, caderet pars rectae inter ea duo puncta posita intra circulum; Quare circulum secaret, quod est contra hypothesin. 
   
PROPOSITION 3. 
THEOR. 2. PROPOS. 3. 
第三題 
If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it. 
SI in circulo recta quaedam linea per centrum extensa quandam non per centrum extensam bifariam secet; et ad angulos rectos ipsam secabit. Et si ad angulos rectos eam secet, bifariam quoque eam secabit. 
直線過圜心。分他直線為兩平分。其分處必為兩直角。為兩直角。必兩平分。 
Let ABC be a circle, and in it let a straight line CD through the centre bisect a straight line AB not through the centre at the point F;  I say that it also cuts it at right angles. 
PER A, centrum circuli BCD, recta CE, extensa dividat rectam BD, non per centrum extensam, bifariam in F.  Dico rectam AF, esse ad angulos rectos ipsi BD. 
解曰。乙丙丁圜。有丙戊線。過甲心。分乙丁線為兩平分於己。  題言甲己必是垂線。而己旁為兩直角。又言己旁旣為兩直角。則甲己分乙丁。必兩平分。 
For let the centre of the circle ABC be taken, and let it be E; let EA, EB be joined. 
Ductis enim rectis AB, AD, 
先論曰。試從甲作甲乙、甲丁、兩線。 
Then, since AF is equal to FB, and FE is common, two sides are equal to two sides;  and the base EA is equal to the base EB;  therefore the angle AFE is equal to the angle BFE. [I. 8]  But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]  therefore each of the angles AFE, BFE is right.  Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles. 
erunt duo latera AF, FB, trianguli AFB, duobus AF, FD, trianguli AFD, aequalia;  et bases AB, AD, aequales.  Igitur anguli AFB, AFD, aequales erunt,99     hoc est, recti.  Quod erat primo propositum. SIT iam AF, ad angulos rectos ipsi BD. Dico rectam BD, bifariam secari in F, a recta CE. 
卽甲乙己角形之乙己。與甲丁己角形之丁己。兩邊等。甲己同邊。  甲乙、甲丁、兩線、俱從心至界。又等。卽兩形等。  則其對等邊之甲己乙、甲己丁。亦等。一卷八    而為兩直角矣。   
Again, let CD cut AB at right angles;  I say that it also bisects it, that is, that AF is equal to FB. 
 
   
For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5]  But the right angle AFE is equal to the right angle BFE,  therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles;  therefore they will also have the remaining sides equal to the remaining sides; [I. 26]  therefore AF is equal to FB. 
Ductis enim iterum rectis AB, AD; cum latera AB, AD, trianguli ABD, sint aequalia, erunt anguli ABD, ADB, aequales.100  
100. 5. primi. 
  Quoniam igitur duo anguli AFB, ABF, trianguli ABF, aequales sunt duobus angulis AFD, ADF, trianguli ADF; et latera AB, AD, quae rectis angulis aequalibus opponuntur, aequalia quoque:    erunt latera FB, FD, aequalia. Quod secundo proponebatur.101  
後論曰。如前作甲乙、甲丁、兩線。甲乙丁角形之甲乙、甲丁、兩邊旣等。則甲乙丁、甲丁乙、兩角亦等。一卷五    又甲乙己角形之甲己乙、甲乙己、兩角。與甲丁己角形之甲己丁、甲丁己、兩角。各等。而對直角之甲乙、甲丁、兩邊又等。    則己乙、己丁、兩邊亦等。一卷廿六 
Therefore etc.  Q. E. D. 
Si igitur in circulo recta quaedam linea per centrum extensa, etc. 
Si igitur in circulo recta quaedam linea per centrum extensa, etc. Quod demonstrandum erat.

FACILE quoque demonstrari poterat secunda haec pars; quae quidem conuersa est primae partis, hac ratione. Si enim AF, perpendicularis est ad BD, erit tam quadratum rectae AB, aequale quadratis rectarum AF, FB, quam quadratum rectae AD, quadratis rectarum AF, FD.102 Cum igitur quadratum rectae AB, aequale sit quadrato rectae AD; erunt et quadrata rectarum AF, FB, aequalia quadratis rectarum AF, FD. Quare dempto communi quadrato rectae AF, remanebunt quadrata rectarum FB, FD, aequalia; atque idcirco et rectae FB, FD, aequales erunt. 
  欲顯次論之旨。

又有一說。如甲丁上直角方形。與甲己、己丁、上兩直角方形幷、等。一卷四七 而甲乙上直角方形。與甲己、乙己、上兩直角方形、幷。亦等、卽甲己、己乙、上兩直角方形、幷。與甲己。己丁、上兩直角方形幷、亦等。此二率者。每減一甲己上直角方形。則所存乙己、己丁、上兩直角方形、自相等。而兩邊亦等。 
PROPOSITION 4. 
THEOR. 3. PROPOS. 4. 
第四題 
If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another. 
SI in circulo duae rectae lineae sese mutuo secent non per centrum extensae; sese mutuo bifariam non secabunt. 
圜內不過心兩直線、相交。不得俱為兩平分。 
Let ABCD be a circle, and in it let the two straight lines AC, BD, which are not through the centre, cut one another at E;  I say that they do not bisect one another. 
   
   
For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED;  let the centre of the circle ABCD be taken [III. 1], and let it be F; let FE be joined. 
   
   
Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3]  therefore the angle FEA is right.  Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3]  therefore the angle FEB is right.  But the angle FEA was also proved right;  therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible.  Therefore AC, BD do not bisect one another. 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 5. 
THEOR. 4. PROPOS. 5. 
第五題 
If two circles cut one another, they will not have the same centre. 
SI duo circuli sese mutuò secent; non erit illorum idem centrum. 
兩圜相交。必不同心。 
For let the circles ABC, CDG cut one another at the points B, C;  I say that they will not have the same centre. 
   
   
For, if possible, let it be E; let EC be joined, and let EFG be drawn through at random.  Then, since the point E is the centre of the circle ABC, EC is equal to EF. [I. Def. 15]  Again, since the point E is the centre of the circle CDG, EC is equal to EG.  But EC was proved equal to EF also;  therefore EF is also equal to EG, the less to the greater : which is impossible.  Therefore the point E is not the centre of the circles ABC, CDG. 
           
           
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 6. 
THEOR. 5. PROPOS. 6. 
第六題 
If two circles touch one another, they will not have the same centre. 
SI duo circuli sese mutuò interius tangant; eorum non erit idem centrum. 
兩圜內相切。必不同心。 
For let the two circles ABC, CDE touch one another at the point C;  I say that they will not have the same centre. 
   
   
For, if possible, let it be F; let FC be joined, and let FEB be drawn through at random. 
 
 
Then, since the point F is the centre of the circle ABC, FC is equal to FB.  Again, since the point F is the centre of the circle CDE, FC is equal to FE.  But FC was proved equal to FB;  therefore FE is also equal to FB, the less to the greater: which is impossible.  Therefore F is not the centre of the circles ABC, CDE. 
         
         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 7. 
THEOR. 6. PROPOS. 7. 
第七題 
If on the diameter of a circle a point be taken which is not the centre of the circle, and from the point straight lines fall upon the circle, that will be greatest on which the centre is, the remainder of the same diameter will be least, and of the rest the nearer to the straight line through the centre is always greater than the more remote, and only two equal straight lines will fall from the point on the circle, one on each side of the least straight line. 
SI in diametro circuli quodpiam sumatur punctum, quod circuli centrum non sit, ab eoque puncto in circulum quædam rectæ lineæ cadant: Maxima quidem erit ea, in qua centrum, minima verò reliqua; aliarũ verò propinquior illi, quæper centrum ducitur, remotiore semper maior est: Duæ autem solùm rectæ lineæ æquales ab eodem puncto in circulum cadunt, ad vtrasque partes minimæ, vel maximæ. 
圜徑離心。任取一點。從點至圜界。任出幾線。其過心線、最大。不過心線、最小。餘線愈大。愈近不過心線者、愈小。而諸線中、止兩線等。 
Let ABCD be a circle, and let AD be a diameter of it;  on AD let a point F be taken which is not the centre of the circle, let E be the centre of the circle, and from F let straight lines FB, FC, FG fall upon the circle ABCD;  I say that FA is greatest, FD is least, and of the rest FB is greater than FC, and FC than FG. 
     
     
For let BE, CE, GE be joined.  Then, since in any triangle two sides are greater than the remaining one, [I. 20] EB, EF are greater than BF.  But AE is equal to BE;  therefore AF is greater than BF.  Again, since BE is equal to CE, and FE is common, the two sides BE, EF are equal to the two sides CE, EF.  But the angle BEF is also greater than the angle CEF; therefore the base BF is greater than the base CF. [I. 24]  For the same reason CF is also greater than FG. 
             
             
Again, since GF, FE are greater than EG, and EG is equal to ED, GF, FE are greater than ED.  Let EF be subtracted from each;  therefore the remainder GF is greater than the remainder FD.  Therefore FA is greatest, FD is least, and FB is greater than FC, and FC than FG. 
       
       
I say also that from the point F only two equal straight lines will fall on the circle ABCD, one on each side of the least FD.  For on the straight line EF, and at the point E on it, let the angle FEH be constructed equal to the angle GEF [I. 23], and let FH be joined.  Then, since GE is equal to EH, and EF is common,  the two sides GE, EF are equal to the two sides HE, EF;  and the angle GEF is equal to the angle HEF;  therefore the base FG is equal to the base FH. [I. 4]  I say again that another straight line equal to FG will not fall on the circle from the point F.  For, if possible, let FK so fall.  Then, since FK is equal to FG, and FH to FG,  FK is also equal to FH,  the nearer to the straight line through the centre being thus equal to the more remote: which is impossible.  Therefore another straight line equal to GF will not fall from the point F upon the circle;  therefore only one straight line will so fall. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 8. 
THEOR. 7. PROPOS. 8. 
第八題 
If a point be taken outside a circle and from the point straight lines be drawn through to the circle, one of which is through the centre and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the centre is greatest, while of the rest the nearer to that through the centre is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote, and only two equal straight lines will fall on the circle from the point, one on each side of the least. 
SI extra circulum sumatur punctum quodquiam, ab eoque puncto ad circulum deducantur rectæ quædam lineæ, quarum vna quidem per centrum protendatur, reliquæ vero vt libet: In cauam peripheriam cadentium rectarum linearum maxima quidem est illa, quæ per centrum ducitur; aliarum autem propinquior ei, quæ per centrum transit, remotiore semper maior est; In conuexam verò peripheriam cadentium rectarum linearum minima quidem est illa, quæ inter punctum, & diametrum interponitur; aliarum autem ea, quæ propinquior est minimæ, remotiore semper minor est. Duæ autem tantum rectæ lineæ æquales ab eo puncto in ipsum circulum cadunt, ad vtrasque partes minimæ, vel maximæ. 
圜外任取一點。從點任出幾線。其至規內。則過圜心線、最大。餘線愈離心、愈小。其至規外。則過圜心線、為徑之餘者、最小。餘線愈近徑餘、愈小。而諸線中止兩線等。 
Let ABC be a circle, and let a point D be taken outside ABC; let there be drawn through from it straight lines DA, DE, DF, DC, and let DA be through the centre;  I say that, of the straight lines falling on the concave circumference AEFC, the straight line DA through the centre is greatest, while DE is greater than DF and DF than DC.; but, of the straight lines falling on the convex circumference HLKG, the straight line DG between the point and the diameter AG is least; and the nearer to the least DG is always less than the more remote, namely DK than DL, and DL than DH. 
   
   
For let the centre of the circle ABC be taken [III. 1], and let it be M; let ME, MF, MC, MK, ML, MH be joined. 
 
 
Then, since AM is equal to EM, let MD be added to each;  therefore AD is equal to EM, MD.  But EM, MD are greater than ED; [I. 20]  therefore AD is also greater than ED.  Again, since ME is equal to MF,  and MD is common,  therefore EM, MD are equal to FM, MD;  and the angle EMD is greater than the angle FMD;  therefore the base ED is greater than the base FD. [I. 24]  Similarly we can prove that FD is greater than CD;  therefore DA is greatest, while DE is greater than DF, and DF than DC. 
                     
                     
Next, since MK, KD are greater than MD, [I. 20]  and MG is equal to MK,  therefore the remainder KD is greater than the remainder GD,  so that GD is less than KD.  And, since on MD, one of the sides of the triangle MLD, two straight lines MK, KD were constructed meeting within the triangle, therefore MK, KD are less than ML, LD; [I. 21]  and MK is equal to ML;  therefore the remainder DK is less than the remainder DL.  Similarly we can prove that DL is also less than DH;  therefore DG is least, while DK is less than DL, and DL than DH. 
                 
                 
I say also that only two equal straight lines will fall from the point D on the circle, one on each side of the least DG.  On the straight line MD, and at the point M on it, let the angle DMB be constructed equal to the angle KMD, and let DB be joined.  Then, since MK is equal to MB, and MD is common,  the two sides KM, MD are equal to the two sides BM, MD respectively;  and the angle KMD is equal to the angle BMD;  therefore the base DK is equal to the base DB. [I. 4]  I say that no other straight line equal to the straight line DK will fall on the circle from the point D.  For, if possible, let a straight line so fall, and let it be DN.  Then, since DK is equal to DN,  while DK is equal to DB,  DB is also equal to DN,  that is, the nearer to the least DG equal to the more remote: which was proved impossible.  Therefore no more than two equal straight lines will fall on the circle ABC from the point D, one on each side of DG the least. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 9. 
THEOR. 8. PROPOS. 9. 
第九題 
If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the point taken is the centre of the circle. 
SI in circulo acceptum fuerit punctum aliquod, & ab eo puncto ad circulum cadant plures, quam duæ, rectæ lineæ æquales; acceptum punctum centrum est ipsius circuli. 
圜內從一點至界。作三線以上、皆等。卽此點必圜心。 
Let ABC be a circle and D a point within it, and from D let more than two equal straight lines, namely DA, DB, DC, fall on the circle ABC;  I say that the point D is the centre of the circle ABC. 
   
   
For let AB, BC be joined and bisected at the points E, F, and let ED, FD be joined and drawn through to the points G, K, H, L. 
 
 
Then, since AE is equal to EB, and ED is common,  the two sides AE, ED are equal to the two sides BE, ED;  and the base DA is equal to the base DB;  therefore the angle AED is equal to the angle BED. [I. 8]  Therefore each of the angles AED, BED is right; [I. Def. 10]  therefore GK cuts AB into two equal parts and at right angles.  And since, if in a circle a straight line cut a straight line into two equal parts and at right angles,  the centre of the circle is on the cutting straight line, [III. 1, Por.]  the centre of the circle is on GK.  For the same reason the centre of the circle ABC is also on HL.  And the straight lines GK, HL have no other point common but the point D;  therefore the point D is the centre of the circle ABC. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 10. 
THEOR. 9. PROPOS. 10. 
第十題 
A circle does not cut a circle at more points than two. 
CIRCVLVS circulum in pluribus, quam duobus, punctis non secat. 
兩圜相交。止於兩點。 
For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H;  let BH, BG be joined and bisected at the points K, L,  and from K, L let KC, LM be drawn at right angles to BH, BG and carried through to the points A, E. 
     
     
Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles,  the centre of the circle ABC is on AC. [III. 1, Por.]  Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles,  the centre of the circle ABC is on NO.  But it was also proved to be on AC, and the straight lines AC, NO meet at no point except at P;  therefore the point P is the centre of the circle ABC.  Similarly we can prove that P is also the centre of the circle DEF;  therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible. [III. 5] 
               
               
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 11. 
THEOR. 10. PROPOS. 11. 
第十一題 
If two circles touch one another internally, and their centres be taken, the straight line joining their centres, if it be also produced, will fall on the point of contact of the circles. 
SI duo circuli sese intus contingant, atque accepta fuerint eorum centra; ad eorum centra adiuncta recta linea, & producta, in contactum circulorum cadet. 
兩圜內相切。作直線聯兩心。引出之。必至切界。 
For let the two circles ABC, ADE touch one another internally at the point A,  and let the centre F of the circle ABC, and the centre G of ADE, be taken;  I say that the straight line joined from G to F and produced will fall on A. 
     
     
For suppose it does not, but, if possible, let it fall as FGH, and let AF, AG be joined. 
 
 
Then, since AG, GF are greater than FA, that is, than FH,  let FG be subtracted from each;  therefore the remainder AG is greater than the remainder GH.  But AG is equal to GD;  therefore GD is also greater than GH, the less than the greater: which is impossible.  Therefore the straight line joined from F to G will not fall outside;  therefore it will fall at A on the point of contact. 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 12. 
THEOR. 11. PROPOS. 12. 
第十二題 
If two circles touch one another externally, the straight line joining their centres will pass through the point of contact. 
SI duo circuli sese exterius contingant, linea recta, quæ ad centra eorum adiungitur, per contactum transibit. 
兩圜外相切。以直線聯兩心。必過切界。 
For let the two circles ABC, ADE touch one another externally at the point A,  and let the centre F of ABC, and the centre G of ADE, be taken  I say that the straight line joined from F to G will pass through the point of contact at A. 
CIRCVLI duo ABC, DBE, tangant se exterius in B,  et centrum circuli ABC, sit F, circuli vero DBE, centrum sit G.  Dico rectam extensam per F, et G, transire per contactum B. 
解曰。甲乙丙、丁乙戊、兩圜。外相切於乙。  其甲乙丙心為己。丁乙戊心為庚。  題言作己庚直線。必過乙。 
For suppose it does not, but, if possible, let it pass as FCDG, and let AF, AG be joined. 
Si enim non transit, 
論曰。如云不然。 
Then, since the point F is the centre of the circle ABC, FA is equal to FC.  Again, since the point G is the centre of the circle ADE, GA is equal to GD.  But FA was also proved equal to FC;  therefore FA, AG are equal to FC, GD,  so that the whole FG is greater than FA, AG;  but it is also less [I. 20]: which is impossible.  Therefore the straight line joined from F to G will not fail to pass through the point of contact at A;  therefore it will pass through it. 
secet circunferentias in C, et E, ducanturque a centris F, G, ad B, contactum rectae FB, GB. Quoniam igitur in triangulo FBG, latera duo BF, BG, 103 maiora sunt latere FG:  
Est autem recta BF, rectae FC, aequalis: (quod F, ponatur centrum circuli ABC,) et recta GB, rectae GE, aequalis; (quod G, ponatur centrum circuli DBE,)   erunt et rectae FC, GE,    maiores quam recta FG, pars quam totum, (cum FG, contineat praeter FC, GE, rectam adhuc CE,)  quod est absurdum.      
而己庚線、截兩圜界於戊、於丙。令於切界作乙己、乙庚、兩線。其乙己庚角形之己乙、乙庚、兩邊幷。大 於己庚一邊。而乙庚與庚戊。乙己、與己丙。俱同心所出線。宜各等。卽庚戊、丙己、兩線幷。亦大於庚己一線矣。一卷二十    夫庚己線。分為庚戊、丙己。尚餘丙戊。    而云庚戊、丙己。大於庚己。則分大於全也。    故直線聯己庚。  必過乙。 
Therefore etc.  Q. E. D. 
Si igitur duo circuli sese exterius contingant, etc.   Quod erat demonstrandum. 
   
PROPOSITION 13. 
THEOR. 12. PROPOS. 13. 
第十三題 
A circle does not touch a circle at more points than one, whether it touch it internally or externally. 
CIRCVLVS circulum non tangit in pluribus punctis, quam vno, siue intus, siue extra tangat. 
二支
圜相切。不論內外。止以一點。 
For, if possible, let the circle ABDC touch the circle EBFD, first internally, at more points than one, namely D, B. 
 
 
Let the centre G of the circle ABDC, and the centre H of EBFD, be taken. 
 
 
Therefore the straight line joined from G to H will fall on B, D. [III. 11]  Let it so fall, as BGHD.  Then, since the point G is the centre of the circle ABCD, BG is equal to GD;  therefore BG is greater than HD;  therefore BH is much greater than HD.  Again, since the point H is the centre of the circle EBFD,  BH is equal to HD;  but it was also proved much greater than it: which is impossible.  Therefore a circle does not touch a circle internally at more points than one. 
                 
                 
I say further that neither does it so touch it externally. 
 
 
For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A, C, and let AC be joined. 
 
 
Then, since on the circumference of each of the circles ABDC, ACK two points A, C have been taken at random, the straight line joining the points will fall within each circle; [III. 2]  but it fell within the circle ABCD and outside ACK [III. Def. 3]: which is absurd. 
   
   
Therefore a circle does not touch a circle externally at more points than one.  And it was proved that neither does it so touch it internally. 
   
   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 14. 
THEOR. 13. PROPOS. 14. 
第十四題 
In a circle equal straight lines are equally distant from the centre, and those which are equally distant from the centre are equal to one another. 
IN circulo æquales rectæ lineæ æqualiter distant a centro. Et quæ æqualiter distant a centro, æquales sunt inter se. 
二支
圜內兩直線等。卽距心之遠近等。距心之遠近等。卽兩直線等。 
Let ABDC be a circle, and let AB, CD be equal straight lines in it;  I say that AB, CD are equally distant from the centre. 
   
   
For let the centre of the circle ABDC be taken [III. 1], and let it be E; from E let EF, EG be drawn perpendicular to AB, CD, and let AE, EC be joined. 
 
 
Then, since a straight line EF through the centre cuts a straight line AB not through the centre at right angles, it also bisects it. [III. 3]  Therefore AF is equal to FB;  therefore AB is double of AF.  For the same reason CD is also double of CG;  and AB is equal to CD;  therefore AF is also equal to CG.  And, since AE is equal to EC,  the square on AE is also equal to the square on EC.  But the squares on AF, EF are equal to the square on AE,  for the angle at F is right;  and the squares on EG, GC are equal to the square on EC,  for the angle at G is right; [I. 47]  therefore the squares on AF, FE are equal to the squares on CG, GE,  of which the square on AF is equal to the square on CG,  for AF is equal to CG;  therefore the square on FE which remains is equal to the square on EG,  therefore EF is equal to EG.  But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [III. Def. 4]  therefore AB, CD are equally distant from the centre. 
                                     
                                     
Next, let the straight lines AB, CD be equally distant from the centre; that is, let EF be equal to EG.  I say that AB is also equal to CD. 
   
   
For, with the same construction, we can prove, similarly, that AB is double of AF, and CD of CG.  And, since AE is equal to CE, the square on AE is equal to the square on CE.  But the squares on EF, FA are equal to the square on AE,  and the squares on EG, GC equal to the square on CE. [I. 47]  Therefore the squares on EF, FA are equal to the squares on EG, GC,  of which the square on EF is equal to the square on EG,  for EF is equal to EG;  therefore the square on AF which remains is equal to the square on CG;  therefore AF is equal to CG.  And AB is double of AF, and CD double of CG;  therefore AB is equal to CD. 
                     
                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 15. 
THEOR. 14. PROPOS. 15. 
第十五題 
Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the centre is always greater than the more remote. 
IN circulo maxima quidem linea est diameter; aliarum autem propinquior centro, remotiore semper maior. 
徑為圜內之大線。其餘線者。近心大於遠心。 
Let ABCD be a circle, let AD be its diameter and E the centre; and let BC be nearer to the diameter AD, and FG more remote;  I say that AD is greatest and BC greater than FG. 
   
   
For from the centre E let EH, EK be drawn perpendicular to BC, FG.  Then, since BC is nearer to the centre and FG more remote, EK is greater than EH. [III. Def. 5]  Let EL be made equal to EH,  through L let LM be drawn at right angles to EK and carried through to N, and let ME, EN, FE, EG be joined. 
       
       
Then, since EH is equal to EL,  BC is also equal to MN. [III. 14]  Again, since AE is equal to EM, and ED to EN,  AD is equal to ME, EN.  But ME, EN are greater than MN, [I. 20] and MN is equal to BC;  therefore AD is greater than BC.  And, since the two sides ME, EN are equal to the two sides FE, EG,  and the angle MEN greater than the angle FEG,  therefore the base MN is greater than the base FG. [I. 24]  But MN was proved equal to BC.  Therefore the diameter AD is greatest  and BC greater than FG. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 16. 
THEOR. 15. PROPOS. 16. 
第十六題 
The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilineal angle. 
QVÆ ab extremitate diametri cuiusque circuli ad angulos rectos ducitur, extra ipsum circulum cadet; & in locum inter ipsam rectam lineam, & peripheriam comprehensum, altera recta linea non cadet. & semicirculi quidem angulus, quouis angulo acuto rectilineo maior est; reliquus autem minor. 
三支
圜徑末之直角線。全在圜外。而直線偕圜界、所作切邊角。不得更作一直線。入其內。其半圜分角。大於各直線銳角。切邊角。小於各直線銳角。 
Let ABC be a circle about D as centre and AB as diameter;  I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle. 
   
   
For suppose it does not, but, if possible, let it fall within as CA, and let DC be joined. 
 
 
Since DA is equal to DC,  the angle DAC is also equal to the angle ACD. [I. 5]  But the angle DAC is right;  therefore the angle ACD is also right:  thus, in the triangle ACD, the two angles DAC, ACD are equal to two right angles: which is impossible. [I. 17]  Therefore the straight line drawn from the point A at right angles to BA will not fall within the circle.  Similarly we can prove that neither will it fall on the circumference;  therefore it will fall outside. 
               
               
Let it fall as AE;  I say next that into the space between the straight line AE and the circumference CHA another straight line cannot be interposed. 
   
   
For, if possible, let another straight line be so interposed, as FA, and let DG be drawn from the point D perpendicular to FA.  Then, since the angle AGD is right, and the angle DAG is less than a right angle, AD is greater than DG. [I. 19]  But DA is equal to DH;  therefore DH is greater than DG, the less than the greater: which is impossible.  Therefore another straight line cannot be interposed into the space between the straight line and the circumference. 
         
         
I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilineal angle,  and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilineal angle. 
   
   
For, if there is any rectilineal angle greater than the angle contained by the straight line BA and the circumference CHA,  and any rectilineal angle less than the angle contained by the circumference CHA and the straight line AE,  then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines  which is greater than the angle contained by the straight line BA and the circumference CHA,  and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE.  But such a straight line cannot be interposed;  therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA,  nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE. 
               
               
PORISM.
From this it is manifest that the straight line drawn at right angles to the diameter of a circle from its extremity touches the circle. 
Q. E. D. 
   
   
PROPOSITION 17. 
PROBL. 2. PROPOS. 17. 
第十七題 
From a given point to draw a straight line touching a given circle. 
A DATO puncto rectam lineam ducere, quæ datum tangat circulum. 
設一點、一圜。求從點作切線。 
Let A be the given point, and BCD the given circle;  thus it is required to draw from the point A a straight line touching the circle BCD. 
   
   
For let the centre E of the circle be taken; [III. 1] let AE be joined, and with centre E and distance EA let the circle AFG be described;  from D let DF be drawn at right angles to EA, and let EF, AB be joined;  I say that AB has been drawn from the point A touching the circle BCD. 
     
     
For, since E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB;  therefore the two sides AE, EB are equal to the two sides FE, ED:  and they contain a common angle, the angle at E;  therefore the base DF is equal to the base AB,  and the triangle DEF is equal to the triangle BEA,  and the remaining angles to the remaining angles; [I. 4  therefore the angle EDF is equal to the angle EBA.  But the angle EDF is right; therefore the angle EBA is also right.  Now EB is a radius;  and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16, Por.]  therefore AB touches the circle BCD. 
                     
                     
Therefore from the given point A the straight line AB has been drawn touching the circle BCD.   
   
   
PROPOSITION 18. 
THEOR. 16. PROPOS. 18. 
第十八題 
If a straight line touch a circle, and a straight line be joined from the centre to the point of contact, the straight line so joined will be perpendicular to the tangent. 
SI circulum tangat recta quæpiam linea, a centro autem ad contactum adiungatur recta quædam linea: quæ adiuncta fuerit, ad ipsam contingentem perpendicularis erit. 
直線切圜。從圜心作直線、至切界。必為切線之垂線。 
For let a straight line DE touch the circle ABC at the point C, let the centre F of the circle ABC be taken, and let FC be joined from F to C;  I say that FC is perpendicular to DE. 
   
   
For, if not, let FG be drawn from F perpendicular to DE. 
 
 
Then, since the angle FGC is right,  the angle FCG is acute; [I. 17]  and the greater angle is subtended by the greater side; [I. 19]  therefore FC is greater than FG.  But FC is equal to FB;  therefore FB is also greater than FG, the less than the greater: which is impossible.  Therefore FG is not perpendicular to DE.  Similarly we can prove that neither is any other straight line except FC;  therefore FC is perpendicular to DE. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 19. 
THEOR. 17. PROPOS. 19. 
第十九題 
If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the tangent, the centre of the circle will be on the straight line so drawn. 
SI circulum tetigerit recta quæpiam linea, a contactu autem recta linea ad angulos rectos ipsi tangenti excitetur: In excitata erit centrum circuli. 
直線切圜。圜內作切線之垂線。則圜心必在垂線之內。 
For let a straight line DE touch the circle ABC at the point C, and from C let CA be drawn at right angles to DE;  I say that the centre of the circle is on AC. 
   
   
For suppose it is not, but, if possible, let F be the centre, and let CF be joined. 
 
 
Since a straight line DE touches the circle ABC, and FC has been joined from the centre to the point of contact,  FC is perpendicular to DE; [III. 18]  therefore the angle FCE is right.  But the angle ACE is also right;  therefore the angle FCE is equal to the angle ACE, the less to the greater: which is impossible.  Therefore F is not the centre of the circle ABC.  Similarly we can prove that neither is any other point except a point on AC. 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 20. 
THEOR. 18. PROPOS. 20. 
第二十題 
In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same circumference as base. 
IN circulo, angulus ad centrum duplex est anguli ad peripheriam, cum fuerit eadem peripheria basis angulorum. 
負圜角與分圜角。所負所分之圜分同。則分圜角、必倍大於負圜角。 
Let ABC be a circle, let the angle BEC be an angle at its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base;  I say that the angle BEC is double of the angle BAC. 
IN circulo ABC, cuius centrum D, superbasin BC, consticuatur angulus BDC, ad centrum; et super eandem basin angulus BAC, ad peripheriam.  Dico angulum BDC, duplum esse anguli BAC. 
解曰。甲乙丙圜。其心丁。有乙丁丙分圜角。乙甲丙負圜角。同以乙丙圜分為底。  題言乙丁丙角。倍大於乙甲丙角。 
For let AE be joined and drawn through to F. 
Includant enim primum duae AB, AC, duas DB, DC; et per centrum D, recta extendatur AE. Quoniam igitur rectae DA, DB, aequales sunt,104 erunt anguli DAB, DBA, aequales: Est autem externus angulus BDE,105 aequalis duobus angulis internis DAB, DBA. Quare BDE, plus erit alterius corum, ut anguli DAB. Eodem modo duplus ostendetur angulus CDE, anguli DAC. Quapropter totus BDC, duplus erit totius BAC. Quando enim duae magnitudines duarum sunt duplae, singulae singularum, est quoque aggregatum ex illis aggregati ex his duplum. Constat ergo propositum. 
104. 5. primi.  105. 32. primi. 
先論分圜角、在乙甲、甲丙、之內者曰。如上圖。試從甲、過丁心、作甲戊線。其甲丁乙角形之丁甲、丁乙等。卽丁甲乙、丁乙甲、兩角等。一卷四而乙丁戊外角。與內相對兩角幷等。一卷卅二卽乙丁戊、倍大於乙甲丁矣。依顯丙丁戊、亦倍大於丙甲丁。則乙丁丙全角。亦倍大於乙甲丙全角。 
Then, since EA is equal to EB, the angle EAB is also equal to the angle EBA; [I. 5]  therefore the angles EAB, EBA are double of the angle EAB.  But the angle BEF is equal to the angles EAB, EBA; [I. 32]  therefore the angle BEF is also double of the angle EAB.  For the same reason the angle FEC is also double of the angle EAC.  Therefore the whole angle BEC is double of the whole angle BAC. 
        DEINDE non includant rectae AB, AC, rectas DB, DC, sed AB, per centrum extendatur. Quoniam igitur externus angulus BDC,106 aequalis est duobus internis DAC, DCA. Hiautem duo107 inter se sunt aequales, quod latera DA, DC sint aequalia; erit angulus BDC, duplus alterius corum, nempe anguli BAC. Quod est propositum.   
        次論分圜角、不在乙甲、甲丙之內、而甲乙線過丁心者、曰。如下圖。依前論、推顯乙丁丙外角。等於內相對之丁甲丙、丁丙甲、兩角幷。一卷卅二而丁甲、丁丙、兩腰等。卽甲、丙、兩角亦等。一卷五則乙丁丙角。倍大於乙甲丙角。   
Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G.  Similarly then we can prove that the angle GEC is double of the angle EDC,  of which the angle GEB is double of the angle EDB;  therefore the angle BEC which remains is double of the angle BDC. 
TERTIO recta AB, secet rectam DC, et per centrum D, extendatur recta AE. Quoniam igitur angulus EDC, ad centrum, et angulus EAC, ad peripheriam, habent eandem basin EC, et recta AE, extenditur per centrum; erit angulus EDC, duplus anguli EAC, ut ostensum est in secunda parte. simili modo erit angulus EDB, duplus anguli EAB; habent enim hianguli eandem basin EB. Reliquusigitur angulus BDC, duplus erit reliqui anguli BAC.108 Quando enim totum totius est duplum, et ablatum ablati; est et reliquum reliqui duplum. In circulo igitur angulus ad centrum duplex est, etc. 
     
後論分圜角在負圜角線之外、而甲乙截丁丙者、曰。如下圖。試從甲過丁心、作甲戊線。其戊丁丙分圜角。與戊甲丙負圜角。同以戊乙丙圜分為底。如前次論戊丁丙角。倍大於戊甲丙角。依顯戊丁乙分圜角。亦倍大於戊甲乙負圜角。次於戊丁丙角、減戊丁乙角。戊甲丙角。減戊甲乙角。則所存乙丁丙角。必倍大於乙甲丙角。       
Therefore etc.  Q. E. D. 
  Quod erat demonstrandum. 
   
 
 
增若乙丁、丁丙。不作角於心。或為半圜。或小於半圜。則丁心外餘地亦倍大於同底之負圜角。 
 
 
論曰。試從甲過丁心、作甲戊線。卽丁心外餘地。分為乙丁戊、戊丁丙、兩角。依前論推顯此兩角。倍大於乙甲丁、丁甲丙、兩角。 
PROPOSITION 21. 
THEOR. 19. PROPOS. 21. 
第二十一題 
In a circle the angles in the same segment are equal to one another. 
IN circulo, qui in eodem segmento sunt, anguli, sunt inter se æquales. 
凡同圜分內、所作負圜角。俱等。 
Let ABCD be a circle, and let the angles BAD, BED be angles in the same segment BAED;  I say that the angles BAD, BED are equal to one another. 
   
   
For let the centre of the circle ABCD be taken, and let it be F; let BF, FD be joined. 
 
 
Now, since the angle BFD is at the centre,  and the angle BAD at the circumference,  and they have the same circumference BCD as base,  therefore the angle BFD is double of the angle BAD. [III. 20]  For the same reason the angle BFD is also double of the angle BED;  therefore the angle BAD is equal to the angle BED. 
           
           
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 22. 
THEOR. 20. PROPOS. 22 
第二十二 
The opposite angles of quadrilaterals in circles are equal to two right angles. 
QVADRILATERORVM in circulis descriptorum anguli, qui ex aduerso, duobus rectis sunt æquales. 
圜內切界四邊形。每相對兩角幷。與兩直角等。 
Let ABCD be a circle, and let ABCD be a quadrilateral in it;  I say that the opposite angles are equal to two right angles. 
   
   
Let AC, BD be joined. 
 
 
Then, since in any triangle the three angles are equal to two right angles, [I. 32]  the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles.  But the angle CAB is equal to the angle BDC,  for they are in the same segment BADC; [III. 21]  and the angle ACB is equal to the angle ADB,  for they are in the same segment ADCB;  therefore the whole angle ADC is equal to the angles BAC, ACB.  Let the angle ABC be added to each;  therefore the angles ABC, BAC, ACB are equal to the angles ABC, ADC.  But the angles ABC, BAC, ACB are equal to two right angles;  therefore the angles ABC, ADC are also equal to two right angles.  Similarly we can prove that the angles BAD, DCB are also equal to two right angles. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 23. 
THEOR. 21. PROPOS. 23. 
第二十三題 
On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side. 
SVPER eadem recta linea, duo segmenta circulorum similia, & inæqualia, non constituentur ad easdem partes. 
一直線上、作兩圜分。不得相似而不相等。 
For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB, ADB be constructed on the same side; let ACD be drawn through, and let CB, DB be joined. 
 
 
Then, since the segment ACB is similar to the segment ADB,  and similar segments of circles are those which admit equal angles, [III. Def. 11]  the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16] 
     
     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 24. 
THEOR. 22. PROPOS. 24. 
第二十四題 
Similar segments of circles on equal straight lines are equal to one another. 
SVPER æqualibus rectis lineis, similia circulorum segmenta sunt inter se æqualia. 
相等兩直線上。作相似兩圜分。必等。 
For let AEB, CFD be similar segments of circles on equal straight lines AB, CD;  I say that the segment AEB is equal to the segment CFD. 
   
   
For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD,  the point B will also coincide with the point D, because AB is equal to CD;  and, AB coinciding with CD, the segment AEB will also coincide with CFD.  For, if the straight line AB coincide with CD  but the segment AEB do not coincide with CFD,  it will either fall with it, or outside it; or it will fall awry, as CGD,  and a circle cuts a circle at more points than two: which is impossible. [III. 10]  Therefore, if the straight line AB be applied to CD, the segment AEB will not fail to coincide with CFD also;  therefore it will coincide with it and will be equal to it. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 25. 
PROBL. 3. PROPOS. 25. 
第二十五題 
Given a segment of a circle, to describe the complete circle of which it is a segment. 
CIRCVLI segmento dato, describere circulum, cuius est segmentum. 
有圜之分。求成圜。 
Let ABC be the given segment of a circle;  thus it is required to describe the complete circle belonging to the segment ABC, that is, of which it is a segment. 
   
   
For let AC be bisected at D, let DB be drawn from the point D at right angles to AC, and let AB. be joined;  the angle ABD is then greater than, equal to, or less than the angle BAD. 
   
   
First let it be greater; and on the straight line BA, and at the point A on it, let the angle BAE be constructed equal to the angle ABD; let DB be drawn through to E, and let EC be joined.  Then, since the angle ABE is equal to the angle BAE,  the straight line EB is also equal to EA. [I. 6]  And, since AD is equal to DC, and DE is common, the two sides AD, DE are equal to the two sides CD, DE respectively;  and the angle ADE is equal to the angle CDE,  for each is right;  therefore the base AE is equal to the base CE.  But AE was proved equal to BE;  therefore BE is also equal to CE;  therefore the three straight lines AE, EB, EC are equal to one another.  Therefore the circle drawn with centre E and distance one of the straight lines AE, EB, EC will also pass through the remaining points and will have been completed. [III. 9]  Therefore, given a segment of a circle, the complete circle has been described.  And it is manifest that the segment ABC is less than a semicircle, because the centre E happens to be outside it. 
                         
                         
Similarly, even if the angle ABD be equal to the angle BAD,  AD being equal to each of the two BD, DC, the three straight lines DA, DB, DC will be equal to one another,  D will be the centre of the completed circle, and ABC will clearly be a semicircle. 
     
     
But, if the angle ABD be less than the angle BAD,  and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD,  the centre will fall on DB within the segment ABC,  and the segment ABC will clearly be greater than a semicircle. 
       
       
Therefore, given a segment of a circle, the complete circle has been described.  Q. E. F. 
   
   
PROPOSITION 26. 
THEOR. 23. PROPOS. 26. 
第二十六題 
In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences. 
IN æqualibus circulis, æquales anguli æqualibus peripheriis insistunt, siue ad centra, siue ad peripherias constituti insistant. 
二支
等圜之乘圜分角。或在心。或在界。等。其所乘之圜分亦等。 
Let ABC, DEF be equal circles,  and in them let there be equal angles, namely at the centres the angles BGC, EHF, and at the circumferences the angles BAC, EDF;  I say that the circumference BKC is equal to the circumference ELF. 
     
     
For let BC, EF be joined. 
 
 
Now, since the circles ABC, DEF are equal,  the radii are equal.  Thus the two straight lines BG, GC are equal to the two straight lines EH, HF;  and the angle at G is equal to the angle at H;  therefore the base BC is equal to the base EF. [I. 4]  And, since the angle at A is equal to the angle at D,  the segment BAC is similar to the segment EDF; [III. Def. 11]  and they are upon equal straight lines.  But similar segments of circles on equal straight lines are equal to one another; [III. 24]  therefore the segment BAC is equal to EDF.  But the whole circle ABC is also equal to the whole circle DEF;  therefore the circumference BKC which remains is equal to the circumference ELF. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 27. 
THEOR. 24. PROPOS. 27. 
第二十七題 
In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences. 
IN æqualibus circulis, anguli, qui æqualibus peripheriis insistunt; sunt inter se æquales, siue ad centra, siue ad peripherias constituti insistant. 
二支
等圜之角、所乘圜分等。則其角、或在心。或在界。俱等。 
For in equal circles ABC, DEF, on equal circumferences BC, EF, let the angles BGC, EHF stand at the centres G, H, and the angles BAC, EDF at the circumferences;  I say that the angle BGC is equal to the angle EHF, and the angle BAC is equal to the angle EDF. 
   
   
For, if the angle BGC is unequal to the angle EHF,  one of them is greater.  Let the angle BGC be greater:  and on the straight line BG, and at the point G on it, let the angle BGK be constructed equal to the angle EHF. [I. 23]  Now equal angles stand on equal circumferences, when they are at the centres; [III. 26]  therefore the circumference BK is equal to the circumference EF.  But EF is equal to BC;  therefore BK is also equal to BC, the less to the greater: which is impossible.  Therefore the angle BGC is not unequal to the angle EHF;  therefore it is equal to it.  And the angle at A is half of the angle BGC,  and the angle at D half of the angle EHF; [III. 20]  therefore the angle at A is also equal to the angle at D. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 28. 
THEOR. 25. PROPOS. 28. 
第二十八題 
In equal circles equal straight lines cut off equal circumferences, the greater equal to the greater and the less to the less. 
IN æqualibus circulis, æquales rectæ lineæ æquales peripherias auferunt, maiorem quidem maiori, minorem autem minori. 
等圜內之直線等。則其割本圜之分。大與大。小與小。各等。 
Let ABC, DEF be equal circles, and in the circles let AB, DE be equal straight lines cutting off ACB, DFE as greater circumferences and AGB, DHE as lesser; I say that the greater circumference ACB is equal to the greater circumference DFE, and the less circumference AGB to DHE. 
 
 
For let the centres K, L of the circles be taken, and let AK, KB, DL, LE be joined. 
 
 
Now, since the circles are equal,  the radii are also equal;  therefore the two sides AK, KB are equal to the two sides DL, LE;  and the base AB is equal to the base DE;  therefore the angle AKB is equal to the angle DLE. [I. 8]  But equal angles stand on equal circumferences, when they are at the centres; [III. 26]  therefore the circumference AGB is equal to DHE.  And the whole circle ABC is also equal to the whole circle DEF;  therefore the circumference ACB which remains is also equal to the circumference DFE which remains. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 29. 
THEOR. 26. PROPOS. 29. 
第二十九題 
In equal circles equal circumferences are subtended by equal straight lines. 
IN æqualibus circulis, æquales peripherias, æquales rectæ lineæ subtendunt. 
等圜之圜分等。則其割圜分之直線亦等。 
Let ABC, DEF be equal circles, and in them let equal circumferences BGC, EHF be cut off;  and let the straight lines BC, EF be joined;  I say that BC is equal to EF. 
     
     
For let the centres of the circles be taken, and let them be K, L; let BK, KC, EL, LF be joined. 
 
 
Now, since the circumference BGC is equal to the circumference EHF,  the angle BKC is also equal to the angle ELF. [III. 27]  And, since the circles ABC, DEF are equal, the radii are also equal;  therefore the two sides BK, KC are equal to the two sides EL, LF;  and they contain equal angles;  therefore the base BC is equal to the base EF. [I. 4] 
           
           
Therefore etc.   
   
   
PROPOSITION 30. 
PROBL. 4. PROPOS. 30. 
第三十題 
To bisect a given circumference. 
DATAM peripheriam bifariam secare. 
有圜之分。求兩平分之。 
Let ADB be the given circumference;  thus it is required to bisect the circumference ADB. 
   
   
Let AB be joined and bisected at C;  from the point C let CD be drawn at right angles to the straight line AB,  and let AD, DB be joined. 
     
     
Then, since AC is equal to CB, and CD is common,  the two sides AC, CD are equal to the two sides BC, CD;  and the angle ACD is equal to the angle BCD,  for each is right;  therefore the base AD is equal to the base DB. [I. 4]  But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [III. 28]  and each of the circumferences AD, DB is less than a semicircle;  therefore the circumference AD is equal to the circumference DB. 
               
               
Therefore the given circumference has been bisected at the point D.  Q. E. F. 
   
   
PROPOSITION 31. 
THEOR. 27. PROPOS. 31. 
第三十一題 
In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle. 
IN circulo angulus, qui in semicirculo, rectus est: qui autem in maiore segmento, minor recto: qui vero in minore segmento, maior est recto. Et insuper angulus maioris segmenti, recto quidem maior est: minoris autem segmenti angulus, minor est recto. 
五支
負半圜角、必直角。負大分角、小於直角。負小分角、大於直角。大圜分角、大於直角。小圜分角、小於直角。 
Let ABCD be a circle, let BC be its diameter, and E its centre, and let BA, AC, AD, DC be joined;  I say that the angle BAC in the semicircle BAC is right,  the angle ABC in the segment ABC greater than the semicircle is less than a right angle,  and the angle ADC in the segment ADC less than the semicircle is greater than a right angle. 
       
       
Let AE be joined, and let BA be carried through to F. 
 
 
Then, since BE is equal to EA,  the angle ABE is also equal to the angle BAE. [I. 5]  Again, since CE is equal to EA,  the angle ACE is also equal to the angle CAE. [I. 5]  Therefore the whole angle BAC is equal to the two angles ABC, ACB.  But the angle FAC exterior to the triangle ABC is also equal to the two angles ABC, ACB; [I. 32]  therefore the angle BAC is also equal to the angle FAC;  therefore each is right; [I. Def. 10]  therefore the angle BAC in the semicircle BAC is right. 
                 
                 
Next, since in the triangle ABC the two angles ABC, BAC are less than two right angles, [I. 17]  and the angle BAC is a right angle,  the angle ABC is less than a right angle;  and it is the angle in the segment ABC greater than the semicircle. 
       
       
Next, since ABCD is a quadrilateral in a circle,  and the opposite angles of quadrilaterals in circles are equal to two right angles, [III. 22]  while the angle ABC is less than a right angle,  therefore the angle ADC which remains is greater than a right angle;  and it is the angle in the segment ADC less than the semicircle. 
         
         
I say further that the angle of the greater segment, namely that contained by the circumference ABC and the straight line AC, is greater than a right angle;  and the angle of the less segment, namely that contained by the circumference ADC and the straight line AC, is less than a right angle.  This is at once manifest.  For, since the angle contained by the straight lines BA, AC is right,  the angle contained by the circumference ABC and the straight line AC is greater than a right angle.  Again, since the angle contained by the straight lines AC, AF is right,  the angle contained by the straight line CA and the circumference ADC is less than a right angle. 
             
             
Therefore etc.  Q. E. D. 
   
   
 
 
一系。凡角形之內。一角與兩角幷、等。其一角必直角。何者。其外角與內相對之兩角等。則與外角等之內交角。豈非直角。
二系。大分之角。大於直角。小分之角。小於直角。終無有角等於直角。又從小過大。從大過小。非大卽小。終無相等。係此題四五論、甚明。與本篇十六題增注、互相發也。 
PROPOSITION 32. 
THEOR. 28. PROPOS. 32. 
第三十二題 
If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle. 
SI circulum tetigerit aliqua recta linea, a contactu autem producatur quædam recta linea circulum secans: Anguli, quos ad contingentem facit, æquales sunt ijs, qui in alternis circuli segmentis consistunt, angulis. 
直線切圜。從切界、任作直線、割圜為兩分。分內、各任為負圜角。其切線與割線、所作兩角。與兩負圜角。交(p. 一七四)互相等。 
For let a straight line EF touch the circle ABCD at the point B,  and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it;  I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD, and the angle EBD is equal to the angle constructed in the segment DCB. 
     
     
For let BA be drawn from B at right angles to EF, let a point C be taken at random on the circumference BD, and let AD, DC, CB be joined. 
 
 
Then, since a straight line EF touches the circle ABCD at B, and BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle ABCD is on BA. [III. 19]  Therefore BA is a diameter of the circle ABCD;  therefore the angle ADB, being an angle in a semicircle, is right. [III. 31]  Therefore the remaining angles BAD, ABD are equal to one right angle. [I. 32]  But the angle ABF is also right;  therefore the angle ABF is equal to the angles BAD, ABD.  Let the angle ABD be subtracted from each;  therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle.  Next, since ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [III. 22]  But the angles DBF, DBE are also equal to two right angles;  therefore the angles DBF, DBE are equal to the angles BAD, BCD,  of which the angle BAD was proved equal to the angle DBF;  therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 33. 
PROBL. 5. PROPOS. 33. 
第三十三題 
On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle. 
SVPER data recta linea describere segmentum circuli, quod capiat angulum æqualem dato angulo rectilineo. 
一線上、求作圜分。而負圜分角、與所設直線角等。 
Let AB be the given straight line, and the angle at C the given rectilineal angle;  thus it is required to describe on the given straight line AB a segment of a circle admitting an angle equal to the angle at C. 
   
   
The angle at C is then acute, or right, or obtuse.  First let it be acute, and, as in the first figure, on the straight line AB, and at the point A, let the angle BAD be constructed equal to the angle at C;  therefore the angle BAD is also acute.  Let AE be drawn at right angles to DA, let AB be bisected at F, let FG be drawn from the point F at right angles to AB, and let GB be joined. 
       
       
Then, since AF is equal to FB, and FG is common,  the two sides AF, FG are equal to the two sides BF, FG;  and the angle AFG is equal to the angle BFG;  therefore the base AG is equal to the base BG. [I. 4]  Therefore the circle described with centre G and distance GA will pass through B also.  Let it be drawn, and let it be ABE; let EB be joined.  Now, since AD is drawn from A, the extremity of the diameter AE, at right angles to AE,  therefore AD touches the circle ABE. [III. 16, Por.]  Since then a straight line AD touches the circle ABE,  and from the point of contact at A a straight line AB is drawn across in the circle ABE,  the angle DAB is equal to the angle AEB in the alternate segment of the circle. [III. 32]  But the angle DAB is equal to the angle at C; therefore the angle at C is also equal to the angle AEB. 
                       
                       
Therefore on the given straight line AB the segment AEB of a circle has been described admitting the angle AEB equal to the given angle, the angle at C. 
 
 
Next let the angle at C be right;  and let it be again required to describe on AB a segment of a circle admitting an angle equal to the right angle at C.  Let the angle BAD be constructed equal to the right angle at C,  as is the case in the second figure;  let AB be bisected at F, and with centre F and distance either FA or FB let the circle AEB be described. 
         
         
Therefore the straight line AD touches the circle ABE, because the angle at A is right. [III. 16, Por.]  And the angle BAD is equal to the angle in the segment AEB,  for the latter too is itself a right angle, being an angle in a semicircle. [III. 31]  But the angle BAD is also equal to the angle at C.  Therefore the angle AEB is also equal to the angle at C. 
         
         
Therefore again the segment AEB of a circle has been described on AB admitting an angle equal to the angle at C. 
 
 
Next, let the angle at C be obtuse;  and on the straight line AB, and at the point A, let the angle BAD be constructed equal to it,  as is the case in the third figure;  let AE be drawn at right angles to AD,  let AB be again bisected at F, let FG be drawn at right angles to AB,  and let GB be joined. 
           
           
Then, since AF is again equal to FB, and FG is common,  the two sides AF, FG are equal to the two sides BF, FG;  and the angle AFG is equal to the angle BFG;  therefore the base AG is equal to the base BG. [I. 4]  Therefore the circle described with centre G and distance GA will pass through B also;  let it so pass, as AEB.  Now, since AD is drawn at right angles to the diameter AE from its extremity,  AD touches the circle AEB. [III. 16, Por.]  And AB has been drawn across from the point of contact at A;  therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [III. 32]  But the angle BAD is equal to the angle at C.  Therefore the angle in the segment AHB is also equal to the angle at C. 
                       
                       
Therefore on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C.  Q. E. F. 
   
   
PROPOSITION 34. 
PROBL. 6. PROPOS. 34. 
第三十四題 
From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle. 
A DATO circulo segmentum abscindere capiens angulum æqualem dato angulo rectilineo. 
設圜。求割一分。而負圜分角、與所設直線角等。 
Let ABC be the given circle, and the angle at D the given rectilineal angle;  thus it is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilineal angle, the angle at D. 
   
   
Let EF be drawn touching ABC at the point B,  and on the straight line FB, and at the point B on it, let the angle FBC be constructed equal to the angle at D. [I. 23] 
   
   
Then, since a straight line EF touches the circle ABC,  and BC has been drawn across from the point of contact at B,  the angle FBC is equal to the angle constructed in the alternate segment BAC. [III. 32]  But the angle FBC is equal to the angle at D;  therefore the angle in the segment BAC is equal to the angle at D. 
         
         
Therefore from the given circle ABC the segment BAC. has been cut off admitting an angle equal to the given rectilineal angle, the angle at D.  Q. E. F. 
   
   
PROPOSITION 35. 
THEOR. 29. PROPOS. 35. 
第三十五題 
If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other. 
SI in circulo duæ rectæ lineæ sese mutuo secuerint, rectangulum comprehensum sub segmentis vnius, æquale est ei, quod sub segmentis alterius comprehenditur, rectangulo. 
圜內兩直線。交而相分。各兩分線矩內直角形、等。 
For in the circle ABCD let the two straight lines AC, BD cut one another at the point E;  I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB. 
   
   
If now AC, BD are through the centre, so that E is the centre of the circle ABCD,  it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB. 
   
   
Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F;  from F let FG, FH be drawn perpendicular to the straight lines AC, DB,  and let FB, FC, FE be joined. 
     
     
Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles,  it also bisects it; [III. 3]  therefore AG is equal to GC.  Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E,  the rectangle contained by AE, EC together with the square on EG is equal to the square on GC; [II. 5]  Let the square on GF be added;  therefore the rectangle AE, EC together with the squares on GE, GF is equal to the squares on CG, GF.  But the square on FE is equal to the squares on EG, GF, and the square on FC is equal to the squares on CG, GF; [I. 47]  therefore the rectangle AE, EC together with the square on FE is equal to the square on FC.  And FC is equal to FB;  therefore the rectangle AE, EC together with the square on EF is equal to the square on FB.  For the same reason, also, the rectangle DE, EB together with the square on FE is equal to the square on FB.  But the rectangle AE, EC together with the square on FE was also proved equal to the square on FB;  therefore the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE.  Let the square on FE be subtracted from each;  therefore the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB. 
                               
                               
Therefore etc.   
   
   
PROPOSITION 36. 
THEOR. 30. PROPOS. 36. 
第三十六題 
If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent. 
SI extra circulum sumatur punctum aliquod, ab eoque in circulum cadant duæ rectæ lineæ, quarum altera quidem circulum secet, altera vero tangat: Quod sub tota secante, & exterius inter punctum & conuexam peripheriam assumpta comprehenditur rectangulum, æquale erit ei, quod a tangente describitur, quadrato. 
圜外、任取一點。從點、出兩直線。一切圜。一割圜。其割圜之全線、偕規外線、矩內直角形。與切圜線上直角方形等。 
For let a point D be taken outside the circle ABC, and from D let the two straight lines DCA, DB fall on the circle ABC;  let DCA cut the circle ABC and let BD touch it;  I say that the rectangle contained by AD, DC is equal to the square on DB. 
     
     
Then DCA is either through the centre or not through the centre.  First let it be through the centre, and let F be the centre of the circle ABC; let FB be joined;  therefore the angle FBD is right. [III. 18]  And, since AC has been bisected at F, and CD is added to it,  the rectangle AD, DC together with the square on FC is equal to the square on FD. [II. 6]  But FC is equal to FB;  therefore the rectangle AD, DC together with the square on FB is equal to the square on FD.  And the squares on FB, BD are equal to the square on FD; [I. 47]  therefore the rectangle AD, DC together with the square on FB is equal to the squares on FB, BD.  Let the square on FB be subtracted from each;  therefore the rectangle AD, DC which remains is equal to the square on the tangent DB. 
                     
                     
Again, let DCA not be through the centre of the circle ABC;  let the centre E be taken, and from E let EF be drawn perpendicular to AC;  let EB, EC, ED be joined.  Then the angle EBD is right. [III. 18]  And, since a straight line EF through the centre cuts a straight line AC not through the centre at right angles,  it also bisects it; [III. 3]  therefore AF is equal to FC.  Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle contained by AD, DC together with the square on FC is equal to the square on FD. [II. 6]  Let the square on FE be added to each;  therefore the rectangle AD, DC together with the squares on CF, FE is equal to the squares on FD, FE.  But the square on EC is equal to the squares on CF, FE, for the angle EFC is right; [I. 47]  and the square on ED is equal to the squares on DF, FE;  therefore the rectangle AD, DC together with the square on EC is equal to the square on ED.  And EC is equal to EB;  therefore the rectangle AD, DC together with the square on EB is equal to the square on ED.  But the squares on EB, BD are equal to the square on ED,  for the angle EBD is right; [I. 47]  therefore the rectangle AD, DC together with the square on EB is equal to the squares on EB, BD.  Let the square on EB be subtracted from each;  therefore the rectangle AD, DC which remains is equal to the square on DB. 
                                       
                                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 37. 
THEOR. 31. PROPOS. 37. 
第三十七題 
If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle. 
SI extra circulum sumatur punctum aliquod, ab eoque puncto in circulum cadant duæ rectæ lineæ, quarum altera circulum secet, altera in eum incidat; sit autem, quod sub tota secante, & exterius inter punctum, & conuexam peripheriam assumpta, comprehenditur rectangulum, æquale ei, quod ab incidente describitur, quadrato; Incidens ipsa circulum tanget. 
圜外任於一點、出兩直線。一至規外。一割圜、至規內。而割圜全線、偕割圜之規外線、矩內直角形。與至規外之線上直角方形、等。則至規外之線必切圜。 
For let a point D be taken outside the circle ABC;  from D let the two straight lines DCA, DB fall on the circle ACB;  let DCA cut the circle and DB fall on it; and let the rectangle AD, DC be equal to the square on DB.  I say that DB touches the circle ABC. 
       
       
For let DE be drawn touching ABC;  let the centre of the circle ABC be taken, and let it be F; let FE, FB, FD be joined.  Thus the angle FED is right. [III. 18]  Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36]  But the rectangle AD, DC was also equal to the square on DB;  therefore the square on DE is equal to the square on DB;  therefore DE is equal to DB.  And FE is equal to FB;  therefore the two sides DE, EF are equal to the two sides DB, BF;  and FD is the common base of the triangles;  therefore the angle DEF is equal to the angle DBF. [I. 8]  But the angle DEF is right;  therefore the angle DBF is also right.  And FB produced is a diameter;  and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16, Por.]  therefore DB touches the circle.  Similarly this can be proved to be the case even if the centre be on AC. 
                                 
                                 
Therefore etc.  Q. E. D. 
   
   
BOOK IV. 
EVCLIDIS ELEMENTVM QVARTVM. 
幾本原本第四卷之首 
DEFINITIONS. 
DEFINITIONES 
界說七則 
1. A rectilineal figure is said to be inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed. 
I. FIGVRA rectilinea in figura rectilinea inscribi dicitur, cum singuli eius figuræ, quæ inscribitur, anguli singula latera eius, in qua inscribitur, tangunt. 
第一界
直線形。居他直線形內。而此形之各角。切他形之各邊。為形內切形。
此卷將論切形在圜之內、外。及作圜在形之內、外。故解形之切在形內、及切在形外者。先以直線形為例。如前圖丁戊己角形之丁、戊、己、三角。切甲乙丙角形之甲乙、乙丙、丙甲、三邊。則丁戊己為甲乙丙之形內切形。如後圖。癸子丑角形。難癸、子、兩角。切庚辛壬角形之庚辛、壬庚、兩邊。而丑角、不切辛壬邊。則癸子丑、不可謂庚辛壬之形內切形。 
2. Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed. 
II. SIMILITER & figura circum figuram describi dicitur, cum singula eius, quæ circumscribitur, latera singulos eius figuræ angulos tetigerint, circum quam illa describitur. 
第二界
一直線形。居他直線形外。而此形之各邊。切他形之各角。為形外切形。
(p. 一八八)如第一界圖、甲乙丙、為丁己戊之形外切形。 其餘各形。倣此二例。 
3. A rectilineal figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle. 
III. FIGVRA rectilinea in circulo inscribi dicitur, cum singuli eius figuræ, quæ inscribitur, anguli tetigerint circuli peripheriam. 
第三界
直線形。之各角。切圜之界。為圜內切形。
甲乙丙形之三角。各切圜界於甲、於乙、於丙、是也。 
4. A rectilineal figure is said to be circumscribed about a circle, when each side of the circumscribed figure touches the circumference of the circle. 
IV. FIGVRA vero rectilinea circa circulum describi dicitur, cum singula latera eius, quæ circumscribitur, circuli peripheriam tangunt. 
第四界
直線形之各邊。切圜之界。為圜外切形。
甲乙丙形之三邊、切圜界於丁、於己、於戊、是也。 
5. Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed. 
V. SIMILITER & circulus in figura rectilinea inscribi dicitur, cum circuli peripheria singula latera tangit eius figuræ, cui inscribitur. 
第五界
圜之界。切直線形之各邊。為形內切圜。
同第四界圖 
6. A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed. 
VI. CIRCVLVS autem circum figuram describi dicitur, cum circuli peripheria singulos tangit eius figuræ, quam circumscribit, angulos. 
第六界
圜之界。切直線形之各角。為形外切圜。
同第三界圖 
7. A straight line is said to be fitted into a circle when its extremities are on the circumference of the circle. 
VII. RECTA linea in circulo accommodari, seu coaptari dicitur, cum eius extrema in circuli peripheria fuerint. 
第七界
直線之兩界。各抵圜界。為合圜線。
甲乙線兩界。各抵甲乙丙圜之界。為合圜線。若丙抵圜而丁不至。及戊之兩俱不至。不為合圜線。 
PROPOSITION 1. 
PROBL. 1. PROPOS. 1. 
幾何原本第四卷 本篇論圜內外形 計十六題
第一題 
Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle. 
IN dato circulo rectam lineam accommodare æqualem datæ rectæ lineæ, quæ circuli diametro non sit maior. 
有圜。求作合圜線。與所設線等。此設線。不大於圜之徑線。 
Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle;  thus it is required to fit into the circle ABC a straight line equal to the straight line D. 
   
   
Let a diameter BC of the circle ABC be drawn.  Then, if BC is equal to D, that which was enjoined will have been done;  for BC has been fitted into the circle ABC equal to the straight line D.  But, if BC is greater than D, let CE be made equal to D,  and with centre C and distance CE let the circle EAF be described;  let CA be joined. 
           
           
Then, since the point C is the centre of the circle EAF,  CA is equal to CE.  But CE is equal to D;  therefore D is also equal to CA. 
       
       
Therefore into the given circle ABC there has been fitted CA equal to the given straight line D.   
   
   
PROPOSITION 2. 
PROBL. 2. PROPOS. 2. 
第二題 
In a given circle to inscribe a triangle equiangular with a given triangle. 
IN dato circulo triangulum describere dato triangulo æquiangulum. 
有圜。求作圜內三角切形。與所設三角形等角。 
Let ABC be the given circle, and DEF the given triangle;  thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF. 
   
   
Let GH be drawn touching the circle ABC at A [III. 16, Por.];  on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF,  and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23]  let BC be joined. 
       
       
Then, since a straight line AH touches the circle ABC,  and from the point of contact at A the straight line AC is drawn across in the circle,  therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32]  But the angle HAC is equal to the angle DEF;  therefore the angle ABC is also equal to the angle DEF.  For the same reason the angle ACB is also equal to the angle DFE;  therefore the remaining angle BAC is also equal to the remaining angle EDF. [I. 32]   
               
               
Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle.  Q. E. F. 
   
   
PROPOSITION 3. 
PROBL. 3. PROPOS. 3. 
第三題 
About a given circle to circumscribe a triangle equiangular with a given triangle. 
CIRCA datum circulum triangulum describere dato triangulo æquiangulum. 
有圜。求作圜外三角切形。與所設三角形等角。 
Let ABC be the given circle, and DEF the given triangle;  thus it is required to circumscribe about the circle ABC a triangle equiangular with the triangle DEF. 
   
   
Let EF be produced in both directions to the points G, H,  let the centre K of the circle ABC be taken [III. 1],  and let the straight line KB be drawn across at random;  on the straight line KB, and at the point K on it,  let the angle BKA be constructed equal to the angle DEG, and the angle BKC equal to the angle DFH; [I. 23]  and through the points A, B, C let LAM, MBN, NCL be drawn touching the circle ABC. [III. 16, Por.] 
           
           
Now, since LM, MN, NL touch the circle ABC at the points A, B, C, and KA, KB, KC have been joined from the centre K to the points A, B, C,  therefore the angles at the points A, B, C are right. [III. 18]  And, since the four angles of the quadrilateral AMBK are equal to four right angles,  inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM, KBM are right,  therefore the remaining angles AKB, AMB are equal to two right angles.  But the angles DEG, DEF are also equal to two right angles; [I. 13]  therefore the angles AKB, AMB are equal to the angles DEG, DEF,  of which the angle AKB is equal to the angle DEG;  therefore the angle AMB which remains is equal to the angle DEF which remains.  Similarly it can be proved that the angle LNB is also equal to the angle DFE;  therefore the remaining angle MLN is equal to the angle EDF. [I. 32]  Therefore the triangle LMN is equiangular with the triangle DEF;  and it has been circumscribed about the circle ABC. 
                         
                         
Therefore about a given circle there has been circumscribed a triangle equiangular with the given triangle.  Q. E. F. 
   
   
PROPOSITION 4. 
PROBL. 4. PROPOS. 4. 
第四題 
In a given triangle to inscribe a circle. 
IN dato triangulo circulum inscribere. 
三角形。求作形內切圜。 
Let ABC be the given triangle;  thus it is required to inscribe a circle in the triangle ABC. 
   
   
Let the angles ABC, ACB be bisected by the straight lines BD, CD [I. 9],  and let these meet one another at the point D;  from D let DE, DF, DG be drawn perpendicular to the straight lines AB, BC, CA. 
     
     
Now, since the angle ABD is equal to the angle CBD,  and the right angle BED is also equal to the right angle BFD,  EBD, FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that subtending one of the equal angles, which is BD common to the triangles;  therefore they will also have the remaining sides equal to the remaining sides; [I. 26]  therefore DE is equal to DF.  For the same reason DG is also equal to DF.  Therefore the three straight lines DE, DF, DG are equal to one another;  therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right.  For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle: which was proved absurd; [III. 16]  therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA;  therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5]  Let it be inscribed, as FGE. 
                       
                       
Therefore in the given triangle ABC the circle EFG has been inscribed.  Q. E. F. 
   
   
PROPOSITION 5. 
PROBL. 5. PROPOS. 5. 
第五題 
About a given triangle to circumscribe a circle. 
CIRCA datum triangulum circulum describere. 
三角形。求作形外切圜。 
Let ABC be the given triangle;  thus it is required to circumscribe a circle about the given triangle ABC. 
SIT circulus describendus circa datum triangulum ABC.    
法曰。甲乙丙角形。求作形外切圜。   
Let the straight lines AB, AC be bisected at the points D, E [I. 10],  and from the points D, E let DF, EF be drawn at right angles to AB, AC;  they will then meet within the triangle ABC, or on the straight line BC, or outside BC. 
Diuidanturduo latera AB, AC, (quae in triangulo rectangulo, vel obtusangulo sumenda sunt facilitatis gratia, circa rectum, vel obtusum angulum, quamvis hoc non sit omnino necessarium, sed duo quaeuis latera bifariam possint secari) bifariam in D, et E, punctis, ex quibus educantur DF, EF, perpendiculares ad dicta latera, coeuntes in F, (Quod enim coeaht, patet. Nam si ducta esset recta DE, fierent anguli FDE, FED, duobus rectis minores.) eritque F, vel intra triangulum, vel in latere BC, vel extra triangulum.     
先平分兩邊。若形是直角、鈍角。則分直角鈍角、之兩旁邊。於丁於戊。次於丁、戊、上各作垂線。為己丁、己戊。而相遇於己。若自丁至戊、作直線。卽己丁戊角形之己丁戊、己戊丁、兩角。小於兩直角。故丁己、戊己、兩線必相遇。其己點或在形內。或在形外。     
First let them meet within at F  and let FB, FC, FA be joined.  Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4]  Similarly we can prove that CF is also equal to AF;  so that FB is also equal to FC;  therefore the three straight lines FA, FB, FC are equal to one another.  Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points,  and the circle will have been circumscribed about the triangle ABC.  Let it be circumscribed, as ABC. 
Ducantur rectae FA, FB, FC. Quoniam igitur latera AD, DF, trianguli ADF, aequalia sunt lateribus BD, DF, trianguli BDF, et anguli ad D, recti; erunt bases FA, FB, aequales. 109  
109. 4. primi. 
    Eodem modo erunt FA, FC, aequales.  Cum ergo tres rectae FA, FB, FC, sint aequales,   circulus descriprus ex F, ad interuallum FA, transibit quoque per puncta B, et C.       
俱作己甲、己乙、己丙、三線。或在乙丙邊上。止作己甲線。其甲丁己角形之甲丁。與乙丁己角形之乙丁。兩腰等。丁己同腰。而丁之兩旁角、俱直角。卽甲己、己乙、兩底必等。一卷四  依顯甲己戊、丙己戊、兩形之甲己、己丙、兩底亦等。則己甲、己乙、己丙、三線俱等。末作圜。以己為心。甲為界。必切丙乙、而為角形之形外切圜。               
Next, let DF, EF meet on the straight line BC at F,  as is the case in the second figure;  and let AF be joined.  Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC. 
       
       
Again, let DF, EF meet outside the triangle ABC at F,  as is the case in the third figure,  and let AF, BF, CF be joined.  Then again, since AD is equal to DB, and DF is common and at right angles,  therefore the base AF is equal to the base BF. [I. 4]  Similarly we can prove that CF is also equal to AF;  so that BF is also equal to FC;  therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points,  and will have been circumscribed about the triangle ABC. 
                 
                 
Therefore about the given triangle a circle has been circumscribed.  Q. E. F. 
Circa datum ergo triangulum circulum descripsimus.  Quod erat faciendum. 
   
And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31] 
HINC manifestum est, si centrum intra triangulum cadat,110 omnes angulos esse acutos, quoniam omnes sunt in maiori segmento circuli: si vero sit in latere BC,111 angulum BAC, ei lateri oppositum, esse rectum, quod sit in semicirculo: si denique cadat extra triangulum,112 angulum oppositum BAC, obtusum esse, cum sit in minori segmento circuli. 
一系。若圜心在三角形內。卽三角形為銳角形。何者。每角在圜大分之上故。若在一邊之上。卽為直角 形。若在形外。卽為鈍角形。 
 
CONTRA vero perspicuum est, si triangulum fuerit acutangulum, cadere intra triangulum: si rectangulum, in latus recto angulo oppositum: si denique obtusangulum fuerit, extra triangulum. Quod quidem facile ostendetur, ducendo ad incommodum aliquod, sive absurdum. Quiasi in acutangulo caderet centrum in vnum latus, esset angulus ei oppofitus rectus: si vero extra, esset idem angulus obtusus. Item si in rectangulo centrum caderet intra, essent omnes anguli acuti, si vero extra, esset angulus oppositus obtusus. Denique si in triangulo obtusangulo caderet in vnum latus, esset angulus ei oppositus, rectus, si vero intra, omnes anguli essent acuti. Quae omnia ex priori parte huius corollarium colliguntur, et pugnant cum hypothesi. 
二系。若三角形為銳角形。卽圜心必在形內。若直角形。必在一邊之上。若鈍角形。必在形外。 
 
COLLIGITVR etiam ex hoc problemate, quanam arte describendus sit circulus, qui per duta tria puncta non in vna recta linea existentia transeat. Nam si data puncta tribus rectis iungantur, ut constituatur triangulum, facile ciica ipsum circulus describetur, ut hac propositione traditum est. Quod tamen facilius efficietur praexiilla, quam tradidimus propos.25.lib.3. Sint enim datis tria puncta A, B, C; Ex A, et B, quouis interuallo eodem duo arcus describantur se intersecantes in D, et E, punctis, per quae recta linea ducatur DH. Item ex A, et C, quouis alio interuallo eodem, vel etiam, si placet, priori illo, alij duo arcus delineentur secantes sese in F, et G, punctis, per quae recta ducatur FH, secans rectam DH, in H. Dico H, esse centrum circuli transeuntis per data puncta A, B, et C. Nam si ducerentur rectae AB, AC, BC, diuiderentur latera AB, AC, trianguli ABC, bifariam a rectis DH, FH, ceu demonstratum est in praxi illa propos.25.lib.3. Quare ut in hoc 5. problemate Euclides ostendit, H, erit centrum circuli circa triangulum ABC, descripti. Quod est propositum. 
增。從此推得一法。任設三點。不在一直線。可作一過三點之圜。其法、先以三點作三直線相聯。成三角形。次依前作。
其用法。甲、乙、丙、三點。先以甲、乙、兩點各自為心。相向、各任作圜分。令兩圜分相交於丁、於戊。次甲、丙、兩點。亦如之。令兩圜分相交於己、於庚。末作丁戊、己庚、兩線、各引長之。令相交於辛。卽辛為圜之心。 論見三卷二十五增。 
PROPOSITION 6. 
PROBL. 6. PROPOS. 6. 
第六題 
In a given circle to inscribe a square. 
IN dato circulo quadratum describere. 
有圜。求作內切圜直角方形。 
Let ABCD be the given circle;  thus it is required to inscribe a square in the circle ABCD. 
   
   
Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and let AB, BC, CD, DA be joined. 
 
 
Then, since BE is equal to ED,  for E is the centre,  and EA is common and at right angles,  therefore the base AB is equal to the base AD. [I. 4]  For the same reason each of the straight lines BC, CD is also equal to each of the straight lines AB, AD;  therefore the quadrilateral ABCD is equilateral.  I say next that it is also right-angled.  For, since the straight line BD is a diameter of the circle ABCD,  therefore BAD is a semicircle;  therefore the angle BAD is right. [III. 31]  For the same reason each of the angles ABC, BCD, CDA is also right;  therefore the quadrilateral ABCD is right-angled.  But it was also proved equilateral;  therefore it is a square; [I. Def. 22]  and it has been inscribed in the circle ABCD. 
                             
                             
Therefore in the given circle the square ABCD has been inscribed.  Q. E. F. 
   
   
PROPOSITION 7. 
PROBL. 7. PROPOS. 7. 
第七題 
About a given circle to circumscribe a square. 
CIRCA datum circulum quadratum describere. 
有圜。求作外切圜直角方形。 
Let ABCD be the given circle;  thus it is required to circumscribe a square about the circle ABCD. 
   
   
Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and through the points A, B, C, D let FG, GH, HK, KF be drawn touching the circle ABCD. [III. 16, Por.] 
 
 
Then, since FG touches the circle ABCD,  and EA has been joined from the centre E to the point of contact at A,  therefore the angles at A are right. [III. 18]  For the same reason the angles at the points B, C, D are also right.  Now, since the angle AEB is right,  and the angle EBG is also right,  therefore GH is parailel to AC. [I. 28]  For the same reason AC is also parallel to FK,  so that GH is also parallel to FK. [I. 30]  Similarly we can prove that each of the straight lines GF, HK is parallel to BED.  Therefore GK, GC, AK, FB, BK are parallelograms;  therefore GF is equal to HK, and GH to FK. [I. 34]  And, since AC is equal to BD,  and AC is also equal to each of the straight lines GH, FK,  while BD is equal to each of the straight lines GF, HK, [I. 34]  therefore the quadrilateral FGHK is equilateral.  I say next that it is also right-angled.  For, since GBEA is a parallelogram,  and the angle AEB is right,  therefore the angle AGB is also right. [I. 34]  Similarly we can prove that the angles at H, K, F are also right.  Therefore FGHK is right-angled.  But it was also proved equilateral;  therefore it is a square;  and it has been circumscribed about the circle ABCD. 
                                                 
                                                 
Therefore about the given circle a square has been circumscribed.  Q. E. F. 
   
   
PROPOSITION 8. 
PROBL. 8. PROPOS. 8. 
第八題 
In a given square to inscribe a circle. 
IN dato quadrato circulum describere. 
直角方形。求作形內切圜。 
Let ABCD be the given square;  thus it is required to inscribe a circle in the given square ABCD. 
   
   
Let the straight lines AD, AB be bisected at the points E, F respectively [I. 10],  through E let EH be drawn parallel to either AB or CD,  and through F let FK be drawn parallel to either AD or BC; [I. 31]  therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are evidently equal. [I. 34]  Now, since AD is equal to AB, and AE is half of AD, and AF half of AB,  therefore AE is equal to AF,  so that the opposite sides are also equal;  therefore FG is equal to GE.  Similarly we can prove that each of the straight lines GH, GK is equal to each of the straight lines FG, GE;  therefore the four straight lines GE, GF, GH, GK are equal to one another.  Therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will pass also through the remaining points.  And it will touch the straight lines AB, BC, CD, DA, because the angles at E, F, H, K are right.  For, if the circle cuts AB, BC, CD, DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle : which was proved absurd; [III. 16]  therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will not cut the straight lines AB, BC, CD, DA.  Therefore it will touch them, and will have been inscribed in the square ABCD. 
                             
                             
Therefore in the given square a circle has been inscribed.  Q. E. F. 
   
   
PROPOSITION 9. 
PROBL. 9. PROPOS. 9. 
第九題 
About a given square to circumscribe a circle. 
CIRCA datum quadratum circulum describere. 
直角方形。求作形外切圜。 
Let ABCD be the given square;  thus it is required to circumscribe a circle about the square ABCD. 
   
   
For let AC, BD be joined, and let them cut one another at E. 
 
 
Then, since DA is equal to AB, and AC is common, therefore the two sides DA, AC are equal to the two sides BA, AC;  and the base DC is equal to the base BC;  therefore the angle DAC is equal to the angle BAC. [I. 8]  Therefore the angle DAB is bisected by AC.  Similarly we can prove that each of the angles ABC, BCD, CDA is bisected by the straight lines AC, DB.  Now, since the angle DAB is equal to the angle ABC,  and the angle EAB is half the angle DAB, and the angle EBA half the angle ABC,  therefore the angle EAB is also equal to the angle EBA;  so that the side EA is also equal to EB. [I. 6]  Similarly we can prove that each of the straight lines EA, EB is equal to each of the straight lines EC, ED.  Therefore the four straight lines EA, EB, EC, ED are equal to one another.  Therefore the circle described with centre E and distance one of the straight lines EA, EB, EC, ED will pass also through the remaining points; and it will have been circumscribed about the square ABCD.  Let it be circumscribed, as ABCD. 
                         
                         
Therefore about the given square a circle has been circumscribed.  Q. E. F. 
   
   
PROPOSITION 10. 
PROBL. 10. PROPOS. 10. 
第十題 
To construct an isosceles triangle having each of the angles at the base double of the remaining one. 
ISOSCELES triangulum constituere; quod habeat vtrumque eorum, qui ad basin sunt, angulorum, duplum reliqui. 
求作兩邊等三角形。而底上兩角。各倍大於腰間角。 
Let any straight line AB be set out, and let it be cut at the point C so that the rectangle contained by AB, BC is equal to the square on CA; [II. 11]  with centre A and distance AB let the circle BDE be described,  and let there be fitted in the circle BDE the straight line BD equal to the straight line AC which is not greater than the diameter of the circle BDE. [IV. 1]  Let AD, DC be joined, and let the circle ACD be circumscribed about the triangle ACD. [IV. 5] 
       
       
Then, since the rectangle AB, BC is equal to the square on AC,  and AC is equal to BD,  therefore the rectangle AB, BC is equal to the square on BD.  And, since a point B has been taken outside the circle ACD,  and from B the two straight lines BA, BD have fallen on the circle ACD  and one of them cuts it, while the other falls on it,  and the rectangle AB, BC is equal to the square on BD,  therefore BD touches the circle ACD. [III. 37]  Since, then, BD touches it, and DC is drawn across from the point of contact at D,  therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32]  Since, then, the angle BDC is equal to the angle DAC,  let the angle CDA be added to each;  therefore the whole angle BDA is equal to the two angles CDA, DAC.  But the exterior angle BCD is equal to the angles CDA, DAC; [I. 32]  therefore the angle BDA is also equal to the angle BCD.  But the angle BDA is equal to the angle CBD,  since the side AD is also equal to AB; [I. 5]  so that the angle DBA is also equal to the angle BCD.  Therefore the three angles BDA, DBA, BCD are equal to one another.  And, since the angle DBC is equal to the angle BCD,  the side BD is also equal to the side DC. [I. 6]  But BD is by hypothesis equal to CA;  therefore CA is also equal to CD,  so that the angle CDA is also equal to the angle DAC; [I. 5]  therefore the angles CDA, DAC are double of the angle DAC.  But the angle BCD is equal to the angles CDA, DAC;  therefore the angle BCD is also double of the angle CAD.  But the angle BCD is equal to each of the angles BDA, DBA;  therefore each of the angles BDA, DBA is also double of the angle DAB. 
                                                         
                                                         
Therefore the isosceles triangle ABD has been constructed having each of the angles at the base DB double of the remaining one.  Q. E. F. 
   
   
PROPOSITION 11. 
PROBL. 11. PROPOS. 11. 
第十一題 
In a given circle to inscribe an equilateral and equiangular pentagon. 
IN dato circulo, pentagonum æquilaterum, & æquiangulum inscribere. 
有圜。求作圜內五邊切形。其形等邊、等角。 
Let ABCDE be the given circle;  thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon. 
   
   
Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10]  let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH,  so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2]  therefore each of the angles ACD, CDA is also double of the angle CAD.  Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9],  and let AB, BC, DE, EA be joined. 
           
           
Then, since each of the angles ACD, CDA is double of the angle CAD,  and they have been bisected by the straight lines CE, DB,  therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another.  But equal angles stand on equal circumferences; [III. 26]  therefore the five circumferences AB, BC, CD, DE, EA are equal to one another.  But equal circumferences are subtended by equal straight lines; [III. 29]  therefore the five straight lines AB, BC, CD, DE, EA are equal to one another;  therefore the pentagon ABCDE is equilateral.  I say next that it is also equiangular.  For, since the circumference AB is equal to the circumference DE, let BCD be added to each;  therefore the whole circumference ABCD is equal to the whole circumference EDCB.  And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB;  therefore the angle BAE is also equal to the angle AED. [III. 27]  For the same reason each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED;  therefore the pentagon ABCDE is equiangular.  But it was also proved equilateral; 
                               
                               
therefore in the given circle an equilateral and equiangular pentagon has been inscribed.  Q. E. F. 
   
   
PROPOSITION 12. 
PROBL. 12. PROPOS. 12. 
第十二題 
About a given circle to circumscribe an equilateral and equiangular pentagon. 
CIRCA datum circulum, pentagonum æquilaterum, & æquiangulum describere. 
有圜。求作圜外五邊切形。其形等邊、等角。 
Let ABCDE be the given circle;  thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE. 
   
   
Let A, B, C, D, E be conceived to be the angular points of the inscribed pentagon,  so that the circumferences AB, BC, CD, DE, EA are equal; [IV. 11]  through A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle; [III. 16, Por.]  let the centre F of the circle ABCDE be taken [III. 1],  and let FB, FK, FC, FL, FD be joined. 
         
         
Then, since the straight line KL touches the circle ABCDE at C,  and FC has been joined from the centre F to the point of contact at C,  therefore FC is perpendicular to KL; [III. 18]  therefore each of the angles at C is right.  For the same reason the angles at the points B, D are also right.  And, since the angle FCK is right,  therefore the square on FK is equal to the squares on FC, CK.  For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK;  so that the squares on FC, CK are equal to the squares on FB, BK,  of which the square on FC is equal to the square on FB;  therefore the square on CK which remains is equal to the square on BK.  Therefore BK is equal to CK.  And, since FB is equal to FC, and FK common,  the two sides BF, FK are equal to the two sides CF, FK;  and the base BK equal to the base CK;  therefore the angle BFK is equal to the angle KFC, [I. 8]  and the angle BKF to the angle FKC.  Therefore the angle BFC is double of the angle KFC,  and the angle BKC of the angle FKC.  For the same reason the angle CFD is also double of the angle CFL,  and the angle DLC of the angle FLC.  Now, since the circumference BC is equal to CD,  the angle BFC is also equal to the angle CFD. [III. 27]  And the angle BFC is double of the angle KFC,  and the angle DFC of the angle LFC;  therefore the angle KFC is also equal to the angle LFC.  But the angle FCK is also equal to the angle FCL;  therefore FKC, FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them;  therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26]  therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC.  And, since KC is equal to CL, therefore KL is double of KC.  For the same reason it can be proved that HK is also double of BK.  And BK is equal to KC;  therefore HK is also equal to KL.  Similarly each of the straight lines HG, GM, ML can also be proved equal to each of the straight lines HK, KL;  therefore the pentagon GHKLM is equilateral.  I say next that it is also equiangular.  For, since the angle FKC is equal to the angle FLC,  and the angle HKL was proved double of the angle FKC,  and the angle KLM double of the angle FLC,  therefore the angle HKL is also equal to the angle KLM.  Similarly each of the angles KHG, HGM, GML can also be proved equal to each of the angles HKL, KLM;  therefore the five angles GHK, HKL, KLM, LMG, MGH are equal to one another.  Therefore the pentagon GHKLM is equiangular.  And it was also proved equilateral;  and it has been circumscribed about the circle ABCDE. 
                                                                                           
                                                                                           
No Egnlish  Q. E. F. 
   
   
PROPOSITION 13. 
PROBL. 13. PROPOS. 13. 
第十三題 
In a given pentagon, which is equilateral and equiangular, to inscribe a circle. 
IN dato pentagono æquilatero & æquiangulo circulum inscribere. 
五邊等邊、等角形。求作形內切圜。 
Let ABCDE be the given equilateral and equiangular pentagon;  thus it is required to inscribe a circle in the pentagon ABCDE. 
   
   
For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively;  and from the point F, at which the straight lines CF, DF meet one another,  let the straight lines FB, FA, FE be joined.  Then, since BC is equal to CD, and CF common, the two sides BC, CF are equal to the two sides DC, CF;  and the angle BCF is equal to the angle DCF;  therefore the base BF is equal to the base DF,  and the triangle BCF is equal to the triangle DCF,  and the remaining angles will be equal to the remaining angles,  namely those which the equal sides subtend. [I. 4]  Therefore the angle CBF is equal to the angle CDF.  And, since the angle CDE is double of the angle CDF,  and the angle CDE is equal to the angle ABC, while the angle CDF is equal to the angle CBF;  therefore the angle CBA is also double of the angle CBF;  therefore the angle ABF is equal to the angle FBC;  therefore the angle ABC has been bisected by the straight line BF.  Similarly it can be proved that the angles BAE, AED have also been bisected by the straight lines FA, FE respectively.  Now let FG, FH, FK, FL, FM be drawn from the point F perpendicular to the straight lines AB, BC, CD, DE, EA.  Then, since the angle HCF is equal to the angle KCF,  and the right angle FHC is also equal to the angle FKC,  FHC, FKC are two triangles having two angles equal to two angles and one side equal to one side,  namely FC which is common to them and subtends one of the equal angles;  therefore they will also have the remaining sides equal to the remaining sides; [I. 26]  therefore the perpendicular FH is equal to the perpendicular FK.  Similarly it can be proved that each of the straight lines FL, FM, FG is also equal to each of the straight lines FH, FK;  therefore the five straight lines FG, FH, FK, FL, FM are equal to one another.  Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will pass also through the remaining points; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right.  For, if it does not touch them, but cuts them,  it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [III. 16]  Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will not cut the straight lines AB, BC, CD, DE, EA;  therefore it will touch them.  Let it be described, as GHKLM. 
                                                             
                                                             
Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed.  Q. E. F. 
   
   
PROPOSITION 14. 
PROBL. 14. PROPOS. 14. 
第十四 
About a given pentagon, which is equilateral and equiangular, to circumscribe a circle. 
CIRCA datum pentagonum æquilaterum, & æquiangulum circulum describere. 
五邊等邊、等角形。求作形外切圜。 
Let ABCDE be the given pentagon, which is equilateral and equiangular;  thus it is required to circumscribe a circle about the pentagon ABCDE. 
   
   
Let the angles BCD, CDE be bisected by the straight lines CF, DF respectively,  and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E.  Then in manner similar to the preceding  it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively.  Now, since the angle BCD is equal to the angle CDE,  and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE,  therefore the angle FCD is also equal to the angle CDF,  so that the side FC is also equal to the side FD. [I. 6]  Similarly it can be proved that each of the straight lines FB, FA, FE is also equal to each of the straight lines FC, FD;  therefore the five straight lines FA, FB, FC, FD, FE are equal to one another.  Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points, and will have been circumscribed.  Let it be circumscribed, and let it be ABCDE. 
                       
                       
Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed.  Q. E. F. 
   
   
PROPOSITION 15. 
PROBL. 15. PROPOS. 15. 
第十五題 
In a given circle to inscribe an equilateral and equiangular hexagon. 
IN dato circulo, hexagonum & æquilaterum & æquiangulum inscribere. 
有圜。求作圜內六邊切形。其形等邊等角。 
Let ABCDEF be the given circle;  thus it is required to inscribe an equilateral and equiangular hexagon in the circle ABCDEF. 
   
   
Let the diameter AD of the circle ABCDEF be drawn;  let the centre G of the circle be taken,  and with centre D and distance DG let the circle EGCH be described;  let EG, CG be joined and carried through to the points B, F,  and let AB, BC, CD, DE, EF, FA be joined.  I say that the hexagon ABCDEF is equilateral and equiangular. 
           
           
For, since the point G is the centre of the circle ABCDEF,  GE is equal to GD.  Again, since the point D is the centre of the circle GCH,  DE is equal to DG.  But GE was proved equal to GD;  therefore GE is also equal to ED;  therefore the triangle EGD is equilateral;  and therefore its three angles EGD, GDE, DEG are equal to one another,  inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [I. 5]  And the three angles of the triangle are equal to two right angles; [I. 32]  therefore the angle EGD is one-third of two right angles.  Similarly, the angle DGC can also be proved to be onethird of two right angles.  And, since the straight line CG standing on EB makes the adjacent angles EGC, CGB equal to two right angles,  therefore the remaining angle CGB is also one-third of two right angles.  Therefore the angles EGD, DGC, CGB are equal to one another;  so that the angles vertical to them, the angles BGA, AGF, FGE are equal. [I. 15]  Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another.  But equal angles stand on equal circumferences; [III. 26]  therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another.  And equal circumferences are subtended by equal straight lines; [III. 29]  therefore the six straight lines are equal to one another;  therefore the hexagon ABCDEF is equilateral.  I say next that it is also equiangular.  For, since the circumference FA is equal to the circumference ED,  let the circumference ABCD be added to each;  therefore the whole FABCD is equal to the whole EDCBA;  and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA;  therefore the angle AFE is equal to the angle DEF. [III. 27]  Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED;  therefore the hexagon ABCDEF is equiangular.  But it was also proved equilateral;  and it has been inscribed in the circle ABCDEF. 
                                                               
                                                               
Therefore in the given circle an equilateral and equiangular hexagon has been inscribed.  Q. E. F. 
   
   
PORISM.
From this it is manifest that the side of the hexagon is equal to the radius of the circle.
And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon.
And further by means similar to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it.
Q. E. F. 
 
一系。凡圜之半徑。為六分圜之一之分弦。何者。庚丁、與丁丙等、故。故一開規為圜不動而可六平分之。
二系。依前十二、十三、十四題。可作六邊等邊等角形。在圜之外。又六邊等邊等角形內。可作切圜。又六邊等邊等角形外。可作切圜。 
PROPOSITION 16. 
PROBL. 16. PROPOS. 16. 
第十六題 
In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular. 
IN dato circulo, quintidecagonum & æquilaterum, & æquiangulum describere. 
有圜。求作圜內十五邊切形。其形等邊。等角。 
Let ABCD be the given circle;  thus it is required to inscribe in the circle ABCD a fifteenangled figure which shall be both equilateral and equiangular. 
   
   
In the circle ABCD let there be inscribed a side AC of the equilateral triangle inscribed in it, and a side AB of an equilateral pentagon;  therefore, of the equal segments of which there are fifteen in the circle ABCD,  there will be five in the circumference ABC which is one-third of the circle,  and there will be three in the circumference AB which is one-fifth of the circle;  therefore in the remainder BC there will be two of the equal segments.  Let BC be bisected at E; [III. 30]  therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD. 
             
             
If therefore we join BE, EC and fit into the circle ABCD straight lines equal to them and in contiguity,  a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it.  Q. E. F. 
     
     
And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle,  there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular.  And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteenangled figure and circumscribe one about it.  Q. E. F. 
       
      一系。依前十二、十三、十四題。可作外切圜十五邊形。又十五邊形內。可作切圜。又十五邊形外。可作切圜。
注曰。依此法。可設一法作無量數形。如本題圖。甲乙圜分。為三分圜之一。卽命三。甲戊圜分。為五分圜之一。卽命五。三與五相乘。得十五。卽知此兩分法。可作十五邊形。又如甲乙命三。甲戊命五。三與五較得二。卽知戊乙得十五分之二。因分戊乙為兩平分。得壬乙線為十五分之一。可作內切圜十五邊形也。以此法為例。作後題。
增題。若圜內從一點、設切圜兩不等等邊等角形之各一邊。此兩邊。一為若干分圜之一。一為若干分圜之一。此兩若干分相乘之數。卽後作形之邊數。此兩若干分之較數。卽兩邊相距之圜分、所得後作形邊數內之分。
法曰。甲乙丙丁戊圜內。從甲點、作數形之各一邊。如甲乙為六邊形之一邊。甲丙為五邊形之一邊。甲丁為四邊形之一邊。甲戊為三邊形之一邊。甲乙命六。甲丙命五。較數一。卽乙丙圜分、為所作三十邊等邊等角形之一邊。何者。五六相乘為三十。故當作三十邊也。較數一。故當為一邊也。(p. 二○八)
論曰。甲乙圜分。為六分圜之一。卽得三十分圜之五。而甲丙為五分圜之一。卽得三十分圜之六。則乙丙得三十分圜之一也。依顯乙丁為二十四邊形之二邊也。何者。甲乙命六。甲丁命四。六乘四得二十四也。又較數二也。依顯乙戊為十八邊形之三邊也。丙丁為二十邊形之一邊也。丙戊為十五邊形之二邊也。丁戊為十二邊形之一邊也。
二系。凡作形於圜之內。等邊、則等角。何者。形之角。所乘之圜分皆等、故。三卷 \\ 廿七凡作形於圜之外。卽從圜心、作直線、抵各角。依本篇十二題可推顯各角等。
三系。凡等邊形。旣可作在圜內。卽依圜內形。可作在圜外。卽形內可作圜。卽形外亦可作圜。皆依本篇十二、十三、十四題。
四系。凡圜內有一形。欲作他形。其形邊、倍於此形邊。卽分此形一邊所合之圜分、為兩平分。而每分各作一合線。卽三邊可作六邊。四邊可作八邊倣此以至無窮。
又補題。圜內有同心圜。求作一多邊形。切大圜不至小圜。其多邊、為偶數、而等。
法曰。甲乙丙、丁戊、兩圜。同以己為心。求於甲乙丙大圜內、作多邊切形。不至丁戊小圜。其多邊、為偶數、(p. 二○九)而等。先從己心、作甲丙徑線、截丁戊圜於戊。次從戊、作庚辛、為甲戊之垂線。卽庚辛線、切丁戊圜於戊也。三卷十 \\ 六之系夫甲庚丙圜分。雖大於丙庚。若于甲庚丙、減其半甲乙。存乙丙。又減其半乙壬。存壬丙。又減其半壬癸。如是遞減。至其減餘丙癸。必小於丙庚。如下 \\ 補論卽得丙癸圜分。小於丙庚。而作丙癸合圜線。卽丙癸為所求切圜形之一邊也。次分乙壬圜分。其分數、與丙壬之分數等。次分甲乙。與乙丙分數等。分丙甲。與甲乙丙分數等。則得所求形。三卷 \\ 廿九而不至丁戊小圜。
論曰。試從癸、作癸子。為甲丙之垂線。遇甲丙於丑。其庚戊丑、癸丑戊、兩皆直角。卽庚辛、癸子、為平行線。一卷 \\ 廿八庚辛線之切丁戊圜。旣止一點。卽癸子線、更在其外。必不至丁戊矣。何況丙癸更遠於丑癸乎。依顯其餘與丙癸等邊、同度距心者。三 // 卷十 \\ 四俱不至丁戊圜也。此係十二卷第十六題。因六卷今增 \\ 題、宜藉此論。故先類附於此。
補論。其題曰。兩幾何、不等。若於大率、遞減其大半。必可使其減餘、小於元設小率。(p. 二一○)
解曰。甲乙大率。丙小率。題言於甲乙遞減其大半。至可使其減餘、小於丙。
論曰。試以丙、倍之。又倍之。至僅大於甲乙而止。為丁戊。丁戊之分。為丁己、己庚、庚戊。各與丙等也。次於甲乙減其大半甲辛。存辛乙。又減其大半辛壬。存壬乙。如是遞減。至甲乙與丁戊之分數等。夫甲辛、辛壬、壬乙。與丁己、己庚、庚戊。分數旣等。丁戊、又大於甲乙。若兩率各為兩分。而大丁戊之減丁己、止於半。小甲乙之減甲辛、為大半。卽丁戊之減餘。必大於甲乙之減餘也。若各為多分。而己戊尚多於丙者。卽又於己戊、減己庚。於辛乙、減其大半辛壬。如是遞減。卒至丁戊之末分庚戊。大於甲乙之末分壬乙也。而庚戊元與丙等。是壬乙小於丙也。
又論曰。若於甲乙遞減其半。亦同前論。何者。大丁戊所減。不大於半。則丁戊之減餘。每大於甲乙之減餘。以至末分。亦大於末分。此係十卷第一題。借 \\ 用於此。以足上論。 
BOOK V. 
 
幾何原本第五卷之首 
DEFINITIONS. 
 
界說十九則
前四卷所論。皆獨幾何也。此下二卷所論。皆自兩以上、多幾何、同例相比者也。而本卷則總說完幾何之同例相比者也。諸卷中。獨此卷以虛例相比。絕不及線、面、體、諸類也。第六卷。則論線、論角、論圜界、諸類、及諸形之同例相比者也。今先解向後所用名目。為界說十九。 
1. A magnitude is a part of a magnitude, the less of the greater, when it measures the greater. 
 
第一界
分者。幾何之幾何也。小能度大。以小為大之分。
以小幾何、度大幾何。謂之分。曰幾何之幾何者。謂非此小幾何。不能為此大幾何之分也。如一點無分。亦非幾何。卽不能為線之分也。一線無廣狹之分。非廣狹之幾何。卽不能為面之分也。一面無厚薄之分。非厚薄之幾何。卽不能為體之分也。曰能度大者。謂小幾何、度大幾何。能書大之分者也。如甲、為乙、為丙、之分。則甲為乙三分之一。為丙六分之一。無贏、不足也。若戊為丁之一、卽贏。為二、卽不足。己為丁之三、卽贏。為四、卽不足。是小不書大。則丁不能為戊、己、之分也。以數明之。若四於八、於十二、於十六、於(p. 二一二)二十、諸數。皆能盡分。無贏、不足也。若四於六、於七、於九、於十、於十八、於三十八、諸數。或贏、或不足。皆不能盡分者也。本書所論。皆指能盡分者。故稱為分。若不盡分者。當稱幾分幾何之幾。如四於六。為三分六之二。不得正名為分。不稱小度大也。不為大幾何內之小幾何也。 
2. The greater is a multiple of the less when it is measured by the less. 
 
第二界
若小幾何能度大者。則大為小之幾倍。
如第一界圖。甲與乙。能度丙。則丙為甲與乙之幾倍。若丁、戊、不能盡己之分。則己不為丁、戊、之幾倍。 
3. A ratio is a sort of relation in respect of size between two magnitudes of the same kind. 
 
第三界
比例者。兩幾何以幾何相比之理。
兩幾何者。或兩數。或兩線。或兩面。或兩體。各以同類大小相比。謂之比例。若線與面、或數與線、相比。此異類。不為比例。又若白線與黑線、熱線與冷線、相比。雖同類。不以幾何相比。亦不為比例也。
比例之說在幾何為正用。亦有借用者。如時。如音。如聲。如所如動。如稱之屬。皆以比例論之。
凡兩幾何相比。以此幾何比他幾何。則此幾何為前率。所比之他幾何為後率。如以六尺之線、比三尺之線。則六尺為前率。三尺為後率也。反用之。以三尺之線。比六尺之線。則三尺為前率。六尺為後率也。比例為用甚廣。故詳論之。如左。
凡比例有二種。有大合。有小合。以數可明者、為大合。如二十尺之線、比十尺之線、是也其非數可明者、(p. 二一三)為小合。如直角方形之兩邊、與其對角線。可以相比、而非數可明者、是也。
如上二種。又有二名。其大合者、為有兩度之線。如二十尺、比八尺、兩線為大合。則二尺、四尺、皆可兩度之者、是也。如此之類。凡數之比例。皆大合也。何者。有數之屬。或無他數可兩度者。無有一數不可兩度者。若七比九。無他數可兩度之。以一、則可兩度之也。其小合線、為無兩度之線。如直角方形之兩邊、與其對角線、為小合。卽分至萬分、以及無數。終無小線、可以盡分、能度兩率者、是也。此論詳見 \\ 十卷末題
小合之比例。至十卷詳之。本篇所論。皆大合也。
凡大合有兩種。有等者。如二十比二十。十尸之線、比十尺之線。是也。有不等者。如二十比十。八比四十。六尺之線比二尺之線。是也。
如上等者。為相同之比例。其不等者。又有兩種。有以大不等。如二十比十是也。有以小不等。如十比二十是也。大合比例之以大不等者。又有五種。一為幾倍大。二為等帶一分。三為等帶幾分。四為幾倍大帶一分。五為幾倍大帶幾分。
一為幾倍大者。謂大幾何內。有小幾何或二、或三、或十、或八也。如二十與四。是二十內。為四者五。如三十尺之線、與五尺之線。是三十尺內。為五尺者六。則二十與四。名為五倍大之比例也。三十尺與五尺。名為六倍大之比例也。倣此為名。可至無窮也。
二為等帶一分者。謂大幾何內。旣有小之一。別帶一分。此一分。或元一之半。或三分之一四分之一。以(p. 二一四)至無窮者。是也。如三與二。是三內旣有二。別帶一。一為二之半。如十二尺、之線。是十二內旣有九。別帶三。三為九三分之一。則三與二。名為等帶半也。十二尺與九尺。名為等帶三分之一也。
三為等帶幾分者。謂大幾何內。旣有小之一。別帶幾分。而此幾分、不能合為一盡分者。是也。如八與五。是八內旣有五。別帶三一。每一各為五之分。而三一不能合而為五之分也。他如十與八。其十內旣有八。別帶二一。雖每一各為八之分。與前例相似。而二一卻能為八四分之一。是為帶一分。屬在第二。不屬三也。則八與五。名為等帶三分也。又如二十二、與十六。卽名為等帶六分也。○四為幾倍大帶一分者。謂大幾何內。旣有小幾何之二、之三、之四、等。別帶一分。此一分。或元一之半。或三分、四分、之一、以至無窮者。是也。如九與四。是九內旣有二四。別帶一。一為四四分之一。則九與四。名為二倍大帶四分之一也。
五為幾倍大帶幾分者。謂大幾何內。旣有小幾何之二、之三、之四、等。別帶幾分。而此幾分。不能合為一盡分者。是也。如十一與三。是十一內旣有三三。別帶二一。每一各為三之分。而二一。不能合而為三之分也。則十一與三。名為三倍大帶二分也。
大合比例之以小不等者。亦有五種。俱與上以大不等五種。相反為名。一為反幾倍大。二為反等帶一分。三為反等帶幾分。四為反幾倍大帶一分。五為反幾倍大帶幾分。
凡比例諸種。如前所設諸數。俱有書法。書法中。有全數。有分數。全數者。如一、二、三、十、百、等。是也。分數者。(p. 二一五)如分一以二、以三、以四、等是也。書全數。依本數書之。不必立法。書分數。必有兩數。一為命分數。一為得分數。卽如分一以三而取其二。則為三分之二。卽三為命分數。二為得分數也。分一為十九而取其七。則為十九分之七。卽十九為命分數。七為得分數也。
書以大、小、不等各五種之比例。其一幾倍大以全數書之。如二十與四。為五倍大之比例。卽書五、是也。若四倍、卽書四。六倍、卽書六也。其反幾倍大。卽用分數書之。而以大比例之數、為命分之數。以一為得分之數。如大為五倍大之比例。則此書五之一、是也。若四倍、卽書四之一。六倍、卽書六之一也。
其二等帶一分之比例。有兩數。一全數。一分數。其全數恆為一。其分數。則以分率之數、為命分數。恆以一為得分數如三與二。名為等帶半。卽書一。別書二之一也。其反等帶一分。則全用分數。而以大比例之命分數、為此之得分數。以大比例之命分數、加一。為此之命分數。如大為等帶二之一。卽此書三之二也。又如等帶八分之一。反書之。卽書九之八也。又如等帶一千分之一。反書之。卽書一千○○一之一千也。
其三等帶幾分之比例。亦有兩數。一全數。一分數。其全數亦恆為一。其分數。亦以分率之數、為命分數。以所分之數、為得分數。如十與七。名為等帶三分。卽書一。別書七之三也。其反等帶幾分。亦全用分數。而以大比例之命分數、為此之得分數。以大比例之命分數、加大之得分數。為此之命分數。如大為等帶七之三。命數七。得數三。七加三為十。卽書十之七也。又如等帶二十之三。反書之。二十加三。卽書二(p. 二一六)十三之二十也。
其四幾倍大帶一分之比例。則以幾倍大之數、為全數。以分率之數、為命分數。恆以一為得分數。如二十二與七。二十二內。旣有三七。別帶一。一為七七分之一。名為三倍大帶七分之一。卽以三為全數。七為命分數。一為得分數。書三。別書七之一也。其反幾倍大帶一分。則以大比例之命分數、為此之得分數。以大之命分數、乘大之倍數。加一。為此之命分數。如大為三帶七之一。卽以七乘三、得二十一。又加一。為命分數。書二十二之七也。又如五帶九之一。反書之。九乘五、得四十五。加一、為四十六。卽書四十六之九也。
其五幾倍大帶幾分之比例。亦以幾倍大之數、為全數。以分率之數、為命分數。以所分之數、為得分數。如二十九與八。二十九內。旣有三八。別帶五一。名為三倍大帶五分。卽以三為全數。八為命分數。五為得分數。書三。別書八之五也。其反幾倍大帶幾分。則以大比例之命分數、為此之得分數。以大比例之命分數、乘大之倍數。加大之得分數。為此之命分數。如大為三帶八之五。卽以八乘三、得二十四。加五、為二十九。書二十九之八也。又如四帶五之二。卽書二十二之五也。
己上大小十種。足盡比例之凡。不得加一、減一。
第四界
兩比例之理相似。為同理之比例。(p. 二一七)
兩幾何相比。謂之比例。兩比例相比謂之同理之比例如甲與乙、兩幾何之比例。偕丙與丁、兩幾何之比例。其理相似。為同理之比例。又若戊與己、兩幾何之比例。偕己與庚、兩幾何之比例。其理相似。亦同理之比例。
凡同理之比例。有三種。有數之比例。有量法之比例。有樂律之比例。本篇所論。皆量法之比例也。量法比例。又有二種。一為連比例。連比例者。相續不斷。其中率、與前、後、兩率。遞相為比例。而中率旣為前率之後。又為後率之前。如後圖。戊與己比。己又與庚比。是也。二為斷比例。斷比例者。居中兩率一取不再用。如前圖。甲自與乙比。丙自與丁比。是也。 
4. Magnitudes are said to have a ratio to one another which are capable, when multiplied, of exceeding one another. 
 
第五界
兩幾何。倍其身而能相勝者。為有比例之幾何。
上文言為比例之幾何。必同類。然同類中。亦有無比例者。故此界顯有比例之幾何也。曰倍其身而能相勝者。如三尺之線、與八尺之線。三尺之線。三倍其身。卽大於八尺之線。是為有比例之線也。又如直角方形之一邊、與其對角線。雖非大合之比例。可以數明。而直角方形之一邊。一倍之。卽大於對角線。兩邊等三角形。其兩邊幷。 \\ 必大於一邊。見一卷二十。是亦有小合比例之線也。又圜之徑。四倍之、卽大於圜之界。則圜之徑與界。(p. 二一八)亦有小合比例之線也。圜之界、當三徑七分徑 \\ 之一弱。別見圜形書。又曲線與直線。亦有比例。如以大小兩曲線相合。為初月形。別作一直角方形。與之等六卷三十三 \\ 一增題今附卽曲直兩線相視。有大、有小。亦有比例也。又方形與圜。雖自古至今。學士無數。不能為相等之形。然兩形相視。有大、有小。亦不可謂無比例也。又直線角與曲線角。亦有比例。如上圖。直角、鈍角、銳角。皆有與曲線角等者。若第一圖。甲乙丙直角。在甲乙、乙丙、兩直線內。而其間設有甲乙丁、與丙乙戊、兩圜分角等。卽於甲乙丁角、加甲乙戊角。則丁乙戊曲線角。與甲乙丙直角等矣。依顯壬庚癸曲線角。與己庚辛鈍角等也。又依顯卯丑辰曲線角。與子丑寅銳角。各減同用之子丑、丑辰、內圜小分。卽兩角亦等也。此五者。皆疑無比例。而實有比例者也。他若有窮之線、與無窮之線。雖則同類。實無比例。何者。有窮之線。畢世倍之。不能勝無窮之線故也。又線與面。面與體。各自為類。亦無比例。何者。畢世倍線。不能及面。畢世倍面。不能及體。故也。又切圜角、與直線銳角。亦無比例。何者。依三卷十六題所說。畢世倍切邊角。不能勝至小之銳角。故也此後諸篇中。每有倍此幾何。令至勝彼幾何者。故備著其理。以需後論也。 
5. Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever be taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order. 
 
第六界
四幾何。若第一與二。偕第三與四。為同理之比例。則第一、第三、之幾倍。偕第二、第四、之幾倍。其相視。或等。或俱為大。俱為小。恆如是。
兩幾何。曷顯其能為比例乎。上第五界所說是也。兩比例。曷顯其能為同理之比例乎。此所說是也。其術通大合、小合。皆以加倍法求之。如一甲、二乙、三丙、四丁、四幾何。於一甲三丙。任加幾倍。為戊、為己。戊倍甲己倍丙。其數自相等。次於二乙四丁。任加幾倍。為庚、為辛。庚倍乙。辛倍丁。其數自相等。而戊與己。偕庚與辛。相視。或等。或俱大。或俱小。如是等、大、小、累試之恆如是。卽知一甲與二乙。偕三丙與四丁。為同理之比例也。
如初試之。甲幾倍之戊。小於乙幾倍之庚。而丙幾倍之己。亦小於丁幾倍之辛。又試之。倍甲之戊。與倍乙之庚等。而倍丙之己。亦與倍丁之辛等。三試之。倍甲之戊。大於倍乙之庚。而倍丙之己。亦大於(p. 二二○)倍丁之辛。此之謂或相等。或雖不等、而俱為大。俱為小。若累合一差。卽元設四幾何。不得為同理之比例。如下第八界所指是也。
下文所論。若言四幾何為同理之比例。卽當推顯第一、第三、之幾倍。與第二、第四、之幾倍。或等。或俱大、俱小。若許其四幾何、為同理之比例。亦如之。
>以數明之。如有四幾何。第一為三。第二為二。第三為六。第四為四。今以第一之三。第三之六。同加四倍。為十二。為二十四。次以第二之二。第四之四。同加七倍。為十四。為二十八。其倍第一之十二。旣小於倍第二之十四。而倍第三之二十四。亦小於倍第四之二十八也。又以第一之三。第三之六。同加六倍。為十八。為三十六。次以第二之二。第四之四。同加九倍。為十八。為三十六。其倍第一之十八。旣等於倍第二之十八。而倍第三之三十六。亦等於倍第四之三十六也。又以第一之三。第三之六。同加三倍。為九。為十八。次以第二之二。第四之四。同加二倍。為四。為八。其倍第一之九。旣大於倍第二之四。而倍第三之十八。亦大於倍第(p. 二二一)四之八也。若爾。或俱大、俱小。或等。累試之、皆合。則三與二。偕六與四。得為同理之比例也。
以上論四幾何者。斷比例之法也。其連比例法倣此。但連比例之中率。兩用之。旣為第二。又為第三。視此異耳。 
6. Let magnitudes which have the same ratio be called proportional. 
 
第七界
同理比例之幾何。為相稱之幾何。
甲與乙。若丙與丁。是四幾何、為同理之比例。卽四幾何、為相稱之幾何。又戊與己。若己與庚。卽三幾何、亦相稱之幾何。 
7. When, of the equimultiples, the multiple of the first magnitude exceeds the multiple of the second, but the multiple of the third does not exceed the multiple of the fourth, then the first is said to have a greater ratio to the second than the third has to the fourth. 
 
第八界
四幾何。若第一之幾倍。大於第二之幾倍。而第三之幾倍。不大於第四之幾倍。則第一與二之比例。大於第三與四之比例。
此反上第六界。而釋不同理之兩比例。其相視。曷顯為大。曷顯為小也。謂第一、第三、之幾倍。與第二、第四、之幾倍。依上累試之。其間有第一之幾倍。大(p. 二二二)於第二之幾倍。而第三之幾倍。乃或等、或小、於第四之幾倍。卽第一與二之比例。大於第三與四之比例也。如上圖。甲一、乙二、丙三、丁四。甲與丙。各三倍、為戊、己。乙與丁。各四倍、為庚辛。其甲三倍之戊。大於乙四倍之庚。而丙三倍之己。乃小於丁四倍之辛。卽甲與乙之比例。大於丙與丁也。若第一之幾倍。小於第二之幾倍。而第三之幾倍。乃或等、或大、於第四之幾倍。卽第一與二之比例。小於第三與四之比例。如是等、大、小、相戾者。但有其一。不必再試。
以數明之。中設三、二、四三、四幾何。先有第一之倍。大於第二之倍。而第三之倍。亦大於第四之倍。後復有第一之倍。大於第二之倍。而第三之倍。乃或等、或小於第四之倍。卽第一與二之比例。大於第三與四也。若以上圖之數反用之。以第一為二。第二為一。第三為四。第四為三。則第一與二之比例。小於第三與四。 
8. A proportion in three terms is the least possible. 
 
第九界
同理之比例。至少必三率。
同理之比例。必兩比例相比。如甲與乙。若丙與丁。是四率。斷比例也。若連比例之戊與己。若己與庚。則(p. 二二三)中率己、旣為戊之後。又為庚之前。是以三率當四率也。 
9. When three magnitudes are proportional, the first is said to have to the third the duplicate ratio of that which it has to the second. 
 
第十界
三幾何。為同理之連比例。則第一與三。為再加之比例。 
10. When four magnitudes are proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually, whatever be the proportion. 
 
四幾何。為同理之連比例。則第一與四。為三加之比例。倣此以至無窮。
甲、乙、丙、丁、戊、五幾何。為同理之連比例。其甲與乙。若乙與丙。乙與丙。若丙與丁。丙與丁。若丁與戊。卽一甲與三丙。視一甲與二乙。為再加之比例。又一甲與四丁。視一甲與二乙。為三加之比例。何者。甲、丁、之中。有乙、丙、兩幾何。為同理之比例、如甲與乙。故也。又一甲與五戊。視一甲與二乙。為四加之比例也。若反用之。以戊為首。則一戊與三丙為再加。與四乙為三加。與五甲為四加也。
下第六卷二十題。言此直角方形、與彼直角方形。為此形之一邊。與彼形之一邊再加之比例。何者。(p. 二二四)若作三幾何、為同理之連比例。則此直角方形、與彼直角方形。若第一幾何、與第三幾何。故也。以數明之。如此直角方形之邊、三尺。而彼直角方形之邊、一尺。卽此形邊、與彼形邊。若九、與一也夫九與一之間。有三。為同理之比例。則九、三、一、三幾何之連比例。旣有三與一、為比例。又以九比三。三比一。為再加之比例也。則彼直角方形。當為此形九分之一。不止為此形三分之一也。大略第一與二之比例。若線相比。第一與三。若平面相比。第一與四。若體相比也。第一與五。若算家三乘方。與六。若四乘 \\ 方。與七。若五乘方。倣此以至無窮。 
11. The term corresponding magnitudes is used of antecedents in relation to antecedents, and of consequents in relation to consequents. 
 
第十一界
>同理之幾何。前與前相當。後與後相當。
上文巳解同理之比例。此又解同理之幾何者。蓋一比例之兩幾何。有前、後。而同理之兩比例四幾何。有兩前、兩後。故特解言比例之論。常以前與前相當。後與後相當也。如上甲與乙。丙與丁。兩比例同理。則甲與丙相當。乙與丁相當也。戊己、己庚、兩比例同理。則己旣為前。又為後。兩相當也。如下文有兩三角形之邊相比。亦常以同理之兩邊相當。不可混也。
上文第六、第八、界說幾何之幾倍。常以一與三同倍。二與四同倍。則以第一、第三、為兩前。第二、第四、為兩後。各同理故。(p. 二二五) 
12. Alternate ratio means taking the antecedent in relation to the antecedent and the consequent in relation to the consequent. 
 
第十二界
有屬理。更前與前。更後與後。
此下說比例六理。皆後論所需也。
四幾何。甲與乙之比例。若丙與丁。今更推甲與丙。若乙與丁、為屬理。 下言屬理。皆省曰更。
此論未證證。見本卷十六。
此界之理。可施於四率同類之比例。若兩線、兩面。或兩面、兩數等。不為同類。卽不得相更也。 
13. Inverse ratio means taking the consequent as antecedent in relation to the antecedent as consequent. 
 
第十三界
有反理。取後為前。取前為後。(p. 二二六)
>甲與乙之比例。若丙與丁。今反推乙與甲。若丁與丙。為反理。
>證見本篇四之系。
此界之理。亦可施於異類之比例。 
14. Composition of a ratio means taking the antecedent together with the consequent as one in relation to the consequent by itself. 
 
第十四界
有合理。合前與後為一、而比其後。
甲乙與乙丙之比例。若丁戊與戊己。今合甲丙為一、而比乙丙。合丁己為一、而比戊己。卽推甲丙與乙丙。若丁己與戊己。是合兩前、後、率、為兩一率。而比兩後率也。
證見本卷十八。(p. 二二七) 
15. Separation of a ratio means taking the excess by which the antecedent exceeds the consequent in relation to the consequent by itself. 
 
第十五界
有分理。取前之較、而比其後。
甲乙與丙乙之比例。若丁戊與己戊。今分推甲乙之較甲丙、與丙乙。若丁戊之較丁己、與己戊。
證見本卷十七。 
16. Conversion of a ratio means taking the antecedent in relation to the excess by which the antecedent exceeds the consequent. 
 
第十六界
有轉理。以前為前。以前之較為後。
甲乙與丙乙之比例。若丁戊與己戊。今轉推甲乙與甲丙。若丁戊與丁己。(p. 二二八)
>證見本卷十九。 
17. A ratio ex aequali arises when, there being several magnitudes and another set equal to them in multitude which taken two and two are in the same proportion, as the first is to the last among the first magnitudes, so is the first to the last among the second magnitudes;
Or, in other words, it means taking the extreme terms by virtue of the removal of the intermediate terms. 
An ordered proportion arises when, as antecedent is to consequent, so is consequent to something else.1  
   
第十七界
有平理。彼此幾何。各自三以上。相為同理之連比例。則此之第一與三。若彼之第一與三。又曰。去其中。取其首尾。
甲、乙、丙、三幾何。丁、戊、己、三幾何。等數。相為同理之連比例者。甲與乙、若丁與戊。乙與丙、若戊與己也。今平推首甲、與尾丙。若首丁、與尾己。(p. 二二九)
平理之分。又有二種。如後二界。 
第十八界
有平理之序者。此之前與後。若彼之前與後。而此之後與他率。若彼之後與他率。
甲與乙。若丁與戊。而後乙、與他率丙。若後戊、與他率己。是序也今平推甲與丙。若丁與己也。此與十七界 \\ 同‧重宣序義‧以別 \\ 後界也(p. 二三○)
證見本卷廿二。 
18. A perturbed proportion arises when, there being three magnitudes and another set equal to them in multitude, as antecedent is to consequent among the first magnitudes, so is antecedent to consequent among the second magnitudes, while, as the consequent is to a third among the first magnitudes, so is a third to the antecedent among the second magnitudes. 
 
第十九界
有平理之錯者。此數幾何。彼數幾何。此之前與後。若彼之前與後。而此之後與他率。若彼之他率與其前。
甲、乙、丙、數幾何。丁、戊、己、數幾何。其甲與乙。若戊與己。又此之後乙、與他率丙。若彼之他率丁、與前戊。是錯也。今平推甲與丙、若丁與己也。十八、十九、界推法。於十七界 \\ 中通論之。故兩題中不再著也。(p. 二三一)
證見本卷廿三。
增。一幾何。有一幾何、相與為比例。卽此幾何。必有彼幾何、相與為比例。而兩比例等。一幾何。有一幾何、相與為比例。卽必有彼幾何、與此幾何為比例。而兩比例等。此例同理。省 \\ 曰比例等。
甲幾何。與乙幾何、為比例。卽此幾何丙。亦必有彼幾何、如丁。相與為比例。若甲與乙也。丙幾何。與丁幾何、為比例。卽必有彼幾何、如戊。與此幾何丙、為比例。若丙與丁也。此理推廣無礙。於理有之。不必舉其率也。舉率之理。備見後卷。 
PROPOSITION I. 
 
幾何原本第五卷本篇論比例 計三十四題
第一題 
If there be any number of magnitudes whatever which are, respectively, equimultiples of any magnitudes equal in multitude, then, whatever multiple one of the magnitudes is of one, that multiple also will all be of all. ma+mb+mc...=m(a+b+c...). 
 
此數幾何。彼數幾何。此之各率。同幾倍於彼之各率。則此之幷率。亦幾倍於彼之幷率。 
Let any number of magnitudes whatever AB, CD be respectively equimultiples of any magnitudes E, F equal in multitude;  I say that, whatever multiple AB is of E, that multiple will AB, CD also be of E, F. 
   
   
For, since AB is the same multiple of E that CD is of F, as many magnitudes as there are in AB equal to E, so many also are there in CD equal to F.  Let AB be divided into the magnitudes AG, GB equal to E, and CD into CH, HD equal to F;  then the multitude of the magnitudes AG, GB will be equal to the multitude of the magnitudes CH, HD.  Now, since AG is equal to E, and CH to F, therefore AG is equal to E, and AG, CH to E, F.  For the same reason GB is equal to E, and GB, HD to E, F;  therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F;  therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F. 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 2. 
 
第二題 
If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth. (ma+na) is the same multiple of as that (mb+nb) is of n. 
 
六幾何。其第一倍第二之數。等於第三倍第四之數。而第五倍第二之數。等於第六倍第四之數。則第一、第五、幷、倍第二之數。等於第三、第六、幷、倍第四之數。 
Let a first magnitude, AB, be the same multiple of a second, C, that a third, DE, is of a fourth, F, and let a fifth, BG, also be the same multiple of the second, C, that a sixth, EH, is of the fourth F;  I say that the sum of the first and fifth, AG, will be the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F. 
   
   
For, since AB is the same multiple of C that DE is of F, therefore, as many magnitudes as there are in AB equal to C, so many also are there in DE equal to F.  For the same reason also, as many as there are in BG equal to C, so many are there also in EH equal to F;  therefore, as many as there are in the whole AG equal to C, so many also are there in the whole DH equal to F.  Therefore, whatever multiple AG is of C, that multiple also is DH of F.  Therefore the sum of the first and fifth, AG, is the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F. 
         
         
Therefore etc.  Q.E.D. 
   
   
PROPOSITION 3. 
 
第三題 
If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will be equimultiples respectively, the one of the second and the other of the fourth. 
 
四幾何。其第一之倍於第二。若第三之倍於第四。次倍第一。又倍第三。其數等。則第一所倍之與第二。若第三所倍之與第四。 
Let a first magnitude A be the same multiple of a second B that a third C is of a fourth D, and let equimultiples EF, GH be taken of A, C;  I say that EF is the same multiple of B that GH is of D. 
   
   
For, since EF is the same multiple of A that GH is of C,  therefore, as many magnitudes as there are in EF equal to A, so many also are there in GH equal to C.  Let EF be divided into the magnitudes EK, KF equal to A, and GH into the magnitudes GL, LH equal to C;  then the multitude of the magnitudes EK, KF will be equal to the multitude of the magnitudes GL, LH.  And, since A is the same multiple of B that C is of D,  while EK is equal to A, and GL to C,  therefore EK is the same multiple of B that GL is of D.  For the same reason KF is the same multiple of B that LH is of D.  Since, then, a first magnitude EK is the same multiple of a second B that a third GL is of a fourth D, and a fifth KF is also the same multiple of the second B that a sixth LH is of the fourth D, therefore the sum of the first and fifth, EF, is also the same multiple of the second B that the sum of the third and sixth, GH, is of the fourth D. [V. 2] 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 4. 
 
第四題其系為反理 
If a first magnitude have to a second the same ratio as a third to a fourth, any equimultiples whatever of the first and third will also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order. If A:B=C:D, then mA:nB=mC:nD. 
 
四幾何。其第一與二。偕第三與四。比例等。第一、第三、同任為若干倍。第二、第四、同任為若干倍。則第一所倍、與第二所倍。第三所倍、與第四所倍。比例亦等。 
For let a first magnitude A have to a second B the same ratio as a third C to a fourth D;  and let equimultiples E, F be taken of A, C, and G, H other, chance, equimultiples of B, D;  I say that, as E is to G, so is F to H. 
     
     
For let equimultiples K, L be taken of E, F, and other, chance, equimultiples M, N of G, H. 
 
 
Since E is the same multiple of A that F is of C,  and equimultiples K, L of E, F have been taken,  therefore K is the same multiple of A that L is of C. [V. 3]  For the same reason M is the same multiple of B that N is of D.  And, since, as A is to B, so is C to D, and of A, C equimultiples K, L have been taken, and of B, D other, chance, equimultiples M, N,  therefore, if K is in excess of M, L also is in excess of N, if it is equal, equal, and if less, less. [V. Def. 5]  And K, L are equimultiples of E, F, and M, N other, chance, equimultiples of G, H;  therefore, as E is to G, so is F to H. [V. Def. 5] 
               
               
Therefore etc.  Q. E. D. 
   
  本卷界 \\ 說六
一系。凡四幾何。第一與二。偕第三與四。比例等。卽可反推第二與一。偕第四與三。比例亦等。何者。如上倍甲之壬、與倍乙之子。偕倍丙之癸、與倍丁之丑。等、大、小、俱同類。而顯甲與乙、若丙與丁。卽可反說。倍乙之子、與倍甲之壬。偕倍丁之丑、與倍丙之癸。等、大、小、俱同類。而乙與甲。亦若丁與丙。本卷界 \\ 說六
二系。別有一論。亦本書中所恆用也。曰。若甲與乙、偕丙與丁。比例等。則甲之或二或三倍、與乙之或二、或三倍。偕丙之或二、或三倍、與丁之或二、或三倍。比例俱等。倣此以至無窮。 
PROPOSITION 5. 
THEOR. 5 PROPOS. 5 
第五題 
If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole. 
SI magnitudo magnitudinis aeque fuerit multiplex, atque ablata ablatae : Etiam reliqua reliquae ita multiplex erit, ut tota totius. 
大小兩幾何。此全所倍於彼全。若此全截取之分、所倍於彼全截取之分。則此全之分餘、所倍於彼全之分餘。亦如之。 
For let the magnitude AB be the same multiple of the magnitude CD that the part AE subtracted is of the part CF subtracted;  I say that the remainder EB is also the same multiple of the remainder FD that the whole AB is of the whole CD. 
Ita multiplex fit tota AB, totius CD, ut est multiplex AE, ablata ablate CF :  Dico reliquam EB, ita esse multipticem reliquae CD, ut est tota AB, totius CD. 
解曰。甲乙大㡬何。丙丁小㡬何。甲乙所倍于丙丁。若甲乙之截分甲戊、所倍于丙丁之截分、丙己。  題言甲戊之分餘、戊乙、所倍于丙己之分餘,己丁。亦如其數。 
For, whatever multiple AE is of CF, let EB be made that multiple of CG. 
Ponatur enim EB, ita multiplex cuiuspiam magnitudinis ut delicet ipsius GC, ut est AE, ipsus CF. 
論曰。試作一他幾何。為庚丙。令戊乙之倍庚丙。若甲戊之倍丙己也。本卷界說增 
Then, since AE is the same multiple of CF that EB is of GC,  therefore AE is the same multiple of CF that AB is of GF. [V. 1]  But, by the assumption, AE is the same multiple of CF that AB is of CD.  Therefore AB is the same multiple of each of the magnitudes GF, CD;  therefore GF is equal to CD.  Let CF be subtracted from each;  therefore the remainder GC is equal to the remainder FD.  And, since AE is the same multiple of CF that EB is of GC, and GC is equal to DF,  therefore AE is the same multiple of CF that EB is of FD.  But, by hypothesis, AE is the same multiple of CF that AB is of CD;  therefore EB is the same multiple of FD that AB is of CD.  That is, the remainder EB will be the same multiple of the remainder FD that the whole AB is of the whole CD. 
Quoniam igitur AE, EB, aeque sunt multiplices ipsarum CF, GC,113  
erit tota AB, totius GF, ita multiplex, ut AE, ipsus CF, hoc est, omnes omnium, ut una unius.  Sed tam multiplex etiam ponitur AB, ipsius CD, quam est multiplex AE, ipsius CF.  Igitur AB, tam est multiplex ipsius GF, quam multiplex est ipsius CD;114   atque idcirco aequales sunt GF, CD.  Ablata igitur communi CF, aequales erunt GC, FD.  Tam multiplex igitur erit EB, ipsius FD, quam multiplex est ipsius GC.  Sed ita multiplex posita fuit EB, ipsius GC, ut AE, ipsius CF,   hoc est, ut tota AB, totius CD. Quare tam multiplex est reliqua EB, reliquae FD,  quam est tota AB, totius CD:  quod est propositum.   
甲戊、戊乙、之倍丙己、庚丙。其數等。卽其兩幷、甲乙之倍庚己。亦若甲戊之倍丙己也。本篇一而甲乙之倍丙丁。元若甲戊之倍丙己。則丙丁與庚己等也。次每減同用之丙己。卽庚丙與己丁、亦等。而戊乙之倍己丁。亦若戊乙之倍庚丙矣。夫戊乙之倍庚丙。旣若甲戊之倍丙己。則戊乙、為甲戊之分餘。所倍於己丁、為丙己之分餘者。亦若甲乙之倍丙丁也。                       
Therefore etc.  Q. E. D. 
 
114. 6. pron. 
 
   
 
ALITER. Sit ita multiplex tota AB, totius CD, ut ablata AE, ablatae CF. Dico reliquam EB, reliquae FD, esse sic multiplicem, ut est tota totius. Posita enim GA, ita multiplici ipsius FD, ut est AE, ipsius CF, vel ut tota AB, totius CD: quoniam AE, GA, aeque multiplices sunt ipsarum CE, FD,115 erit tota GE, sic multiplex totius CD, ut AE, ipsius CF: Sed ita quoque multiplex est AB, eiusdem CD, ut AE, ipsius C F, ex hypothesi. Aeque multiplices sunt igitur GE, AB, ipsius CD;116 atque adeo inter se aequales. Quare dempta communi AE, aequalet erunt GA, EB: Ideoque aequemultiplices ipsius FD; cum GA, sit multiplex posita ipsius FD: Atqui ita est multiplex posita GA, ipsius FD, ut AB, ipsius CD. Igitur et EB, reliqua sic erit multiplex ipsius FD, reliquae, ut AB, tota totius CD; quod est propositum. Si magnitudo itaque magnitudinis aeque fuerit multiplex, etc. Quod erat demonstrandum.  
115. 1. quinti.  116. 6. pron. 
又論曰。試作一他幾何、為庚甲。令庚甲之倍己丁。若甲戊之倍丙己。本卷界說二十卽其兩幷、庚戊之倍丙丁。亦若甲戊之倍丙己也。本篇一而甲乙之倍丙丁。元若甲戊之倍丙己。是庚戊與甲乙等矣。次每減同用之甲戊。卽庚甲與戊乙等也。而庚甲之倍己丁。若甲乙之倍丙丁也。則戊乙之倍己丁。亦若甲乙之倍丙丁也。 
PROPOSITION 6. 
 
第六題 
If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted from them be equimultiples of the same, the remainders also are either equal to the same or equimultiples of them. mA-nA (n<m) is the same multiple of A that mB-nB is of B. 
 
兩幾何名倍于彼 兩幾何其數等于此兩幾何每滅一分其一分之各倍于所當彼幾何其數等則其餘或各與彼幾何等。或尚各倍於彼幾何。其數亦等。 
For let two magnitudes AB, CD be equimultiples of two magnitudes E, F, and let AG, CH subtracted from them be equimultiples of the same two E, F;  I say that the remainders also, GB, HD, are either equal to E, F or equimultiples of them. 
   
   
For, first, let GB be equal to E; I say that HD is also equal to F. 
 
 
For let CK be made equal to F.  Since AG is the same multiple of E that CH is of F, while GB is equal to E and KC to F,  therefore AB is the same multiple of E that KH is of F. [V. 2]  But, by hypothesis, AB is the same multiple of E that CD is of F;  therefore KH is the same multiple of F that CD is of F.  Since then each of the magnitudes KH, CD is the same multiple of F,  therefore KH is equal to CD.  Let CH be subtracted from each;  therefore the remainder KC is equal to the remainder HD.  But F is equal to KC;  therefore HD is also equal to F.  Hence, if GB is equal to E,  HD is also equal to F. 
                         
                         
Similarly we can prove that, even if GB be a multiple of E, HD is also the same multiple of F. 
 
 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 7. 
 
第七題二支 
Equal magnitudes have to the same the same ratio, as also has the same to equal magnitudes. 
 
此兩幾何等。則與彼幾何各為比例、必等。而彼幾何、與此相等之兩幾何。各為比例、亦等。 
Let A, B be equal magnitudes and C any other, chance, magnitude;  I say that each of the magnitudes A, B has the same ratio to C, and C has the same ratio to each of the magnitudes A, B. 
   
   
For let equimultiples D, E of A, B be taken, and of C another, chance, multiple F. 
 
 
Then, since D is the same multiple of A that E is of B, while A is equal to B,  therefore D is equal to E.  But F is another, chance, magnitude.  If therefore D is in excess of F, E is also in excess of F, if equal to it, equal; and, if less, less.  And D, E are equimultiples of A, B, while F is another, chance, multiple of C;  therefore, as A is to C, so is B to C. [V. Def. 5] 
           
           
I say next that C also has the same ratio to each of the magnitudes A, B. 
 
 
For, with the same construction, we can prove similarly that D is equal to E;  and F is some other magnitude.  If therefore F is in excess of D, it is also in excess of E, if equal, equal; and, if less, less.  And F is a multiple of C, while D, E are other, chance, equimultiples of A, B;  therefore, as C is to A, so is C to B. [V. Def. 5] 
         
         
Therefore etc. 
 
 
PORISM.
From this it is manifest that, if any magnitudes are proportional, they will also be proportional inversely. 
Q. E. D. 
   
後論與本篇第四題之系。同用反理。如甲與丙。若乙與丙。反推之。丙與甲。亦若丙與乙也。   
PROPOSITION 8. 
 
第八題 
Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater. 
 
大小兩幾何。各與他幾何為比例。則大與他之比例。大於小與他之比例。而他與小之比例。大於他與大之比例 
Let AB, C be unequal magnitudes, and let AB be greater; let D be another, chance, magnitude;  I say that AB has to D a greater ratio than C has to D, and D has to C a greater ratio than it has to AB. 
   
   
For, since AB is greater than C, let BE be made equal to C;  then the less of the magnitudes AE, EB, if multiplied, will sometime be greater than D. [V. Def. 4] 
   
   
[Case I.] First, let AE be less than EB; let AE be multiplied,  and let FG be a multiple of it which is greater than D;  then, whatever multiple FG is of AE,  let GH be made the same multiple of EB and K of C;  and let L be taken double of D, M triple of it, and successive multiples increasing by one,  until what is taken is a multiple of D and the first that is greater than K.  Let it be taken, and let it be N which is quadruple of D and the first multiple of it that is greather than K. 
             
             
Then, since K is less than N first, therefore K is not less than M.  And, since FG is the same multiple of AE that GH is of EB,  therefore FG is the same multiple of AE that FH is of AB. [V. 1]  But FG is the same multiple of AE that K is of C;  therefore FH is the same multiple of AB that K is of C;  therefore FH, K are equimultiples of AB, C.  Again, since GH is the same multiple of EB that K is of C,  and EB is equal to C,  therefore GH is equal to K.  But K is not less than M;  therefore neither is GH less than M.  And FG is greater than D;  therefore the whole FH is greater than D, M together.  But D, M together are equal to N,  inasmuch as M is triple of D, and M, D together are quadruple of D,  while N is also quadruple of D;  whence M, D together are equal to N.  But FH is greater than M, D;  therefore FH is in excess of N,  while K is not in excess of N.  And FH, K are equimultiples of AB, C, while N is another, chance, multiple of D;  therefore AB has to D a greater ratio than C has to D. [V. Def. 7] 
                                           
                                           
I say next, that D also has to C a greater ratio than D has to AB. 
 
 
For, with the same construction, we can prove similarly that N is in excess of K, while N is not in excess of FH.  And N is a multiple of D, while FH, K are other, chance, equimultiples of AB, C;  therefore D has to C a greater ratio than D has to AB. [V. Def. 7] 
     
     
[Case 2.] Again, let AE be greater than EB.  Then the less, EB, if multiplied, will sometime be greater than D. [V. Def. 4]  Let it be multiplied, and let GH be a multiple of EB and greater than D;  and, whatever multiple GH is of EB, let FG be made the same multiple of AE, and K of C.  Then we can prove similarly that FH, K are equimultiples of AB, C;  and, similarly, let N be taken a multiple of D but the first that is greater than FG,  so that FG is again not less than M.  But GH is greater than D;  therefore the whole FH is in excess of D, M, that is, of N.  Now K is not in excess of N,  inasmuch as FG also, which is greater than GH, that is, than K, is not in excess of N.  And in the same manner, by following the above argument, we complete the demonstration. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 9. 
 
第九題二支
 
Magnitudes which have the same ratio to the same are equal to one another; and magnitudes to which the same has the same ratio are equal. 
 
兩幾何、與一幾何。各為比例、而等。則兩幾何必等。一幾何、與兩幾何。各為比例、而等。則兩幾何亦等。
先解曰。甲、乙、兩幾何。各與丙為比例、等。題言甲與乙等。 
For let each of the magnitudes A, B have the same ratio to C;  I say that A is equal to B. 
   
   
For, otherwise, each of the magnitudes A, B would not have had the same ratio to C; [V. 8]  but it has; therefore A is equal to B. 
   
   
Again, let C have the same ratio to each of the magnitudes A, B;  I say that A is equal to B. 
   
   
For, otherwise, C would not have had the same ratio to each of the magnitudes A, B; [V. 8]  but it has; therefore A is equal to B. 
   
   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 10. 
 
第十題二支 
Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less. 
 
彼此兩幾何。此幾何、與他幾何之比例。大於彼與他之比例。則此幾何、大於彼。他幾何、與彼幾何之比例。大於他與此之比例。則彼幾何、小於此。 
For let A have to C a greater ratio than B has to C;  I say that A is greater than B. 
   
   
For, if not, A is either equal to B or less.  Now A is not equal to B;  for in that case each of the magnitudes A, B would have had the same ratio to C; [V. 7]  but they have not; therefore A is not equal to B.  Nor again is A less than B;  for in that case A would have had to C a less ratio than B has to C; [V. 8]  but it has not; therefore A is not less than B.  But it was proved not to be equal either;  therefore A is greater than B. 
                 
                 
Again, let C have to B a greater ratio than C has to A;  I say that B is less than A. 
   
   
For, if not, it is either equal or greater.  Now B is not equal to A;  for in that case C would have had the same ratio to each of the magnitudes A, B; [V. 7]  but it has not; therefore A is not equal to B.  Nor again is B greater than A;  for in that case C would have had to B a less ratio than it has to A; [V. 8]  but it has not; therefore B is not greater than A.  But it was proved that it is not equal either;  therefore B is less than A. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 11. 
 
第十一題 
Ratios which are the same with the same ratio are also the same with one another. 
 
此兩幾何之比例。與他兩幾何之比例、等。而彼兩幾何之比例。與他兩幾何之比例、亦等。則彼兩幾何之比例。與此兩幾何之比例、亦等。
解曰。甲乙偕丙丁之比例。各與戊己之比例等。題言甲乙與丙丁之比例、亦等。 
For, as A is to B, so let C be to D, and, as C is to D, so let E be to F;  I say that, as A is to B, so is E to F. 
   
   
For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N. 
 
 
Then since, as A is to B, so is C to D,  and of A, C equimultiples G, H have been taken,  and of B, D other, chance, equimultiples L, M, therefore,  if G is in excess of L, H is also in excess of M,  if equal, equal, and if less, less.  Again, since, as C is to D, so is E to F,  and of C, E equimultiples H, K have been taken,  and of D, F other, chance, equimultiples M, N,  therefore, if H is in excess of M, K is also in excess of N,  if equal, equal, and if less, less.  But we saw that, if H was in excess of M, G was also in excess of L;  if equal, equal; and if less, less;  so that, in addition, if G is in excess of L, K is also in excess of N,  if equal, equal, and if less, less.  And G, K are equimultiples of A, E, while L, N are other, chance, equimultiples of B, F;  therefore, as A is to B, so is E to F. 
                               
                               
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 12. 
 
第十二題 
If any number of magnitudes be proportional, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents. If A:a=B:b=C:c etc., each ratio is equal to the ratio (A+B+C...):(a+b+c...) 
 
數幾何。所為比例皆等。則幷前率、與幷後率、之比例。若各前率、與各後率、之比例。 
Let any number of magnitudes A, B, C, D, E, F be proportional, so that, as A is to B, so is C to D and E to F;  I say that, as A is to B, so are A, C, E to B, D, F. 
   
   
For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N. 
 
 
Then since, as A is to B, so is C to D, and E to F,  and of A, C, E equimultiples G, H, K have been taken, and of B, D, F other, chance, equimultiples L, M, N,  therefore, if G is in excess of L, H is also in excess of M, and K of N,  if equal, equal, and if less, less;  so that, in addition, if G is in excess of L, then G, H, K are in excess of L, M, N,  if equal, equal, and if less, less.  Now G and G, H, K are equimultiples of A and A, C, E,  since, if any number of magnitudes whatever are respectively equimultiples of any magnitudes equal in multitude,  whatever multiple one of the magnitudes is of one, that multiple also will all be of all. [V. 1]  For the same reason L and L, M, N are also equimultiples of B and B, D, F;  therefore, as A is to B, so are A, C, E to B, D, F. [V. Def. 5] 
                     
                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 13. 
 
第十三題 
If a first magnitude have to a second the same ratio as a third to a fourth, and the third have to the fourth a greater ratio than a fifth has to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth. If a:b=c:d and c:d>e:f, then a:b>e:f. 
 
數幾何。第一與二之比例。若第三與四之比例。而第三與四之比例。大於第五與六之比例。則第一與二之比例亦大於第五與六之比例。 
For let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F;  I say that the first A will also have to the second B a greater ratio than the fifth E to the sixth F. 
   
   
For, since there are some equimultiples of C, E, and of D, F other, chance, equimultiples,  such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, [V. Def. 7] let them be taken,  and let G, H be equimultiples of C, E, and K, L other, chance, equimultiples of D, F,  so that G is in excess of K, but H is not in excess of L;  and, whatever multiple G is of C, let M be also that multiple of A,  and, whatever multiple K is of D, let N be also that multiple of B. 
           
           
Now, since, as A is to B, so is C to D,  and of A, C equimultiples M, G have been taken, and of B, D other, chance, equimultiples N, K,  therefore, if M is in excess of N, G is also in excess of K,  if equal, equal, and if less, less. [V. Def. 5]  But G is in excess of K;  therefore M is also in excess of N.  But H is not in excess of L;  and M, H are equimultiples of A, E, and N, L other, chance, equimultiples of B, F;  therefore A has to B a greater ratio than E has to F. [V. Def. 7] 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 14. 
 
第十四題 
If a first magnitude have to a second the same ratio as a third has to a fourth, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less. If a:b=c:d, then, accordingly as a>=<b, also c>=<d. 
 
四幾何。第一與二之比例。若第三與四之比例。而第一幾何大於第三。則第二幾何亦大於第四。第一或等、或小、於第三。則第二亦等、亦小、於第四。 
For let a first magnitude A have the same ratio to a second B as a third C has to a fourth D; and let A be greater than C;  I say that B is also greater than D. 
   
   
For, since A is greater than C, and B is another, chance, magnitude,  therefore A has to B a greater ratio than C has to B. [V. 8]  But, as A is to B, so is C to D;  therefore C has also to D a greater ratio than C has to B. [V. 13]  But that to which the same has a greater ratio is less; [V. 10]  therefore D is less than B;  so that B is greater than D. 
             
             
Similarly we can prove that, if A be equal to C, B will also be equal to D; and, if A be less than C, B will also be less than D. 
 
 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 15. 
 
 
Parts have the same ratio as the same multiples of them taken in corresponding order. a:b=ma:mb. 
 
兩分之比例。與兩多分幷之比例、等。 
For let AB be the same multiple of C that DE is of F;  I say that, as C is to F, so is AB to DE. 
   
   
For, since AB is the same multiple of C that DE is of F,  as many magnitudes as there are in AB equal to C, so many are there also in DE equal to F.  Let AB be divided into the magnitudes AG, GH, HB equal to C, and DE into the magnitudes DK, KL, LE equal to F;  then the multitude of the magnitudes AG, GH, HB will be equal to the multitude of the magnitudes DK, KL, LE.  And, since AG, GH, HB are equal to one another, and DK, KL, LE are also equal to one another,  therefore, as AG is to DK, so is GH to KL, and HB to LE. [V. 7]  Therefore, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [V. 12]  therefore, as AG is to DK, so is AB to DE.  But AG is equal to C and DK to F;  therefore, as C is to F, so is AB to DE. 
                   
                   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 16. 
THEOR. 16. PROPOS. 16. 
第十六題 更理 
If four magnitudes be proportional, they will also be proportional alternately. If a:b=c:d, then a:c=b:d. 
SI Quatuor magnitudines proportionales fuerint, et vicissim proportionales erunt. 
四幾何、為兩比例、等。卽更推前與前、後與後為比例。亦等。 
Let A, B, C, D be four proportional magnitudes, so that, as A is to B, so is C to D;  I say that they will also be so alternately, that is, as A is to C, so is B to D. 
HIC demonstratur Alterna, sive Permutata proportio, seu ratio, quae definitione 12. explicata est. Sit enim A, ad B, ut C, ad D.  Dico vicissim, seu permutando, esse quoque A, ad C, ut B, ad D. 
解曰。甲、乙、丙、丁、四幾何。甲與乙之比例。若丙與丁。  題言更推之。甲與丙之比例。亦若乙與丁。 
For of A, B let equimultiples E, F be taken, and of C, D other, chance, equimultiples G, H. 
Sumantur enim ipsarum A, B, primae ac secundae, aequemultiplices E, F; Item ipsarum C, D, tertiae et quartae aequemultiplices G, H; 
論曰。試以甲與乙。同任倍之為戊、為己。別以丙與丁。同任倍之為庚、為辛。 
Then, since E is the same multiple of A that F is of B, and parts have the same ratio as the same multiples of them, [V. 15] therefore, as A is to B, so is E to F.    But as A is to B, so is C to D; therefore also, as C is to D, so is E to F. [V. 11]  Again, since G, H are equimultiples of C, D, therefore, as C is to D, so is G to H. [V. 15] But, as C is to D, so is E to F; therefore also, as E is to F, so is G to H. [V. 11]  But, if four magnitudes be proportional, and the first be greater than the third, the second will also be greater than the fourth;  if equal, equal; and if less, less. [V. 14]  Therefore, if E is in excess of G, F is also in excess of H, if equal, equal, and if less, less.  Now E, F are equimultiples of A, B, and G, H other, chance, equimultiples of C, D; therefore, as A is to C, so is B to D. [V. Def. 5] 
117 eritque E, ad F, ut A, ad B; cum E, et F, sint pariter multiplices partium A, et B. 
Eadem ratione erit G, ad H, ut C, ad D.  Cum igitur proportiones E, ad F, et C, ad D, sint eaedem proportioni A, ad B; 118 erunt et ipsae inter se eaedem.  Rursus quia proportiones E, ad F, et G, ad H, eaedem sunt proportioni C, ad D; 119 erunt et ipsae eaedem inter se;  hoc est, ut est E, prima ad F, secundam, ita erit G, tertia ad H, quartam, 120     Quare si E, prima maior est quam G, tertia, vel aequalis, vel minor, erit quoque F, secunda maior quam H, quarta, vel aequalis, vel minor, in quacunque multiplicatione accepta sint aeque multiplicia E, F, et aeque multiplicia G, H.   121 Est igitur A, prima ad C, secundam, ut B, tertia ad D, quartam (cum E, et F, sint aeque multiplices primae A, ac tertiae B; At G, et H, aeque multiplices C, secundae, et D, quartae, et illae ab his una deficiant, vel una aequales sint, vel una excedant, etc.) quod est propositum. 
卽戊與己。若甲與乙也。本篇十五  庚與辛。若丙與丁也。  夫甲與乙。若丙與丁。而戊與己。亦若甲與乙。  卽戊與己。亦若丙與丁矣。依顯庚與辛。若丙與丁。卽戊與己。亦若庚與辛也。本篇十一  次三試之。若戊大於庚則己亦大於辛也。  若等、亦等。若小、亦小。  任作幾許倍。恆如是也。本篇十四  則倍一甲之戊。倍三乙之己。與倍二丙之庚。倍四丁之辛。其等、大、小、必同類也。而甲與丙。若乙與丁矣。 
Therefore etc.  Q. E. D. 
Si quatuorigitur magnitudines proportionales fuerint, et vicissim proportionales erunt. 
Quod ostendendum erat. 
   
PROPOSITION 17. 
 
第十七題 分理 
If magnitudes be proportional componendo, they will also be proportional separando. If a:b=c:d, then (a-b):b=(c-d):d. 
 
相合之兩幾何、為比例等。則分之為比例、亦等。 
Let AB, BE, CD, DF be magnitudes proportional componendo, so that, as AB is to BE, so is CD to DF;  I say that they will also be proportional separando, that is, as AE is to EB, so is CF to DF. 
   
   
For of AE, EB, CF, FD let equimultiples GH, HK, LM, MN be taken, and of EB, FD other, chance, equimultiples, KO, NP. 
 
 
Then, since GH is the same multiple of AE that HK is of EB,  therefore GH is the same multiple of AE that GK is of AB. [V. 1]  But GH is the same multiple of AE that LM is of CF;  therefore GK is the same multiple of AB that LM is of CF.  Again, since LM is the same multiple of CF that MN is of FD,  therefore LM is the same multiple of CF that LN is of CD. [V. 1]  But LM was the same multiple of CF that GK is of AB;  therefore GK is the same multiple of AB that LN is of CD.  Therefore GK, LN are equimultiples of AB, CD.  Again, since HK is the same multiple of EB that MN is of FD,  and KO is also the same multiple of EB that NP is of FD,  therefore the sum HO is also the same multiple of EB that MP is of FD. [V. 2]  And, since, as AB is to BE, so is CD to DF,  and of AB, CD equimultiples GK, LN have been taken, and of EB, FD equimultiples HO, MP,  therefore, if GK is in excess of HO, LN is also in excess of MP, if equal, equal, and if less, less.  Let GK be in excess of HO;  then, if HK be subtracted from each, GH is also in excess of KO.  But we saw that, if GK was in excess of HO, LN was also in excess of MP;  therefore LN is also in excess of MP,  and, if MN be subtracted from each, LM is also in excess of NP;  so that, if GH is in excess of KO, LM is also in excess of NP.  Similarly we can prove that, if GH be equal to KO,  LM will also be equal to NP, and if less, less.  And GH, LM are equimultiples of AE, CF, while KO, NP are other, chance, equimultiples of EB, FD;  therefore, as AE is to EB, so is CF to FD. 
                                                 
                                                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 18. 
 
第十八題 合理 
If magnitudes be proportional separando, they will also be proportional componendo. If a:b=c:d, then (a+b):b=(c+d):d 
 
兩幾何。分之為比例、等。則合之為比例、亦等。 
Let AE, EB, CF, FD be magnitudes proportional separando, so that, as AE is to EB, so is CF to FD;  I say that they will also be proportional componendo, that is, as AB is to BE, so is CD to FD. 
   
   
For, if CD be not to DF as AB to BE, then, as AB is to BE, so will CD be either to some magnitude less than DF or to a greater. 
 
 
First, let it be in that ratio to a less magnitude DG.  Then, since, as AB is to BE, so is CD to DG,  they are magnitudes proportional componendo;  so that they will also be proportional separando. [V. 17]  Therefore, as AE is to EB, so is CG to GD.  But also, by hypothesis, as AE is to EB, so is CF to FD.  Therefore also, as CG is to GD, so is CF to FD. [V. 11]  But the first CG is greater than the third CF;  therefore the second GD is also greater than the fourth FD. [V. 14]  But it is also less: which is impossible.  Therefore, as AB is to BE, so is not CD to a less magnitude than FD.  Similarly we can prove that neither is it in that ratio to a greater;  it is therefore in that ratio to FD itself. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 19 
 
第十九題其系為轉理 
If, as a whole is to a whole, so is a part subtracted to a part subtracted, the remainder will also be to the remainder as whole to whole. If a:b=c:d (c<a and d<b), then (a-c)=(b-d). 
 
兩幾何。各截取一分。其所截取之比例。與兩全之比例等。則分餘之比例。與兩全之比例亦等。 
For, as the whole AB is to the whole CD, so let the part AE subtracted be to the part CF subtracted;  I say that the remainder EB will also be to the remainder FD as the whole AB to the whole CD. 
   
   
For since, as AB is to CD, so is AE to CF, alternately also, as BA is to AE, so is DC to CF. [V. 16]  And, since the magnitudes are proportional componendo, they will also be proportional separando, [V. 17]  that is, as BE is to EA, so is DF to CF,  and, alternately, as BE is to DF, so is EA to FC. [V. 16]  But, as AE is to CF, so by hypothesis is the whole AB to the whole CD.  Therefore also the remainder EB will be to the remainder FD as the whole AB is to the whole CD. [V. 11] 
           
           
Therefore etc. 
 
 
 
 
[PORISM.
From this it is manifest that, if magnitudes be proportional componendo, they will also be proportional convertendo.] Q. E. D. 
 
 
PROPOSITION 20 
 
第二十題 三支 
If there be three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali the first be greater than the third, the fourth will also be greater than the sixth; if equal, equal; and, if less, less. If a:b=d:e and b:c=e:f, then, accordingly as a>=<c, also d>=<f. 
 
有三幾何。又有三幾何。相為連比例。而第一幾何大於第三。則第四亦大於第六。第一或等、或小於第三。則第四亦等、亦小、於第六。 
Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and as B is to C, so is E to F; and let A be greater than C ex aequali;  I say that D will also be greater than F; if A is equal to C, equal; and, if less, less. 
   
   
For, since A is greater than C, and B is some other magnitude, and the greater has to the same a greater ratio than the less has, [V. 8] therefore A has to B a greater ratio than C has to B.  But, as A is to B, so is D to E, and, as C is to B, inversely, so is F to E;  therefore D has also to E a greater ratio than F has to E. [V. 13]  But, of magnitudes which have a ratio to the same, that which has a greater ratio is greater; [V. 10]  therefore D is greater than F.  Similarly we can prove that, if A be equal to C,  D will also be equal to F; and if less, less. 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 21. 
 
第二十一題 三支 
If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then, if ex aequali the first magnitude is greater than the third, the fourth will also be greater than the sixth; if equal, equal; and if less, less. If a:b=e:f and b:c=d:e, then, accordingly as a>=<c, also d>=<f. 
 
有三幾何。又有三幾何。相為連比例而錯。以平理推之。若第一幾何大於第三。則第四亦大於第六。若第一或等、或小、於第三。則第四亦等、亦小、於第六。 
Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, and let the proportion of them be perturbed, so that, as A is to B, so is E to F, and, as B is to C, so is D to E, and let A be greater than C ex aequali;  I say that D will also be greater than F; if A is equal to C, equal; and if less, less. 
   
   
For, since A is greater than C, and B is some other magnitude, therefore A has to B a greater ratio than C has to B. [V. 8]  But, as A is to B, so is E to F, and, as C is to B,  inversely, so is E to D.  Therefore also E has to F a greater ratio than E has to D. [V. 13]  But that to which the same has a greater ratio is less; [V. 10]  therefore F is less than D;  therefore D is greater than F.  Similarly we can prove that, if A be equal to C,  D will also be equal to F; and if less, less. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 22. 
 
第二十二題 平理之序 
If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, they will also be in the same ratio ex aequali. If a:b=d:e and b:c=e:f then a:c=d:f. 
 
有若干幾何。又有若干幾何。其數等。相為連比例。則以平理推。 
Let there be any number of magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two together are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F;  I say that they will also be in the same ratio ex aequali,
   
   
For of A, D let equimultiples G, H be taken, and of B, E other, chance, equimultiples K, L;  and, further, of C, F other, chance, equimultiples M, N. 
   
   
Then, since, as A is to B, so is D to E,  and of A, D equimultiples G, H have been taken,  and of B, E other, chance, equimultiples K, L,  therefore, as G is to K, so is H to L. [V. 4]  For the same reason also, as K is to M, so is L to N.  Since, then, there are three magnitudes G, K, M, and others H, L, N equal to them in multitude, which taken two and two together are in the same ratio,  therefore, ex aequali, if G is in excess of M, H is also in excess of N;  if equal, equal; and if less, less. [V. 20]  And G, H are equimultiples of A, D,  and M, N other, chance, equimultiples of C, F.  Therefore, as A is to C, so is D to F. [V. Def. 5] 
                     
                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 23. 
 
第二十三題 平理之錯 
If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, they will also be in the same ratio ex aequali. 
 
若干幾何。又若干幾何。相為連比例而錯。亦以平理推。 
Let there be three magnitudes A, B, C, and others equal to them in multitude, which, taken two and two together, are in the same proportion, namely D, E, F;  and let the proportion of them be perturbed, so that,  as A is to B, so is E to F,  and, as B is to C, so is D to E;  I say that, as A is to C, so is D to F. 
         
         
Of A, B, D let equimultiples G, H, K be taken, and of C, E, F other, chance, equimultiples L, M, N. 
 
 
Then, since G, H are equimultiples of A, B, and parts have the same ratio as the same multiples of them, [V. 15]  therefore, as A is to B, so is G to H.  For the same reason also, as E is to F, so is M to N.  And, as A is to B, so is E to F;  therefore also, as G is to H, so is M to N. [V. 11]  Next, since, as B is to C, so is D to E,  alternately, also, as B is to D, so is C to E. [V. 16]  And, since H, K are equimultiples of B, D,  and parts have the same ratio as their equimultiples,  therefore, as B is to D, so is H to K. [V. 15]  But, as B is to D, so is C to E;  therefore also, as H is to K, so is C to E. [V. 11]  Again, since L, M are equimultiples of C, E,  therefore, as C is to E, so is L to M. [V. 15]  But, as C is to E, so is H to K;  therefore also, as H is to K, so is L to M, [V. 11]  and, alternately, as H is to L, so is K to M. [V. 16]  But it was also proved that, as G is to H, so is M to N.  Since, then, there are three magnitudes G, H, L,  and others equal to them in multitude K, M, N, which taken two and two together are in the same ratio,  and the proportion of them is perturbed,  therefore, ex aequali, if G is in excess of L, K is also in excess of N;  if equal, equal; and if less, less. [V. 21]  And G, K are equimultiples of A, D, and L, N of C, F.  Therefore, as A is to C, so is D to F. 
                                                 
                                                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 24. 
 
第二十四題 
If a first magnitude have to a second the same ratio as a third has to a fourth, and also a fifth have to the second the same ratio as a sixth to the fourth, the first and fifth added together will have to the second the same ratio as the third and sixth have to the fourth. If a:c=d:f and b:c=e:f then (a+b):c=(d+e):f. 
 
凡第一與二幾何之比例。若第三與四幾何之比例。而第五與二之比例。若第六與四。則第一第五幷。與二之比例。若第三第六幷。與四。 
Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F; and let also a fifth BG have to the second C the same ratio as a sixth EH has to the fourth F;  I say that the first and fifth added together, AG, will have to the second C the same ratio as the third and sixth, DH, has to the fourth F. 
   
   
For since, as BG is to C, so is EH to F,  inversely, as C is to BG, so is F to EH.  Since, then, as AB is to C, so is DE to F,  and, as C is to BG, so is F to EH,  therefore, ex aequali, as AB is to BG, so is DE to EH. [V. 22]  And, since the magnitudes are proportional separando, they will also be proportional componendo; [V. 18]  therefore, as AG is to GB, so is DH to HE.  But also, as BG is to C, so is EH to F;  therefore, ex aequali, as AG is to C, so is DH to F. [V. 22] 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 25. 
 
第二十五題 
If four magnitudes be proportional, the greatest and the least are greater than the remaining two. If a:b=c:d and a is the greatest and d is the least, then a+d>b+c. 
 
四幾何、為斷比例。則最大與最小兩幾何幷。大於餘兩幾何幷。 
Let the four magnitudes AB, CD, E, F be proportional so that, as AB is to CD, so is E to F, and let AB be the greatest of them and F the least;  I say that AB, F are greater than CD, E. 
   
   
For let AG be made equal to E, and CH equal to F. 
 
 
Since, as AB is to CD, so is E to F, and E is equal to AG, and F to CH,  therefore, as AB is to CD, so is AG to CH.  And since, as the whole AB is to the whole CD,  so is the part AG subtracted to the part CH subtracted,  the remainder GB will also be to the remainder HD as the whole AB is to the whole CD. [V. 19]  But AB is greater than CD;  therefore GB is also greater than HD.  And, since AG is equal to E, and CH to F,  therefore AG, F are equal to CH, E.  And if, GB, HD being unequal, and GB greater, AG, F be added to GB and CH, E be added to HD,  it follows that AB, F are greater than CD, E. 
                     
                     
Therefore etc.  Q. E. F.  Proposition 26. If the ratio between a first and a second magnitude is larger than that between a third and a fourth, then, invertendo, the ratio between the second and the first is smaller than that between the fourth and the third. If a:b>c:d then b:a<d:c.2   Proposition 27. If the ratio between a first and a second [quantity] is greater thatn that between a third and a fourth, then, alternando, the ratio of the first and the third is also greater than that of the second and the fourth. If a:b>c:d then a:c>b:d.3   If the ratio bewteen a first and a second [quantity] is greater than that between a third and a fourth, then, componendo, the ration of the first and the second together to the second will also be greater than that of the thrid and the fourth together to the fourth. If a:b>c:d then (a+b):b>(c+d):d.4   Proposition 29. If the ratiio of a first and second [quantity] together to the second is greater than that of a third and a fouth, then the ratio of the first and the second is also greater than that of the third and the fourth. If (a+b):b>(c+d):d then a:b>c:d.5   Proposition 30. If the ratio of a first and a second [quantity] together to the second is greater than that of a third and a fourth together to the fourth, then, invertendo, the ratio of the first and the second together to the first is lesser than that of the third and the fourth together to the third. If (a+b):b>(c+d):d then (a+b):a<(c+d):c.6   Proposition 31. If there be these three quantities, and those three quantities, such as the ratio of the first and the second of these is greater than that between the first and the second of those, while the ratio of the second and the third of these is greater than that of the second and third of those, if arranged in order. then ex equali the ratio between the first and the third of these will also be greater than the ratio between the first and the third of those. If a:b>d:e and b:c>e:f then a:c>d:f.7   Proposition 32. If a:b>e:f and b:c>d:e then a:c>d:f.  Proposition 33. If a:b>c:d then (a-c):(b-d)>a:b.  If A:a, B:b,...K:k then (A+B...+K)>(a+b+...+k):k and also (A+B+...+K):K>(B+...+K):(b+...+k) but (A+B+...+K):K<A:a. 
                     
    第二十六題
第一與二幾何之比例。大於第三與四之比例。反之。則第二與一之比例。小於第四與三之比例。
解曰。一甲與二乙之比例。大於三丙與四丁。題言反之。二乙與一甲之比例。小於四丁與三丙。(p. 二七○)
論曰。試作戊與乙之比例。若丙與丁。卽甲與乙之比例。大於戊與乙。而甲幾何大於戊。本篇 \\ 十則乙與戊之比例。大於乙與甲也。本篇 \\ 八反之。則乙與戊之比例。若丁與丙本篇 \\ 四而乙與甲之比例。小於丁與丙。 
第二十七題
第一與二之比例。大於第三與四之比例。更之。則第一與三之比例。亦大於第二與四之比例。
解曰。一甲與二乙之比例。大於三丙與四丁。題言更之。則一甲與三丙之比例。亦大於二乙與四丁。
論曰。試作戊與乙之比例。若丙與丁。卽甲與乙之比例。大於戊與乙。而甲幾何大於戊。本篇 \\ 十則甲與丙之比例。大於戊於丙也。本篇 \\ 八夫戊與乙之比例。旣若丙與丁。更之。則戊與丙之比例。亦若乙與丁。本篇 \\ 十六而甲與丙之比例。大於乙與丁矣。 
第二十八題
第一與二之比例。大於第三與四之比例。合之。則第一、第二、幷、與二之比例。亦大於第三、第四、幷、與四之比例。
解曰。一甲乙與二乙丙之比例。大於三丁戊與四戊己。題言合之。則甲丙與乙丙之比例。亦大於丁己與戊己。(p. 二七一)
論曰。試作庚乙與乙丙之比例。若丁戊與戊己。卽甲乙與乙丙之比例。大於庚乙與乙丙。而甲乙幾何大於庚乙矣。本篇 \\ 十此二率者。每加一乙丙。卽甲丙亦大於庚丙。而甲丙與乙丙之比例。大於庚丙與乙丙也。本篇 \\ 八夫庚乙與乙丙之比例。旣若丁戊與戊己。合之。則庚丙與乙丙之比例。亦若丁己與戊己也。本篇 \\ 十八而甲丙與乙丙之比例。大於丁己與戊己矣。 
第二十九題
第一合第二、與二之比例。大於第三合第四、與四之比例。分之。則第一與二之比例。亦大於第三與四之比例。
解曰。甲丙與乙丙之比例。大於丁己與戊己。題言分之。則甲乙與乙丙之比例。亦大於丁戊與戊己。
論曰。試作庚丙與乙丙之比例。若丁己與戊己。卽甲丙與乙丙之比例。亦大於庚丙與乙丙。而甲丙幾何大於庚丙矣。本篇 \\ 十此二率者。每減一同用之乙丙。卽甲乙亦大與庚乙。而甲乙與乙丙之比例。大於庚乙與乙丙也。本篇 \\ 八夫庚丙與乙丙之比例。旣若丁己與戊己。分之。則庚乙與乙丙之比例。亦若丁戊與戊己也。本篇 \\ 十七而甲乙與乙丙之比例。大於丁戊與戊己矣。 
第三十題
(p. 二七二)第一合第二、與二之比例。大於第三合第四、與四之比例。轉之。則第一合第二、與一之比例。小於第三合第四、與三之比例。
解曰。甲丙與乙丙之比例。大於丁己與戊己。題言轉之。則甲丙與甲乙之比例。小於丁己與丁戊。
論曰。甲丙與乙丙之比例。旣大於丁己與戊己。分之。卽甲乙與乙丙之比例。亦大於丁戊與戊己也。本篇 \\ 廿九又反之。乙丙與甲乙之比例。小於戊己與丁戊矣。本 \\ 篇廿 \\ 六又合之。甲丙與甲乙之比例。亦小於丁己與丁戊也。本篇 \\ 廿八 
第三十一題
此三幾何。彼三幾何。此第一與二之比例。大於彼第一與二之比例。此第二與三之比例。大於彼第二與三之比例。如是序者。以平理推。則此第一與三之比例。亦大於彼第一與三之比例。
解曰。甲、乙、丙。此三幾何。丁、戊、己。彼三幾何。而甲與乙之比例。大於丁與戊。乙與丙之比例。大於戊與己。如是序者。題言以平理推。則甲與丙之比例。亦大於丁與己。
論曰。試作庚與丙之比例。若戊與己。卽乙與丙之比例。大於庚與丙。而乙幾何大於庚本篇 \\ 十是甲與小庚之比例。大於甲與大乙矣。本篇 \\ 八夫甲與乙之比例。元大於丁與戊。卽(p. 二七三)甲與庚之比例。更大於丁與戊也次作辛與庚之比例若丁與戊卽甲與庚之比例。亦大於辛與庚。而甲幾何大於辛。本篇 \\ 十是大甲與丙之比例。大於小辛與丙矣。本篇 \\ 八夫辛與丙之比例。以平理推之。若丁與己也。本篇 \\ 廿二則甲與丙之比例。大於丁與己也。 
第三十二題
此三幾何。彼三幾何。此第一與二之比例。大於彼第二與三之比例。此第二與三之比例。大於彼第一與二之比例。如是錯者。以平理推。用此第一與三之比例。亦大於彼第一與三之比例。
解曰。甲、乙、丙。此三幾何。丁、戊、己。彼三幾何。而甲與乙之比例。大於戊與己。乙與丙之比例。大於丁與戊。如是錯者。題言以平理推。則甲與丙之比例。亦大於丁與己。
論曰。試作庚與丙之比例。若丁與戊。卽乙與丙之比例。大於庚與丙。而乙幾何大於庚。本篇 \\ 十是甲與小庚之比例。大於甲與大乙矣。本篇 \\ 八夫甲與乙之比例。旣大於戊與己。卽甲與庚之比例。更大於戊與己也。次作辛與庚之比例。若戊與己。卽甲與庚之比例。亦大於辛與庚。而甲幾何大於辛。本篇 \\ 十是大甲與丙之比例。大於小辛與丙矣。本篇 \\ 八(p. 二七四)夫辛與丙之比例。以平理推之。若丁與己也。本篇 \\ 廿三則甲與丙之比例。大於丁與己也。 
第三十三題
此全與彼全之比例。大於此全截分、與彼全截分之比例。則此全分餘、與彼全分餘之比例。大於此全與彼全之比例。
解曰。甲乙全與丙丁全之比例。大於兩截分、甲戊與丙己。題言兩分餘、戊乙與己丁之比例。大於甲乙與丙丁。
論曰。甲乙與丙丁之比例。旣大於甲戊與丙己。更之。卽甲乙與甲戊之比例。亦大於丙丁與丙己也。本篇 \\ 廿七又轉之。甲乙與戊乙之比例。小於丙丁與己丁也。本篇 \\ 三十又更之。甲乙與丙丁之比例。小於戊乙與己丁也。本篇 \\ 廿七戊乙與己丁。分餘也。則分餘之比例。大於甲乙全、與丙丁全矣。依顯兩全之比例。小於截分。則分餘之比例。小於兩全。 
第三十四題 三支
若干幾何。又有若干幾何。其數等。而此第一與彼第一之比例。大於此第二與彼第二之比例。此第二與彼第二之比例。大於此第三與彼第三之比例。以(p. 二七五)後俱如。是則此幷與彼幷之比例。大於此末與彼末之比例。亦大於此幷減第一、與彼幷減第一之比例。而小於此第一與彼第一之比例。
解曰。如甲、乙、丙、三幾何。又有丁、戊、己、三幾何。其甲與丁之比例。大於乙與戊。乙與戊之比例。大於丙與己。題先言甲乙丙幷、與丁戊己幷之比例。大於丙與己。次言亦大於乙丙幷、與戊己幷。後言小於甲與丁。
論曰。甲與丁之比例。旣大於乙與戊。更之。卽甲與乙之比例。大於丁與戊也。本篇 \\ 廿七又合之。甲乙幷與乙之比例。大於丁戊幷與戊也。本篇 \\ 廿八又更之。甲乙幷與丁戊幷之比例。大於乙與戊也。本篇 \\ 廿七是甲乙全與丁戊全之比例。大於減幷乙與減幷戊也。旣爾。卽減餘甲與減餘丁之比例。大於甲乙全與丁戊全也。本篇 \\ 卅三依顯乙與戊之比例。亦大於乙丙全與戊己全。卽甲與丁之比例。更大於乙丙全與戊己全也。又更之。甲與乙丙幷之比例。大於丁與戊己幷也。本篇 \\ 廿七又合之。甲乙丙全與乙丙幷之比例。大於丁戊己全與戊己幷也。本篇 \\ 廿八又更之。甲乙丙全與丁戊己全之比例。大於乙丙幷與戊己幷也。本篇 \\ 廿七則得次解也。又甲乙丙全與丁戊己全之比例。旣大於減幷乙丙與減幷戊己。卽減餘甲與減餘丁之比例。大於甲乙丙全與丁戊己全也本篇 \\ 卅二則得後解也。(p. 二七六)又乙與戊之比例。旣大於丙與己。更之。卽乙與丙之比例。大於戊與己也。本篇 \\ 廿七又合之。乙丙全與丙之比例。大於戊己全與己也。本篇 \\ 廿八又更之。乙丙幷與戊己幷之比例。大於丙與己也。本篇 \\ 廿七而甲乙丙幷與丁戊己幷之比例。旣大於乙丙幷與戊己幷。卽更大於末丙與末己也則得先解也。
若兩率各有四幾何而丙與己之比例。亦大於庚與辛。卽與前論同理。蓋依上文論、乙與戊之比例。大於乙丙庚幷與戊己辛幷。卽甲與丁之比例。更大於乙丙庚幷與戊己辛幷也。更之。卽甲與乙丙庚幷之比例。大於丁與戊己辛幷也。本篇 \\ 十八又合之。甲乙丙庚全與乙丙庚幷之比例。大於丁戊己辛全與戊己辛幷也。又更之。甲乙丙庚全與丁戊己辛全之比例。大於乙丙庚幷與戊己辛幷也。本篇 \\ 廿七則得次解也。又甲乙丙庚全與丁戊己辛全之比例。旣大於減幷乙丙庚與減幷戊己辛。卽減餘甲與減餘丁之比例。大於甲乙丙庚全與丁戊己辛全也。本篇 \\ 卅三則得後解也。又依前論、顯乙丙庚幷與戊己辛幷之比例。旣大於庚與辛。而甲乙丙庚全與丁戊己辛全之比例。大於乙丙庚幷與戊己辛幷。卽更大於末庚與末辛也。則得先解也。自五以上。至於無窮。俱@此論。可@全題之旨。 
 
幾何原本第六卷之首 
DEFINITIONS. 
 
界說六則 
1. Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional. 
 
第一界
凡形相當之各角等。而各等角旁兩線之比例。俱等。為相似之形。
甲乙丙、丁戊己、兩角形之甲角、與丁角等。乙與戊、丙與己、各等。其甲角旁之甲乙、與甲丙、兩線之比例。若丁角旁之丁戊與丁己兩線。而甲乙與乙丙。若丁戊與戊己。甲丙與丙乙。若丁己與己戊。則此兩角形、為相似之形。依顯凡平邊形、皆相似之形。如庚辛壬、癸子丑、俱平邊角形。其各角俱等。而各邊之比例亦等者、是也。四邊、五邊、以上諸形。俱倣此。 
2. [When two sides of one figure together with two sides of another figure form antecedents and consequents in a proportion, the figures are reciprocally related. 
 
第二界
兩形之各兩邊線。互為前後率。相與為比例而等。為互相視之形。
甲乙丙丁、戊己庚辛、兩方形。其甲乙、乙丙、邊。與戊己、己庚、邊。相與為比例等。而彼此互為前、後。如甲乙與。戊己。若己庚與乙丙也。則此兩形為互相視之形。依顯壬癸子、丑寅卯、兩角形之壬子與丑寅。若丑卯與壬癸。或壬癸與丑寅。若丑卯與壬子。亦互相視之形也。 
3. A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less. 
 
第三界
理分中末線者。一線兩分之。其全與大分之比例。若大分與小分之比例。(p. 二七九)
甲乙線。兩分之於丙。而甲乙與大分甲丙之比例。若大分甲丙與小分丙乙。此為理分中末線。其分法。見本卷三十題。而與二卷十一題理同名異。此線為用甚廣、至量體。尤所必須。十三卷諸題多賴之。古人目為神分線也。 
4. The height of any figure is the perpendicular drawn from the vertex to the base. 
 
第四界
度各形之高。皆以垂線之亘為度。
甲乙丙角形。從甲頂、向乙丙底。作甲庚垂線。卽甲庚為甲乙丙之高。又丁戊己角形。作丁辛垂線。卽丁辛為丁戊己之高。若兩形相視。兩垂線等。卽兩形之高、必等。如上兩形在兩平行線之內者是也。若以丙、己、為頂。以甲乙、丁戊、為底。則不等。自餘諸形之度高、俱倣此。(p. 二八○)
凡度物高。以頂底為界。以垂線為度。蓋物之定度。止有一。不得有二。自頂至底。垂線一而己。偏線無數也。 
A ratio is said to be compounded of ratios when the sizes of the ratios multiplied together make som (?ratio, or size).  A parallelogram not “filling” a line is a figure lesser than the line. If it has a surplus and the line is not enough, it is a figure greater than the line.8  
   
第五界
比例以比例相結者。以多比例之命數、相乘、除、而結為一比例之命數。
此各比例、不同理、而相聚為一比例者。則用相結之法。合各比例之命數。求首尾一比例之命數也。曷為比例之命數。謂大幾何、所倍於小幾何若干。或小幾何、在大幾何內若干也。如大幾何、四倍於小。或(p. 二八一)小幾何、為大四分之一。卽各以四為命比例之數也。五卷界 \\ 說三今言以彼多比例之命數、相乘、除、而結為此一比例之命數者。如十二倍之此比例。則以彼二倍、六倍、兩比例相結也。二六相乘為十二、故也。或以彼三倍、四倍、兩比例相結也。三四相乘亦十二、故也。又如三十倍之此比例。則以彼二倍、三倍、五倍、三比例相結也。二乘三為六、六乘五為三十、故也。其曰相結者。相結之理。蓋在中率。凡中率為前比例之後。後比例之前。故以二比例合為一比例。則中率為輳合之因。如兩爿合。此為之膠。如兩襟合。此為之紐矣。第五卷第十界、言數幾何為同理之比例。則第一與第三、為再加之比例。再加者。以前中二率之命數。再加為前後二率之命數。亦以中率為紐也。但彼所言者。多比例同理。故止以第一比例之命數累加之。此題所言。則不同理之多比例。不得以第一比例之命數累加之。故用此乘除相結之理。于不同理之中。求其同理。別為累加之法。其紐結之義。頗相類焉。下文仍發明借象(p. 二八二)之術、以需後用也。
五卷言多比例同理者。第一、與第三為再加。與第四為三加。與第五為四加。以至無窮。今此相結之理。亦以三率為始。三率。則兩比例相乘除、而中率為紐也。若四率。則先以前三率之兩比例、相乘除、而結為一比例。復以此初結之比例、與第三比例乘除、相結為一比例也。若五率。則先以前三率之兩比例、乘除相結。復以此再結之比例、與第三比例、乘除相結。又以三結之比例、與第四比例、乘除相結、為一比例也。或以第一第二第三率之兩比例、乘除相結。以第三第四第五之兩比例、乘除相結。又以此二所結比例、乘除相結、而為一比例也。自六以上。倣此以至無窮。
設三幾何、為二比例、不同理、而合為一比例。則以第一與二、第二與三、兩比例相結也。如上圖。三幾何、(p. 二八三)二比例。皆以大不等者。其甲乙與丙丁為二倍大丙丁與戊己為三倍大。則甲乙與戊己、為六倍大。二乘三為六也。若以小不等。戊己為第一。甲乙為第三。三乘二亦六。則戊己與甲乙、為反六倍大也。
甲乙與丙丁。旣二倍大。試以甲乙二平分之。為甲庚、庚乙。必各與丙丁等。丙丁與戊己。旣三倍大。而甲庚、庚乙、各與丙丁等。卽甲庚亦三倍大于戊己。庚乙亦三倍大於戊己。而甲乙必六倍大於戊己。

又如上圖。三幾何、二比例。前以大不等。後以小不等者。中率小于前後兩率也。其甲乙與丙丁、為三倍大。丙丁與戊己、為反二倍大。反二倍大者。丙 \\ 丁得戊己之半。卽甲乙與戊己、為等帶半。三乘半。得等帶半也。若以戊己為第一。甲乙為第三。反推之。半除三。為反等帶半也。
又如上圖。三幾何、二比例。前以小不等。後以大不等者。中率大於前後二率也。其甲乙與丙丁、為反二倍大。甲乙得丙 \\ 丁之半。丙丁與戊己、為等帶三分之一。卽甲乙與戊己、為反等帶半。甲乙得戊己 \\ 三分之二。何者。如甲乙二。卽丙丁當四。丙丁四。卽戊己當三。是甲乙二。戊己當三也。
後增。其乘除之法。則以命數三。帶得數一。為四。以半除之得二。二比三、為反等帶半也。若以戊己為第(p. 二八四)一。甲乙為第三。三比二、為等帶半也。
設四幾何、為三比例、不同理、而合為一比例。則以第一與二、第二與三、第三與四、三比例相結也。如上圖。甲、乙、丙、丁、四幾何、三比例。先依上論。以甲與乙、乙與丙、二比例、相結。為甲與丙之比例。次以甲與丙、丙與丁、相結。卽得甲與丁之比例也。如是遞結。可至無窮也。
或用此圖、申明本題之旨曰。甲與乙之命數為丁。乙與丙之命數為戊。卽甲與丙之命數為己。何者。三命數、以一丁、二戊、相乘得三己。卽三比例、以一甲與乙、二乙與丙、相乘得三甲與丙、
後增。若多幾何、各帶分、而多寡不。等者。當用通分法。如設前比例、為反五倍帶三之二。後比例、為二倍大帶八之一。卽以前命數三、通其五倍、為十五。得分數從之、為十七。是前比例為三與十七也。以後命數八、通其二倍、為十六。得分數從之、為十七。是後比例為十七與八也。卽首尾二幾何之比例。為三與八。得(p. 二八五)幾二倍大帶三之二也。
曷謂借象之術。如上所說、三幾何、二比例者。皆以中率為前比例之後。後比例之前。乘除相結。略如連比例之同用一中率也。而不同理。別有二比例異中率者。是不同理之斷比例也。無法可以相結。當于其所設幾何之外。別立三幾何、二比例、而同中率者。乘除相結。作為儀式。以彼異中率之四幾何、二比例。依倣求之。卽得。故謂之借象術也。假如所設幾何。十六為首。十二為尾。却云十六與十二之比例。若十六#八#廿四#十六#六#廿四#十六#六#廿四#三#九##九#三六##二#八#二#九##四#三六##四#八十二#四#十八#十二#二#十八#十二#九#十八十六#四#廿四#十六#四#廿四#十六#四#廿四#九#五四##二#十二##六#三六#六#五四##六#十二##二#三六十二#二#十八#十二#九#十八#十二#一#十八八與三、及二與四之比例。八為前比例之前。四為後比例之後。三與二、為前之後、後之前。此所謂異中率也。欲以此二比例、乘除相結。無法可通矣。用是別立三幾何、二比例。如其八與三、二與四、之比例。而務令同中率。如三其八、得二十四。為前比例之前。三其三、得九。為前比例之後。卽以九為後比例之前。又求九與何數為比例、若二與四。得十八。為後比例之後。其二十四與九。若八與三也。九與十八。若二與四也。則十六與十二。若二十四與十八。俱為等帶半之比例矣。是用借象之術。變異中率為同中率。乘除相結。而合二比例為一比例也。其三比例以上。亦如上方所說。展轉借象。遞結之。 詳見本卷二十三題。算家所用借象金法、雙金法、俱本此。  第六界
平行方形不滿一線。為形小于線。若形有餘。線不足。為形大于線。
甲乙線。其上作甲戊丁丙平行方形。不滿甲乙線。而丙乙上無形。卽作己乙線、與丁丙平行。次引戊丁線、遇己乙於己。是為甲戊己乙滿甲乙線平行方形。則甲丁為依甲乙線之有闕平行方形。而丙己平行方形為甲丁之闕形。又甲丙線上、作甲戊己乙平行方形。其甲乙邊、大于元設甲丙線之較、為丙乙、而甲己形、大于甲丙線上之甲丁形。則甲己為依甲丙線之帶餘平行方形。而丙己平行方形、為甲己之餘形。 

PROPOSITION 1. 
 
幾何原本第六卷 本篇論線面之比例 計三十三題
第一題 
Triangles and parallelograms which are under the same height are to one another as their bases. 
 
等高之三角形、方形。自相與為比例。與其底之比例等。 
Let ABC, ACD be triangles and EC, CF parallelograms under the same height;  I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF. 
   
   
For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH be made equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined. 
 
 
Then, since CB, BG, GH are equal to one another,  the triangles ABC, AGB, AHG are also equal to one another. [I. 38]  Therefore, whatever multiple the base HC is of the base BC,  that multiple also is the triangle AHC of the triangle ABC.  For the same reason, whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD;  and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38]  if the base HC is in excess of the base CL,  the triangle AHC is also in excess of the triangle ACL, and, if less, less.  Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD,  equimultiples have been taken of the base BC and the triangle ABC, namely the base HC and the triangle AHC,  and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC;  and it has been proved that, if the base HC is in excess of the base CL,  the triangle AHC is also in excess of the triangle ALC;  if equal, equal; and, if less, less.  Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. Def. 5] 
                             
                             
Next, since the parallelogram EC is double of the triangle ABC, [I. 41]  and the parallelogram FC is double of the triangle ACD,  while parts have the same ratio as the same multiples of them, [V. 15]  therefore, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram FC.  Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD,  and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF,  therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11] 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 2. 
 
第二題 二支 
If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle. 
 
三角形。任依一邊作平行線。卽此線分兩餘邊以為比例。必等。三角形內。有一線分兩邊以為比例、而等。卽此線與餘邊為平行。 
For let DE be drawn parallel to BC, one of the sides of the triangle ABC;  I say that, as BD is to DA, so is CE to EA. 
   
   
For let BE, CD be joined. 
 
 
Therefore the triangle BDE is equal to the triangle CDE;  for they are on the same base DE and in the same parallels DE, BC. [I. 38]  And the triangle ADE is another area.  But equals have the same ratio to the same; [V. 7]  therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE.  But, as the triangle BDE is to ADE, so is BD to DA;  for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1]  For the same reason also, as the triangle CDE is to ADE, so is CE to EA.  Therefore also, as BD is to DA, so is CE to EA. [V. 11] 
                 
                 
Again, let the sides AB, AC of the triangle ABC be cut proportionally,  so that, as BD is to DA, so is CE to EA;  and let DE be joined.  I say that DE is parallel to BC. 
       
       
For, with the same construction,  since, as BD is to DA, so is CE to EA,  but, as BD is to DA, so is the triangle BDE to the triangle ADE,  and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1]  therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11]  Therefore each of the triangles BDE, CDE has the same ratio to ADE.  Therefore the triangle BDE is equal to the triangle CDE; [V. 9]  and they are on the same base DE.  But equal triangles which are on the same base are also in the same parallels. [I. 39]  Therefore DE is parallel to BC. 
                   
                   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 3. 
 
第三題 二支 
If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle. 
 
三角形。任以直線、分一角為兩平分。而分對角邊為兩分。則兩分之比例。若餘兩邊之比例。三角形分角(p. 二九一)之線。所分對角邊之比例。若餘兩邊。則所分角為兩平分。 
Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD;  I say that, as BD is to CD, so is BA to AC. 
   
   
For let CE be drawn through C parallel to DA, and let BA be carried through and meet it at E. 
 
 
Then, since the straight line AC falls upon the parallels AD, EC,  the angle ACE is equal to the angle CAD. [I. 29]  But the angle CAD is by hypothesis equal to the angle BAD;  therefore the angle BAD is also equal to the angle ACE.  Again, since the straight line BAE falls upon the parallels AD, EC,  the exterior angle BAD is equal to the interior angle AEC. [I. 29]  But the angle ACE was also proved equal to the angle BAD;  therefore the angle ACE is also equal to the angle AEC,  so that the side AE is also equal to the side AC. [I. 6]  And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE,  therefore, proportionally, as BD is to DC, so is BA to AE.  But AE is equal to AC; [VI. 2]  therefore, as BD is to DC, so is BA to AC. 
                         
                         
Again, let BA be to AC as BD to DC,  and let AD be joined;  I say that the angle BAC has been bisected by the straight line A.D. 
     
     
For, with the same construction,  since, as BD is to DC, so is BA to AC,  and also, as BD is to DC, so is BA to AE:  for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2]  therefore also, as BA is to AC, so is BA to AE. [V. 11]  Therefore AC is equal to AE, [V. 9]  so that the angle AEC is also equal to the angle ACE. [I. 5]  But the angle AEC is equal to the exterior angle BAD, [I. 29]  and the angle ACE is equal to the alternate angle CAD; [id.]  therefore the angle BAD is also equal to the angle CAD.  Therefore the angle BAC has been bisected by the straight line AD. 
                     
                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 4. 
 
第四題 
In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. 
 
幾何原本 卷六
凡等角三角形。其在等角旁之各兩腰線。相與為比例、必等。而對等角之邊、為相似之邊。 
Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED;  I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. 
   
   
For let BC be placed in a straight line with CE.  Then, since the angles ABC, ACB are less than two right angles, [I. 17]  and the angle ACB is equal to the angle DEC,  therefore the angles ABC, DEC are less than two right angles;  therefore BA, ED, when produced, will meet. [I. Post. 5]  Let them be produced and meet at F. 
           
           
Now, since the angle DCE is equal to the angle ABC,  BF is parallel to CD. [I. 28]  Again, since the angle ACB is equal to the angle DEC,  AC is parallel to FE. [I. 28]  Therefore FACD is a parallelogram;  therefore FA is equal to DC, and AC to FD. [I. 34]  And, since AC has been drawn parallel to FE, one side of the triangle FBE,  therefore, as BA is to AF, so is BC to CE. [VI. 2]  But AF is equal to CD;  therefore, as BA is to CD, so is BC to CE,  and alternately, as AB is to BC, so is DC to CE. [V. 16]  Again, since CD is parallel to BF,  therefore, as BC is to CE, so is FD to DE. [VI. 2]  But FD is equal to AC;  therefore, as BC is to CE, so is AC to DE,  and alternately, as BC is to CA, so is CE to ED. [V. 16]  Since then it was proved that, as AB is to BC, so is DC to CE,  and, as BC is to CA, so is CE to ED;  therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22] 
                                     
                                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 5. 
 
第五題 
If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend. 
 
兩三角形。其各兩邊之比例等。卽兩形為等角形。而對各相似邊之角、各等。 
Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF;  I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF. 
   
   
For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23]  therefore the remaining angle at A is equal to the remaining angle at G. [I. 32] 
   
   
Therefore the triangle ABC is equiangular with the triangle GEF.  Therefore in the triangles ABC, GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [VI. 4]  therefore, as AB is to BC, so is GE to EF.  But, as AB is to BC, so by hypothesis is DE to EF;  therefore, as DE is to EF, so is GE to EF. [V. 11]  Therefore each of the straight lines DE, GE has the same ratio to EF;  therefore DE is equal to GE. [V. 9]  For the same reason DF is also equal to GF.  Since then DE is equal to EG,  and EF is common, the two sides DE, EF are equal to the two sides GE, EF;  and the base DF is equal to the base FG;  therefore the angle DEF is equal to the angle GEF, [I. 8]  and the triangle DEF is equal to the triangle GEF,  and the remaining angles are equal to the remaining angles,  namely those which the equal sides subtend. [I. 4]  Therefore the angle DFE is also equal to the angle GFE,  and the angle EDF to the angle EGF.  And, since the angle FED is equal to the angle GEF,  while the angle GEF is equal to the angle ABC,  therefore the angle ABC is also equal to the angle DEF.  For the same reason the angle ACB is also equal to the angle DFE,  and further, the angle at A to the angle at D;  therefore the triangle ABC is equiangular with the triangle DEF. 
                                             
                                             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 6. 
 
第六題 
If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend. 
 
兩三角形之一角等。而等角旁之各兩邊、比例等。卽兩形為等角形。而對各相似邊之角、各等。 
Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF;  I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE. 
   
   
For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23]  therefore the remaining angle at B is equal to the remaining angle at G. [I. 32] 
   
   
Therefore the triangle ABC is equiangular with the triangle DGF.  Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4]  But, by hypothesis, as BA is to AC, so also is ED to DF;  therefore also, as ED is to DF, so is GD to DF. [V. 11]  Therefore ED is equal to DG;  [V. 9] and DF is common;  therefore the two sides ED, DF are equal to the two sides GD, DF;  and the angle EDF is equal to the angle GDF;  therefore the base EF is equal to the base GF,  and the triangle DEF is equal to the triangle DGF,  and the remaining angles will be equal to the remaining angles,  namely those which the equal sides subtend. [I. 4]  Therefore the angle DFG is equal to the angle DFE,  and the angle DGF to the angle DEF.  But the angle DFG is equal to the angle ACB;  therefore the angle ACB is also equal to the angle DFE.  And, by hypothesis, the angle BAC is also equal to the angle EDF;  therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32]  therefore the triangle ABC is equiangular with the triangle DEF. 
                                     
                                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 7. 
 
第七題 
If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional. 
 
兩三角形之第一角等。而第二相當角、各兩旁之邊、比例等。其第三相當角。或俱小于直角。或俱不小于直角。卽兩形為等角形。而對各相似邊之角、各等。 
Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle;  I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F. 
   
   
For, if the angle ABC is unequal to the angle DEF, one of them is greater.  Let the angle ABC be greater;  and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23] 
     
     
Then, since the angle A is equal to D, and the angle ABG to the angle DEF,  therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32]  Therefore the triangle ABG is equiangular with the triangle DEF.  Therefore, as AB is to BG, so is DE to EF [VI. 4]  But, as DE is to EF, so by hypothesis is AB to BC;  therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11]  therefore BC is equal to BG, [V. 9]  so that the angle at C is also equal to the angle BGC. [I. 5]  But, by hypothesis, the angle at C is less than a right angle;  therefore the angle BGC is also less than a right angle;  so that the angle AGB adjacent to it is greater than a right angle. [I. 13]  And it was proved equal to the angle at F;  therefore the angle at F is also greater than a right angle.  But it is by hypothesis less than a right angle: which is absurd.  Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it.  But the angle at A is also equal to the angle at D;  therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]  Therefore the triangle ABC is equiangular with the triangle DEF. 
                                   
                                   
But, again, let each of the angles at C, F be supposed not less than a right angle;  I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF. 
   
   
For, with the same construction, we can prove similarly that BC is equal to BG;  so that the angle at C is also equal to the angle BGC. [I. 5]  But the angle at C is not less than a right angle;  therefore neither is the angle BGC less than a right angle.  Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17]  Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it.  But the angle at A is also equal to the angle at D;  therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]  Therefore the triangle ABC is equiangular with the triangle DEF. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 8. 
 
第八題 
If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another. 
 
直角三邊形。從直角向對邊。作一垂線。分本形為兩直角三邊形。卽兩形皆與全形相似。亦自相似。 
Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC;  I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another. 
   
   
For, since the angle BAC is equal to the angle ADB,  for each is right,  and the angle at B is common to the two triangles ABC and ABD,  therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32]  therefore the triangle ABC is equiangular with the triangle ABD.  Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD,  so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD,  and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4]  Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional.  Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1]  Similarly we can prove that the triangle ABC is also similar to the triangle ADC;  therefore each of the triangles ABD, ADC is similar to the whole ABC. 
                       
                       
I say next that the triangles ABD, ADC are also similar to one another. 
 
 
For, since the right angle BDA is equal to the right angle ADC,  and moreover the angle BAD was also proved equal to the angle at C,  therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32]  therefore the triangle ABD is equiangular with the triangle ADC.  Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD,  so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B,  and so also is BA to AC, these sides subtending the right angles; [VI. 4]  therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1] 
               
               
Therefore etc. 
 
 
PORISM.
From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. 
Q. E. D. 
   
   
PROPOSITION 9. 
PROBL. 1 PROPOS. 9 
第九題 
From a given straight line to cut off a prescribed part. 
ADATA recta linea imperatam partem auferre. 
一直線。求截所取之分。 
Let AB be the given straight line;  thus it is required to cut off from AB a prescribed part. 
IMPERETVR, ut ex linea AB, auferamus partem tertiam.    
法曰。甲乙直線。求截取三分之一。   
Let the third part be that prescribed.  Let a straight line AC be drawn through from A containing with AB any angle;  let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3]  Let BC be joined, and through D let DF be drawn parallel to it. [I. 31] 
  Ex A, ducatur recta AC, utcunque faciens angulum CAB;   et ex AC, abscindantur tot partes aequales cuiuslibet magnitudinis, quota pars detrahenda est ex AB, ut in proposito exemplo tres AD, DE, EF. Deinde ex F, ad B, recta ducatur FB, cui per D, parallela agatur DG.   
  先從甲、任作一甲丙線、為丙甲乙角。次從甲向丙。任作所命分之平度。如甲丁、丁戊、戊己、為三分也。次作己乙直線。末作丁庚線。與己乙平行。卽甲庚為甲乙三分之一。     
Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC,  therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2]  But CD is double of DA;  therefore BF is also double of FA;  therefore BA is triple of AF. 
Dico AG, esse partem tertiam imperatam rectae AB. Nam cum in triangulo ABF, lateri FB, parallela sit recta DG;122 erit ut FD, ad DA, ita BG, ad GA. 123 Componendo igitur , ut FA, ad DA, ita BA, erit ad GA: sed FA, ipsius AD, est tripla, ex constructione. Igitur et BA, ipsis AG, erit tripla, ideoque AG, tertia pars erit ipsius AB, quae imperabatur. 
       
論曰。甲乙己角形內之丁庚線。旣與乙己邊平行。卽己丁與丁甲之比例。若乙庚與庚甲也。本篇二 合之。己甲與甲丁。若乙甲與庚甲也。五卷十八 而甲丁旣為己甲三分之一。卽庚甲亦為乙甲三分之一也。         
Therefore from the given straight line AB the prescribed third part AF has been cut off.  Q. E. F. 
A data ergo recta linea imperatam partem abstulimus.  Quod faciendum erat. 
   
 
SCHOLIUM. QVOD si ex AB, auferenda sit pars non aliquota, sed quae plures aliquotas non efficientes unam complectatur, nimirum quae contineat quatuor undecimas ipsius AB, sumendae erunt ex AC, undecim partes aquales usque ad D, punctum, ex quo ad B recta ducatur DB; et huic parallela EF, ex E, termino quatuor partium. Nam AF, erit pars imperata. Erit enim rursus ut DA, ad AE, ita BA, ad AF. Quare, et convertendo ut AE, ad AD, ita AF, ad AB: Est autem AE, pars continens quatuor undecimas ipsius AD, ex constructione. Igitur et AF, eadem pars erit rectae AB. Quod est propositum. Non aliter detrahetur ex AB, pars complectens quotcunque partes ipsius aliquotas non facientes unam. 
注曰。甲乙線。欲截取十一分之四。先作甲丙線、為丙甲乙角。從甲向丙。任平分十一分、至丁。次作丁乙線。末從甲取四分、得戊。作戊己線。與丁乙平行。卽甲己為十一分甲乙之四。何者。依上論、丁甲與戊甲之比例。若乙甲與己甲也。反之。甲戊與甲丁。若甲己與甲乙也。五卷四 甲戊為十一分甲丁之四。則甲己亦十一分甲乙之四矣。依此可推不盡分之數。蓋四不為十一之盡分故。 
PROPOSITION 10. 
 
第十題 
To cut a given uncut straight line similarly to a given cut straight line. 
 
一直線。求截各分。如所設之截分 
Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31] 
 
 
Therefore each of the figures FH, HB is a parallelogram;  therefore DH is equal to FG and HK to GB. [I. 34]  Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC,  therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]  But KH is equal to BG, and HD to GF;  therefore, as CE is to ED, so is BG to GF.  Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE,  therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2]  But it was also proved that, as CE is to ED, so is BG to GF;  therefore, as CE is to ED, so is BG to GF,  and, as ED is to DA, so is GF to FA. 
                     
                     
Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC.  Q. E. F. 
   
   
PROPOSITION 11. 
PROBL. 3. PROPOS. 11. 
第十一題 
To two given straight lines to find a third proportional. 
DUABUS datis rectis lineis tertiam proportionalem adinuenire. 
兩直線。求別作一線。相與為連比例。 
Let BA, AC be the two given straight lines, and let them be placed so as to contain any angle;  thus it is required to find a third proportional to BA, AC.  For let them be produced to the points D, E, and let BD be made equal to AC; [I. 3] let BC be joined, and through D let DE be drawn parallel to it. [I. 31] 
SINT duae rectae AB, AC, ita dispositae, ut efficiant angulum A, quemcunque, fitque inuenienda illis tertia proportionalis, sicut quidem AB, ad AC, ita AC, ad tertiam. Producatur AB, quam volumus esse antecedentem, et capiatur BD, aequalis ipsi AC, quae consequens esse debet, sive media. Deinde ducta recta BC, agatur illi ex D, parallela DE, occurrens ipsi AC, productae in E.     
法曰。甲乙、甲丙、兩線。求別作一線。相與為連比例者。合兩線。任作甲角。而甲乙與甲戊之比例。若甲丙與他線也。先于甲乙引長之、為乙丁。與甲丙等。次作丙乙線相聯。次從丁作丁戊線。與丙乙平行。末于甲丙引長之、遇于戊。卽丙戊為所求線。如以甲丙為前率。倣此。     
Since, then, BC has been drawn parallel to DE, one of the sides of the triangle ADE,  proportionally, as AB is to BD, so is AC to CE. [VI. 2]  But BD is equal to AC;  therefore, as AB is to AC, so is AC to CE. 
Dico CE, esse tertiam proportionalem: hoc est, esse ut AB, ad AC, ita AC, ad CE. Cum enim in triangulo ADE, lateri DE, parallela sit recta BC;124 erit ut AB, ad BD, ita AC, ad CE: 125 Sed ut AB, ad BD, ita eadem AB, ad AC, aequalem ipsi BD. Ut igitur AB, ad AC, ita AC, ad CE. quod est propositum.  
124. 4. sexti  125. 7. quinti. 
     
論曰。甲丁戊角形內之丙乙線。旣與戊丁邊平行。卽甲乙與乙丁之比例。若甲丙與丙戊也。本篇二而乙丁、甲丙、元等。卽甲乙與甲丙。若甲丙與丙戊也。五卷七       
Therefore to two given straight lines AB, AC a third proportional to them, CE, has been found.  Q. E. F. 
Duabus ergo datis rectis lineis, tertiam proportionalem adinvenimus.  Quod erat faciendum. 
   
 
SCHOLIUM. ALITER idem demonstrabimus, hoc modo. Duae rectae datae AB, BC, constituantur ad angulum rectum ABC, et coniungatur recta AC. Producta autem AB, antecedente, ducatur ex C, ad AC, perpendicularis CD, occurrens ipsi AB, productae in D. Dico BD, esse tertiam proportionalem.  
注曰。別有一法。以甲乙、乙丙、兩線。列作甲乙丙直角。次以甲丙線聯之。而甲乙引長之。末從丙作丙丁。為甲丙之垂線。遇引長線于丁。卽乙丁為所求線。 
 
Cum enim in triangulo ACD, angulus ACD, sit rectus, et ab eo ad basin AD, deducta perpendicularis CB; erit por corollarium propositio 8 huius liber BC, media proportionalis inter AB, et BD, hoc est, ut AB, ad BC, ita erit BC, ad BD. Quod est propositum.  
論曰。甲丙丁角形之甲丙丁。旣為直角。而從直角至甲。丁底。有丙乙垂線。卽丙乙為甲乙、乙丁、比例之中率。本篇八之系 則甲乙與乙丙。若乙丙與乙丁也。 
 
INVENTA autem tertia linea continue proportionali, si primam omiseris, et alijs duabus tertiam inueneris, habebis quatuor lineas continue proportionales. Ut si lineis A, et B, adinueniatur tertia proportionalis C, et duabus B, et C, tertia proportionalis D, erunt quatuor lineae A, B, C, D, continue proportionales. Eadem artereperietur quinta proportionalis, sexta, septima, octaua; et sic in infinitum.INVENTA autem tertia linea continue proportionali, si primam omiseris, et alijs duabus tertiam inueneris, habebis quatuor lineas continue proportionales. Ut si lineis A, et B, adinueniatur tertia proportionalis C, et duabus B, et C, tertia proportionalis D, erunt quatuor lineae A, B, C, D, continue proportionales. Eadem artereperietur quinta proportionalis, sexta, septima, octaua; et sic in infinitum. 
旣從一二得三。卽從二、三、求四、以上、至于無窮。俱倣此。 
PROPOSITION 12. 
 
第十二題 
To three given straight lines to find a fourth proportional. 
 
三直線。求別作一線。相與為斷比例。 
Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C. 
 
 
Let two straight lines DE, DF be set out containing any angle EDF;  let DG be made equal to A, GE equal to B, and further DH equal to C;  let GH be joined, and let EF be drawn through E parallel to it. [I. 31] 
     
     
Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF,  therefore, as DG is to GE, so is DH to HF. [VI. 2]  But DG is equal to A, GE to B, and DH to C;  therefore, as A is to B, so is C to HF. 
       
       
Therefore to the three given straight lines A, B, C a fourth proportional HF has been found.  Q. E. F. 
   
   
PROPOSITION 13. 
 
第十三題 
To two given straight lines to find a mean proportional. 
 
兩直線。求別作一線。為連比例之中率。 
Let AB, BC be the two given straight lines;  thus it is required to find a mean proportional to AB, BC. 
   
   
Let them be placed in a straight line, and let the semicircle ADC be described on AC;  let BD be drawn from the point B at right angles to the straight line AC,  and let AD, DC be joined. 
     
     
Since the angle ADC is an angle in a semicircle, it is right. [III. 31]  And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base,  therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8, Por.] 
     
     
Therefore to the two given straight lines AB, BC a mean proportional DB has been found.  Q. E. F. 
   
   
PROPOSITION 14. 
 
第十四題 二支 
In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal. 
 
兩平行方形等。一角又等。卽等角旁之兩邊。為互相視之邊。兩平行方形之一角等。而等角旁兩邊、為互相視之邊。卽兩形等。 
Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14]  I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF. 
   
   
For let the parallelogram FE be completed.  Since, then, the parallelogram AB is equal to the parallelogram BC,  and FE is another area,  therefore, as AB is to FE, so is BC to FE. [V. 7]  But, as AB is to FE, so is DB to BE, [VI. 1]  and, as BC is to FE, so is GB to BF. [id.]  therefore also, as DB is to BE, so is GB to BF. [V. 11]  Therefore in the parallelograms AB, BC the sides about the equal angles are reciprocally proportional. 
               
               
Next, let GB be to BF as DB to BE;  I say that the parallelogram AB is equal to the parallelogram BC. 
   
   
For since, as DB is to BE, so is GB to BF,  while, as DB is to BE, so is the parallelogram AB to the parallelogram FE, [VI. 1]  and, as GB is to BF, so is the parallelogram BC to the parallelogram FE, [VI. 1]  therefore also, as AB is to FE, so is BC to FE; [V. 11]  therefore the parallelogram AB is equal to the parallelogram BC. [V. 9] 
         
         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 15. 
 
第十五題二支 
In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. 
 
相等兩三角形之一角等。卽等角旁之各兩邊、互相視。兩三角形之一角等。而等角旁之各兩邊、互相視。卽兩三角形等。 
Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE;  I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB. 
   
   
For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [I. 14]  Let BD be joined. 
   
   
Since then the triangle ABC is equal to the triangle ADE,  and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 7]  But, as CAB is to BAD, so is CA to AD, [VI. 1]  and, as EAD is to BAD, so is EA to AB. [id.]  Therefore also, as CA is to AD, so is EA to AB. [V. 11]  Therefore in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional. 
           
           
Next, let the sides of the triangles ABC, ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD;  I say that the triangle ABC is equal to the triangle ADE. 
   
   
For, if BD be again joined,  since, as CA is to AD, so is EA to AB, while,  as CA is to AD, so is the triangle ABC to the triangle BAD,  and, as EA is to AB, so is the triangle EAD to the triangle BAD, [VI. 1]  therefore, as the triangle ABC is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 11]  Therefore each of the triangles ABC, EAD has the same ratio to BAD.  Therefore the triangle ABC is equal to the triangle EAD. [V. 9] 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 16. 
 
第十六題二支 
If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional. 
 
四直線為斷比例。卽首尾兩線、矩內直角形。與中兩線、矩內直角形、等。首尾兩線、與中兩線、兩矩內直角形等。卽四線為斷比例。 
Let the four straight lines AB, CD, E, F be proportional, so that, as AB is to CD, so is E to F;  I say that the rectangle contained by AB, F is equal to the rectangle contained by CD, E. 
   
   
Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E.  Let the parallelograms BG, DH be completed. 
   
   
Then since, as AB is to CD, so is E to F,  while E is equal to CH, and F to AG,  therefore, as AB is to CD, so is CH to AG.  Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional.  But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [VI. 14]  therefore the parallelogram BG is equal to the parallelogram DH.  And BG is the rectangle AB, F,  for AG is equal to F;  and DH is the rectangle CD, E,  for E is equal to CH;  therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E. 
                     
                     
Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E;  I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F. 
   
   
For, with the same construction,  since the rectangle AB, F is equal to the rectangle CD, E,  and the rectangle AB, F is BG,  for AG is equal to F,  and the rectangle CD, E is DH,  for CH is equal to E,  therefore BG is equal to DH.  And they are equiangular.  But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [VI. 14]  Therefore, as AB is to CD, so is CH to AG.  But CH is equal to E, and AG to F;  therefore, as AB is to CD, so is E to F. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 17 
 
第十七題 二支 
If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional. 
 
三直線為連比例。卽首尾兩線、矩內直角形。與中線上直角方形、等。首尾線矩內直角形、與中線上直角方形、等。卽三線為連比例。 
Let the three straight lines A, B, C be proportional, so that, as A is to B, so is B to C;  I say that the rectangle contained by A, C is equal to the square on B. 
   
   
Let D be made equal to B. 
 
 
Then, since, as A is to B, so is B to C,  and B is equal to D,  therefore, as A is to B, so is D to C.  But, if four straight lines be proportional,  the rectangle contained by the extremes is equal to the rectangle contained by the means. [VI. 16]  Therefore the rectangle A, C is equal to the rectangle B, D.  But the rectangle B, D is the square on B,  for B is equal to D;  therefore the rectangle contained by A, C is equal to the square on B. 
                 
                 
Next, let the rectangle A, C be equal to the square on B;  I say that, as A is to B, so is B to C. 
   
   
For, with the same construction,  since the rectangle A, C is equal to the square on B,  while the square on B is the rectangle B, D,  for B is equal to D,  therefore the rectangle A, C is equal to the rectangle B, D.  But, if the rectangle contained by the extremes be equal to that contained by the means,  the four straight lines are proportional. [VI. 16]  Therefore, as A is to B, so is D to C.  But B is equal to D;  therefore, as A is to B, so is B to C. 
                   
                   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 18. 
 
第十八題 
On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure. 
 
直線上。求作直線形。與所設直線形、相似而體勢等。 
Let AB be the given straight line and CE the given rectilineal figure;  thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE. 
   
   
Let DF be joined,  and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C,  and the angle ABG equal to the angle CDF. [I. 23]  Therefore the remaining angle CFD is equal to the angle AGB; [I. 32]  therefore the triangle FCD is equiangular with the triangle GAB.  Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB.  Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE,  and the angle GBH equal to the angle FDE. [I. 23]  Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32]  therefore the triangle FDE is equiangular with the triangle GBH;  therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4]  But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB;  therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB.  And, since the angle CFD is equal to the angle AGB,  and the angle DFE to the angle BGH,  therefore the whole angle CFE is equal to the whole angle AGH.  For the same reason the angle CDE is also equal to the angle ABH.  And the angle at C is also equal to the angle at A,  and the angle at E to the angle at H.  Therefore AH is equiangular with CE;  and they have the sides about their equal angles proportional;  therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1] 
                                           
                                           
Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE.  Q. E. F. 
   
   
PROPOSITION 19. 
 
第十九 
Similar triangles are to one another in the duplicate ratio of the corresponding sides. 
 
相似三角形之比例。為其相似。邊再加之比例。 
Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [V. Def. 11]  I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF. 
   
   
For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11]  and let AG be joined. 
   
   
Since then, as AB is to BC, so is DE to EF,  therefore, alternately, as AB is to DE, so is BC to EF. [V. 16]  But, as BC is to EF, so is EF to BG;  therefore also, as AB is to DE, so is EF to BG. [V. 11]  Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional.  But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15]  therefore the triangle ABG is equal to the triangle DEF.  Now since, as BC is to EF, so is EF to BG,  and, if three straight lines be proportional,  the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9]  therefore BC has to BG a ratio duplicate of that which CB has to EF.  But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1]  therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF.  But the triangle ABG is equal to the triangle DEF;  therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF. 
                             
                             
Therefore etc. 
 
 
PORISM.
From this it is manifest that, if three straight lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second. 
 
 
Q. E. D. 
 
 
PROPOSITION 20. 
Theor. 20 Propos. 20 
第二十題 三支 
Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side. 
Si sint tres magnitudines, et aliae ipsis aequales numero, quae binae, et in eadem ratione sumantur; ex aequo autem prima quam tertia, maior fuerit; Erit et quarta quam sexta, maior. Quod si prima quam tertia fuerit aequalis, erit et quarta aequalis sextae; sin illa minor, haec quoque minor erit. 
以三角形、分相似之多邊直線形。則分數必等。而相當之各三角形、各相似。其各相當兩三角形之比例。若兩元形之比例。其元形之比例。為兩相似邊再加之比例。 
Let ABCDE, FGHKL be similar polygons, and let AB correspond to FG;  I say that the polygons ABCDE, FGHKL are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG. 
SINT tres magnitudines A, B, C, et totidem D, E, F, sitque A, ad B, ut D, ad E; et B, ad C, ut E, ad F, sit autem primum A, prima maior quam C, tertia. Dico et D, quartam esse maiorem F, sexta. Cum enim A, maior sit quam C, erit maior proportio A, ad B, 8. quinti. quam C, ad B. Estautem ut A, ad B, ita D, ad E. Maior igitur 13. quinti. proportio quoque erit D, ad E, quam C, ad B. At ut C, ad B, ita est F, ad E. (Cum enim sit B, ad C, ut E, ad F, erit conuertendo ut C, ad B, ita F, ad E.) Maiorigitur quoque proportio erit D, ad E, quam 10. quinti. F, ad E. Quare D, maior erit, quam F. Quod est propositum.   
先解曰。此甲乙丙丁戊、彼己庚辛壬癸、兩多邊直線形。其乙甲戊、庚己癸、兩角等。餘相當之各角俱等。而各等角旁各兩邊之比例各等。  題先言各以角形分之。其角形之分數必等。而相當之各角形各相似。 
Let BE, EC, GL, LH be joined. 
 
論曰。試從乙甲戊、庚己癸、兩角。向各對角、俱作直線。為甲丙、甲丁、己辛、己壬。 
Now, since the polygon ABCDE is similar to the polygon FGHKL,  the angle BAE is equal to the angle GFL;  and, as BA is to AE, so is GF to FL. [VI. Def. 1]  Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional,  therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6]  so that it is also similar; [VI. 4 and Def. 1]  therefore the angle ABE is equal to the angle FGL.  But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons;  therefore the remaining angle EBC is equal to the angle LGH.  And, since, because of the similarity of the triangles ABE, FGL, as EB is to BA, so is LG to GF,  and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH,  therefore, ex aequali, as EB is to BC, so is LG to GH; [V. 22]  that is, the sides about the equal angles EBC, LGH are proportional;  therefore the triangle EBC is equiangular with the triangle LGH, [VI. 6]  so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1]  For the same reason the triangle ECD is also similar to the triangle LHK.  Therefore the similar polygons ABCDE, FGHKL have been divided into similar triangles, and into triangles equal in multitude. 
                              SIT deinde A, aequalis ipsi C. Dico et D, aequalem esse ipsi F. Cum enim A, sit ipsi C, aequalis, erit A, ad B, ut C, ad B. Est autem ut A, ad B, ita 7. quinti. D, ad EIgitur erit et D, ad E, ut C, ad B: At ut C, ad B, ita est F, ad E, per inuersam rationem, 11. quinti. uti prius. Quare erit quoque D, ad E, ut F, ad E; Ideoque aequales erunt D, et F. Quod est pro9. quinti. positum. SIT tertio A, minor quam C. Dico et D, minorem esse, quam F. Cum enim A, minor sit quam C, erit minor proportio A, ad 8. quinti. B, quam C, ad B. sed ut A, ad B, ita est D, ad E. Minor ergo quoque proportio est D, ad 13. quinti. E, quam C, ad B. Est autem conuertendo, ut prius, ut C, ad B, ita F, ad E. Igitur minor est quoque proportio D, ad E, quam F, ad E, proptereaque D, minor erit quam F. Quod 10. quinti. est propositum. si sint itaque tres magnitudines, et aliae ipsis aequales numero, etc.   
其元形旣相似。卽角數等。而所分角形之數亦等。  又乙角旣與庚角等。而角旁各兩邊之比例亦等。卽甲乙丙、與己庚辛、兩角形必相似。本篇六乙甲丙、與庚己辛、兩角。甲丙乙、與己辛庚、兩角。各等。而各等角旁、各兩邊之比例、各等。本篇四依顯甲戊丁、己癸壬、兩角形亦相似。又甲丙與丙乙之比例。旣若己辛與辛庚。而丙乙與丙丁。若辛庚與辛壬。兩元形相似故平之。卽甲丙與丙丁。若己辛與辛壬也。五卷廿二又乙丙丁角。旣與庚辛壬角等。而各減一相等之甲丙乙角、己辛庚角。卽所存甲丙丁角、與己辛壬角、必等。則甲丙丁與己辛壬兩角形。亦等角形。亦相似矣。本篇六                               
I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents, and that the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG. 
 
次解曰。題又言各相當角形之比例。若兩元形之比例。 
For let AC, FH be joined.  Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH,  and, as AB is to BC, so is FG to GH,  the triangle ABC is equiangular with the triangle FGH; [VI. 6]  therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF.  And, since the angle BAM is equal to the angle GFN,  and the angle ABM is also equal to the angle FGN,  therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32]  therefore the triangle ABM is equiangular with the triangle FGN.  Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH.  Therefore, proportionally, as AM is to MB, so is FN to NG,  and, as BM is to MC, so is GN to NH; so that,  in addition, ex aequali, as AM is to MC, so is FN to NH.  But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC;  for they are to one another as their bases. [VI. 1]  Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12]  therefore, as the triangle AMB is to BMC, so is ABE to CBE.  But, as AMB is to BMC, so is AM to MC;  therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC.  For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH.  And, as AM is to MC, so is FN to NH;  therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH;  and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH.  Similarly we can prove, if BD, GK be joined,  that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK.  And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH,  and further ECD to LHK,  therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12]  therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL.  But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG;  for similar triangles are in the duplicate ratio of the corresponding sides. [VI. 19]  Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG. 
                                                               
論曰。甲乙丙、己庚辛、兩角形旣相似。卽兩形之比例。為甲丙、己辛、兩相似邊再加之比例。本篇十九依顯甲丙丁、己辛壬、之比例。亦為甲丙、己辛、再加之比例。則甲乙丙與己庚辛兩角形之比例。若甲丙丁與己辛壬兩角形之比例。依顯甲丁戊與己壬癸之比例。亦若甲丙丁與己辛壬之比例。則此形中諸角形之比例。若彼形中諸角形之比例。此諸形為前率。彼諸形為後率。而一前與一後之比例。又若幷前與幷後之比例。五卷十二卽此一角形、與相當彼一角形之比例。若此元形、與彼元形之比例矣。

後解曰。題又言兩多邊元形之比例。為兩相似邊再加之比例。
論曰。甲乙丙、與己庚辛、兩角形之比例。旣若甲乙丙丁戊、與己庚辛壬癸、兩多邊形之比例。而甲乙丙、與己庚辛、兩形之比例。為甲乙、己庚、兩相似邊再加之比例。本篇十九則兩元形亦為甲乙、己庚、再加之比例。
增題。此直線、倍大于彼直線。則此線上方形、與彼線上方形。為四倍大之比例。若此方形、與彼方形、 為四倍大之比例。則此方形邊、與彼方形邊、為二倍大之比例。

先解曰。甲線、倍乙線。題言甲上方形、與乙上方形。為四倍大之比例。

論曰。凡直角方形、俱相似。本卷界說一依本題論。則甲方形與乙方形之比例。為甲線與乙線再加之比例。甲線與乙線。旣為倍大之比例。則兩方形為四倍大之比例矣。何者。四倍大之比例。為二倍大再加之比例。若一、二、四、為連比例故也。

後解曰。若甲上方形、與乙上方形、為四倍大之比例。題言甲邊、與乙邊、為二倍大之比例。
論曰。兩方形四倍大之比例。旣為兩邊再加之比例。則甲邊二倍大于乙邊。                                                               

Therefore etc. 
 
 
PORISM.
Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides.
And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. 
Q. E. D. 
   

系。依此題。可顯三直線為連比例。如甲、乙、丙。則第一線上多邊形、與第二線上相似多邊形之比例。若第一線與第三線之比例。

此系與本篇第十九題之系同論。    

 
PORRO propositione 22. ostendet Euclides, A, et D, magnitudines non solum esse maiores, vel aequales, vel minores duabus magnitudinibus C, et E, ut hic demonstrauit, sed etiam illas ad has eandem habere proportionem ex aequalitate: quod quidem demonstrare non poterat, nisi prius theorema hoc ostendisset, ut ex eadem propositione 22. erit perspicuum. 
 
PROPOSITION 21. 
 
第二十一題 
Figures which are similar to the same rectilineal figure are also similar to one another. 
 
兩直線形。各與他直線形相似。則自相似。 
For let each of the rectilineal figures A, B be similar to C;  I say that A is also similar to B. 
   
   
For, since A is similar to C,  it is equiangular with it and has the sides about the equal angles proportional. [VI. Def. 1]  Again, since B is similar to C,  it is equiangular with it and has the sides about the equal angles proportional.  Therefore each of the figures A, B is equiangular with C and with C has the sides about the equal angles proportional;  therefore A is similar to B.  Q. E. D. 
             
             
PROPOSITION 22. 
 
第二十二題二支 
If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional. 
 
四直線為斷比例。則兩比例線上、各任作自相似之直線形。亦為斷比例。兩比例線上、各任作自相似之直線形、為斷比例。則四直線亦為斷比例。 
Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH;  I say that, as KAB is to LCD, so is MF to NH. 
   
   
For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11]  Then since, as AB is to CD, so is EF to GH,  and, as CD is to O, so is GH to P,  therefore, ex aequali, as AB is to O, so is EF to P. [V. 22]  But, as AB is to O, so is KAB to LCD, [VI. 19, Por.]  and, as EF is to P, so is MF to NH;  therefore also, as KAB is to LCD, so is MF to NH. [V. 11] 
             
             
Next, let MF be to NH as KAB is to LCD;  I say also that, as AB is to CD, so is EF to GH.  For, if EF is not to GH as AB to CD,  let EF be to QR as AB to CD, [VI. 12]  and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH. [VI. 18] 
         
         
Since then, as AB is to CD, so is EF to QR,  and there have been described on AB, CD the similar and similarly situated figures KAB, LCD,  and on EF, QR the similar and similarly situated figures MF, SR,  therefore, as KAB is to LCD, so is MF to SR.  But also, by hypothesis, as KAB is to LCD, so is MF to NH;  therefore also, as MF is to SR, so is MF to NH. [V. 11]  Therefore MF has the same ratio to each of the figures NH, SR;  therefore NH is equal to SR. [V. 9]  But it is also similar and similarly situated to it;  therefore GH is equal to QR.  And, since, as AB is to CD, so is EF to QR,  while QR is equal to GH,  therefore, as AB is to CD, so is EF to GH. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
 
 
 
PROPOSITION 23. 
 
第二十三題 
Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides. 
 
等角兩平行方形之比例。以兩形之各兩邊兩比例相結。 
Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG;  I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides. 
   
   
For let them be placed so that BC is in a straight line with CG;  therefore DC is also in a straight line with CE. 
   
   
Let the parallelogram DG be completed;  let a straight line K be set out,  and let it be contrived that, as BC is to CG, so is K to L,  and, as DC is to CE, so is L to M. [VI. 12] 
       
       
Then the ratios of K to L and of L to M are the same as the ratios of the sides,  namely of BC to CG and of DC to CE.  But the ratio of K to M is compounded of the ratio of K to L and of that of L to M;  so that K has also to M the ratio compounded of the ratios of the sides.  Now since, as BC is to CG, so is the parallelogram AC to the parallelogram CH, [VI. 1]  while, as BC is to CG, so is K to L,  therefore also, as K is to L, so is AC to CH. [V. 11]  Again, since, as DC is to CE, so is the parallelogram CH to CF, [VI. 1]  while, as DC is to CE, so is L to M,  therefore also, as L is to M, so is the parallelogram CH to the parallelogram CF. [V. 11]  Since then it was proved that, as K is to L, so is the parallelogram AC to the parallelogram CH,  and, as L is to M, so is the parallelogram CH to the parallelogram CF,  therefore, ex aequali, as K is to M, so is AC to the parallelogram CF.  But K has to M the ratio compounded of the ratios of the sides;  therefore AC also has to CF the ratio compounded of the ratios of the sides. 
                             
                             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 24. 
 
第二十四題 
In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another. 
 
平行線方形之兩角線方形。自相似。亦與全形相似。 
Let ABCD be a parallelogram, and AC its diameter, and let EG, HK be parallelograms about AC;  I say that each of the parallelograms EG, HK is similar both to the whole ABCD and to the other. 
   
   
For, since EF has been drawn parallel to BC, one of the sides of the triangle ABC,  proportionally, as BE is to EA, so is CF to FA. [VI. 2]  Again, since FG has been drawn parallel to CD, one of the sides of the triangle ACD,  proportionally, as CF is to FA, so is DG to GA. [VI. 2]  But it was proved that, as CF is to FA, so also is BE to EA;  therefore also, as BE is to EA, so is DG to GA,  and therefore, componendo, as BA is to AE, so is DA to AG, [V. 18]  and, alternately, as BA is to AD, so is EA to AG. [V. 16]  Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional.  And, since GF is parallel to DC,  the angle AFG is equal to the angle DCA;  and the angle DAC is common to the two triangles ADC, AGF;  therefore the triangle ADC is equiangular with the triangle AGF.  For the same reason the triangle ACB is also equiangular with the triangle AFE,  and the whole parallelogram ABCD is equiangular with the parallelogram EG.  Therefore, proportionally, as AD is to DC, so is AG to GF,  as DC is to CA, so is GF to FA,  as AC is to CB, so is AF to FE,  and further, as CB is to BA, so is FE to EA.  And, since it was proved that, as DC is to CA, so is GF to FA,  and, as AC is to CB, so is AF to FE,  therefore, ex aequali, as DC is to CB, so is GF to FE. [V. 22]  Therefore in the parallelograms ABCD, EG the sides about the equal angles are proportional;  therefore the parallelogram ABCD is similar to the parallelogram EG. [VI. Def. 1]  For the same reason the parallelogram ABCD is also similar to the parallelogram KH;  therefore each of the parallelograms EG, HK is similar to ABCD.  But figures similar to the same rectilineal figure are also similar to one another; [VI. 21]  therefore the parallelogram EG is also similar to the parallelogram HK. 
                                                       
                                                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 25. 
 
第二十五題 
To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure. 
 
兩直線形、求作他直線形。與一形相似。與一形相等。 
Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal;  thus it is required to construct one and the same figure similar to ABC and equal to D. 
   
   
Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44],  and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45]  Therefore BC is in a straight line with CF, and LE with EM.  Now let GH be taken a mean proportional to BC, CF [VI. 13],  and on GH let KGH be described similar and similarly situated to ABC. [VI. 18] 
         
         
Then, since, as BC is to GH, so is GH to CF,  and, if three straight lines be proportional,  as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19, Por.]  therefore, as BC is to CF, so is the triangle ABC to the triangle KGH.  But, as BC is to CF, so also is the parallelogram BE to the parallelogram EF. [VI. 1]  Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF;  therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16]  But the triangle ABC is equal to the parallelogram BE;  therefore the triangle KGH is also equal to the parallelogram EF.  But the parallelogram EF is equal to D;  therefore KGH is also equal to D.  And KGH is also similar to ABC. 
                       
                       
Therefore one and the same figure KGH has been constructed similar to the given rectilineal figure ABC and equal to the other given figure D.  Q. E. D. 
   
   
PROPOSITION 26. 
 
第二十六題 
If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole 
 
平行方形之內。減一平行方形。其減形與元形。相似而體勢等。又一角同。則減形必依元形之對角線。 
For from the parallelogram ABCD let there be taken away the parallelogram AF similar and similarly situated to ABCD, and having the angle DAB common with it;  I say that ABCD is about the same diameter with AF. 
   
   
For suppose it is not, but, if possible, let AHC be the diameter let GF be produced and carried through to H,  and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31] 
     
     
Since, then, ABCD is about the same diameter with KG,  therefore, as DA is to AB, so is GA to AK. [VI. 24]  But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE;  therefore also, as GA is to AK, so is GA to AE. [V. 11]  Therefore GA has the same ratio to each of the straight lines AK, AE.  Therefore AE is equal to AK [V. 9], the less to the greater: which is impossible.  Therefore ABCD cannot but be about the same diameter with AF;  therefore the parallelogram ABCD is about the same diameter with the parallelogram AF. 
               
               
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 27. 
 
第二十七題 
Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect. 
 
凡依直線之有闕平行方形。不滿線者。其闕形、與半線上之闕形、相似而體勢等。則半線上似闕形之有闕依形。必大于此有闕依形。 
Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB; I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest. For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB;  I say that AD is greater than AF. 
   
   
For, since the parallelogram DB is similar to the parallelogram FB, they are about the same diameter. [VI. 26]  Let their diameter DB be drawn, and let the figure be described. 
   
   
Then, since CF is equal to FE, [I. 43] and FB is common,  therefore the whole CH is equal to the whole KE.  But CH is equal to CG, since AC is also equal to CB. [I. 36]  Therefore GC is also equal to EK.  Let CF be added to each;  therefore the whole AF is equal to the gnomon LMN;  so that the parallelogram DB, that is, AD, is greater than the parallelogram AF. 
             
             
Therefore etc.   
   
   
PROPOSITION 28. 
 
第二十八題 
To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect. 
 
一直線。求作依線之有闕平行方形。與所設直線形等。而其闕形、與所設平行方形相似。其所設直線形。不大于半線上所作平行方形。與所設平行方形相似者。 
Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar;  thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D. 
   
   
Let AB be bisected at the point E,  and on EB let EBFG be described similar and similarly situated to D; [VI. 18]  let the parallelogram AG be completed. 
     
     
If then AG is equal to C, that which was enjoined will have been done;  for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D.  But, if not, let HE be greater than C.  Now HE is equal to GB;  therefore GB is also greater than C.  Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25]  But D is similar to GB;  therefore KM is also similar to GB. [VI. 21]  Let, then, KL correspond to GE, and LM to GF.  Now, since GB is equal to C, KM,  therefore GB is greater than KM;  therefore also GE is greater than KL, and GF than LM.  Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed;  therefore it is equal and similar to KM.  Therefore GQ is also similar to GB; [VI. 21]  therefore GQ is about the same diameter with GB. [VI. 26]  Let GQB be their diameter, and let the figure be described. 
                                 
                                 
Then, since BG is equal to C, KM, and in them GQ is equal to KM,  therefore the remainder, the gnomon UWV, is equal to the remainder C.  And, since PR is equal to OS,  let QB be added to each;  therefore the whole PB is equal to the whole OB.  But OB is equal to TE,  since the side AE is also equal to the side EB; [I. 36]  therefore TE is also equal to PB.  Let OS be added to each;  therefore the whole TS is equal to the whole, the gnomon VWU.  But the gnomon VWU was proved equal to C;  therefore TS is also equal to C. 
                       
                       
Therefore to the given straight line AB there has been applied the parallelogram ST equal to the given rectilineal figure C and deficient by a parallelogrammic figure QB which is similar to D.  Q. E. F. 
   
   
PROPOSITION 29. 
 
第二十九題 
To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one. 
 
一直線。求作依線之帶餘平行方形。與所設直線形等。而其餘形、與所設平行方形相似。 
Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar;  thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D. 
   
   
Let AB be bisected at E;  let there be described on EB the parallelogram BF similar and similarly situated to D;  and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25]  Let KH correspond to FL and KG to FE.  Now, since GH is greater than FB,  therefore KH is also greater than FL,  and KG than FE.  Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed;  therefore MN is both equal and similar to GH.  But GH is similar to EL;  therefore MN is also similar to EL; [VI. 21]  therefore EL is about the same diameter with MN. [VI. 26]  Let their diameter FO be drawn, and let the figure be described. 
                         
                         
Since GH is equal to EL, C,  while GH is equal to MN,  therefore MN is also equal to EL, C.  Let EL be subtracted from each;  therefore the remainder, the gnomon XWV, is equal to C.  Now, since AE is equal to EB,  AN is also equal to NB [I. 36], that is, to LP [I. 43].  Let EO be added to each;  therefore the whole AO is equal to the gnomon VWX.  But the gnomon VWX is equal to C;  therefore AO is also equal to C. 
                     
                     
Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24].  Q. E. F. 
   
   
PROPOSITION 30. 
 
第三十題 
To cut a given finite straight line in extreme and mean ratio. 
 
一直線。求作理分中末線。 
Let AB be the given finite straight line;  thus it is required to cut AB in extreme and mean ratio. 
   
   
On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29] 
 
 
Now BC is a square;  therefore AD is also a square.  And, since BC is equal to CD,  let CE be subtracted from each;  therefore the remainder BF is equal to the remainder AD.  But it is also equiangular with it;  therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14]  therefore, as FE is to ED, so is AE to EB.  But FE is equal to AB, and ED to AE.  Therefore, as BA is to AE, so is AE to EB.  And AB is greater than AE;  therefore AE is also greater than EB. 
                       
                       
Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE.  Q. E. F. 
   
   
PROPOSITION 31. 
 
第三十一題 
In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. 
 
三邊直角形之對直角邊上一形。與直角旁邊上兩形。若相似而體勢等。則一形與兩形幷、等。 
Let ABC be a right-angled triangle having the angle BAC right;  I say that the figure on BC is equal to the similar and similarly described figures on BA, AC. 
   
   
Let AD be drawn perpendicular. 
 
 
Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8]  And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1]  And, since three straight lines are proportional,  as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [VI. 19, Por.]  Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA.  For the same reason also, as BC is to CD, so is the figure on BC to that on CA;  so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC.  But BC is equal to BD, DC;  therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 32. 
 
第三十二題 
If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line. 
 
兩三角形。此形之兩邊。與彼形之兩邊、相似。而平置兩形。成一外角。若备相似之各兩邊、各平行。則其餘各一邊、相聯為一直線。 
Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE;  I say that BC is in a straight line with CE. 
   
   
For, since AB is parallel to DC,  and the straight line AC has fallen upon them,  the alternate angles BAC, ACD are equal to one another. [I. 29]  For the same reason the angle CDE is also equal to the angle ACD;  so that the angle BAC is equal to the angle CDE.  And, since ABC, DCE are two triangles having one angle, the angle at A, equal to one angle, the angle at D,  and the sides about the equal angles proportional,  so that, as BA is to AC, so is CD to DE,  therefore the triangle ABC is equiangular with the triangle DCE; [VI. 6]  therefore the angle ABC is equal to the angle DCE.  But the angle ACD was also proved equal to the angle BAC;  therefore the whole angle ACE is equal to the two angles ABC, BAC.  Let the angle ACB be added to each;  therefore the angles ACE, ACB are equal to the angles BAC, ACB, CBA.  But the angles BAC, ABC, ACB are equal to two right angles; [I. 32]  therefore the angles ACE, ACB are also equal to two right angles.  Therefore with a straight line AC, and at the point C on it, the two straight lines BC, CE not lying on the same side make the adjacent angles ACE, ACB equal to two right angles;  therefore BC is in a straight line with CE. [I. 14] 
                                   
                                   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 33. 
THEOR. 23. PROPOS. 33. 
第三十三題三支 
In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences. 
IN aequalibus circulis, anguli eandem habent rationem cum peripherijs, quibus insistunt, sive ad centra, sive ad peripherias constituti insistant: Insuper vero et sectores, quippe qui ad centra consistunt. 
等圜之乘圜分角。或在心。或在界。其各相當兩乘圜角之比例。皆若所乘兩圜分之比例。而兩分圜形之比例。亦若所乘兩圜分之比例。 
Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences;  I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. 
SINT duo circuli aequales ABC, EFG, quorum centra D, H; sumanturque ex circulis duo arcus quicunque BC, FG, quibus ad centra quidem insistant anguli BDC, FHG; ad circumferentias vero anguli BAC, EFG.  Dico esse ex sententia defin. 6. liber 5. ut arcum BC, ad arcum FG, ira angulum BDC, ad angulum FHG, et angulum BAC, ad angulum EFG; et sectorem insuper BDC, qui rectis BD, DC, et arcu BC, continetur ad sectorem FHG; quem comprehendunt rectae FH, HG, et arcus FG. 
解曰甲乙丙、戊己庚、兩圜等。其心為丁、為辛。兩圜各任割一圜分為乙丙、為己庚。其乘圜角之在心者。為乙丁丙、己辛庚。在界者。為乙甲丙、己戊庚。  題先言乙丙、與己庚、兩圜分之比例。若乙丁丙、與己辛庚、兩角。次言乙甲丙、與己戊庚、兩角之比例。若乙丙、與己庚、兩圜分。後言乙丁、丁丙、兩腰、偕乙丙圜分、內乙丁丙分圜形。 與己辛、辛庚、兩腰、偕己庚圜分、內己辛庚分圜形、之比例。亦若乙丙、與己庚、兩圜分。 
For let any number of consecutive circumferences CK, KL be made equal to the circumference BC,  and any number of consecutive circumferences FM, MN equal to the circumference EF;  and let GK, GL, HM, HN be joined. 
     
     
Then, since the circumferences BC, CK, KL are equal to one another,  the angles BGC, CGK, KGL are also equal to one another; [III. 27]  therefore, whatever multiple the circumference BL is of BC,  that multiple also is the angle BGL of the angle BGC.  For the same reason also, whatever multiple the circumference NE is of EF,  that multiple also is the angle NHE of the angle EHF.  If then the circumference BL is equal to the circumference EN,  the angle BGL is also equal to the angle EHN; [III. 27]  if the circumference BL is greater than the circumference EN,  the angle BGL is also greater than the angle EHN; and, if less, less.  There being then four magnitudes,  two circumferences BC, EF,  and two angles BGC, EHF,  there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL,  and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN.  And it has been proved that, if the circumference BL is in excess of the circumference EN,  the angle BGL is also in excess of the angle EHN;  if equal, equal; and if less, less.  Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF. [V. Def. 5]  But, as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF;  for they are doubles respectively.  Therefore also, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. 
                                           
                                           
Therefore etc.  Q. E. D. 
   
   
BOOK VΙΙ. 
 
 
DEFINITIONS. 
 
 
1. An unit is that by virtue of which each of the things that exist is called one. 
 
 
2. A number is a multitude composed of units. 
 
 
3. A number is a part of a number, the less of the greater, when it measures the greater; 
 
 
4. but parts when it does not measure it. 
 
 
5. The greater number is a multiple of the less when it is measured by the less. 
 
 
6. An even number is that which is divisible into two equal parts. 
 
 
7. An odd number is that which is not divisible into two equal parts, or that which differs by an unit from an even number. 
 
 
8. An even-times even number is that which is measured by an even number according to an even number. 
 
 
9. An even-times odd number is that which is measured by an even number according to an odd number. 
 
 
10. An odd-times odd number is that which is measured by an odd number according to an odd number. 
 
 
11. A prime number is that which is measured by an unit alone. 
 
 
12. Numbers prime to one another are those which are measured by an unit alone as a common measure. 
 
 
13. A composite number is that which is measured by some number. 
 
 
14. Numbers composite to one another are those which are measured by some number as a common measure. 
 
 
15. A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other, and thus some number is produced. 
 
 
16. And, when two numbers having multiplied one another make some number, the number so produced is called plane, and its sides are the numbers which have multiplied one another. 
 
 
17. And, when three numbers having multiplied one another make some number, the number so produced is solid, and its sides are the numbers which have multiplied one another. 
 
 
18. A square number is equal multiplied by equal, or a number which is contained by two equal numbers. 
 
 
19. And a cube is equal multiplied by equal and again by equal, or a number which is contained by three equal numbers. 
 
 
20. Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third is of the fourth. 
 
 
21. Similar plane and solid numbers are those which have their sides proportional. 
 
 
22. A perfect number is that which is equal to its own parts. 
 
 
PROPOSITION I. 
 
 
Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until an unit is left, the original numbers will be prime to one another. 
 
 
For, the less of two unequal numbers AB, CD being continually subtracted from the greater, let the number which is left never measure the one before it until an unit is left;  I say that AB, CD are prime to one another, that is, that an unit alone measures AB, CD. 
   
   
For, if AB, CD are not prime to one another, some number will measure them.  Let a number measure them, and let it be E;  let CD, measuring BF, leave FA less than itself,  let AF, measuring DG, leave GC less than itself,  and let GC, measuring FH, leave an unit HA. 
         
         
Since, then, E measures CD, and CD measures BF, therefore E also measures BF.  But it also measures the whole BA; therefore it will also measure the remainder AF.  But AF measures DG; therefore E also measures DG.  But it also measures the whole DC therefore it will also measure the remainder CG.  But CG measures FH; therefore E also measures FH.  But it also measures the whole FA; therefore it will also measure the remainder, the unit AH, though it is a number: which is impossible.  Therefore no number will measure the numbers AB, CD; therefore AB, CD are prime to one another. [VII. Def. 12]  Q. E. D. 
               
               
PROPOSITION 2. 
 
 
Given two numbers not prime to one another, to find their greatest common measure. 
 
 
Let AB, CD be the two given numbers not prime to one another.  Thus it is required to find the greatest common measure of AB, CD. 
   
   
If now CD measures AB — and it also measures itself —  CD is a common measure of CD, AB.  And it is manifest that it is also the greatest;  for no greater number than CD will measure CD. 
       
       
But, if CD does not measure AB, then, the less of the numbers AB, CD being continually subtracted from the greater, some number will be left which will measure the one before it.  For an unit will not be left; otherwise AB, CD will be prime to one another [VII. 1], which is contrary to the hypothesis.  Therefore some number will be left which will measure the one before it.  Now let CD, measuring BE, leave EA less than itself,  let EA, measuring DF, leave FC less than itself, and let CF measure AE.  Since then, CF measures AE, and AE measures DF, therefore CF will also measure DF.  But it also measures itself; therefore it will also measure the whole CD.  But CD measures BE; therefore CF also measures BE.  But it also measures EA; therefore it will also measure the whole BA.  But it also measures CD; therefore CF measures AB, CD.  Therefore CF is a common measure of AB, CD.  I say next that it is also the greatest.  For, if CF is not the greatest common measure of AB, CD, some number which is greater than CF will measure the numbers AB, CD.  Let such a number measure them, and let it be G.  Now, since G measures CD, while CD measures BE, G also measures BE.  But it also measures the whole BA; therefore it will also measure the remainder AE.  But AE measures DF; therefore G will also measure DF.  But it also measures the whole DC; therefore it will also measure the remainder CF, that is, the greater will measure the less: which is impossible.  Therefore no number which is greater than CF will measure the numbers AB, CD;  therefore CF is the greatest common measure of AB, CD.   
                                         
                                         
PORISM.
From this it is manifest that, if a number measure two numbers, it will also measure their greatest common measure.
Q. E. D. 
 
 
PROPOSITION 3. 
 
 
Given three numbers not prime to one another, to find their greatest common measure. 
 
 
Let A, B, C be the three given numbers not prime to one another;  thus it is required to find the greatest common measure of A, B, C. 
   
   
For let the greatest common measure, D, of the two numbers A, B be taken; [VII. 2]  then D either measures, or does not measure, C.  First, let it measure it.  But it measures A, B also; therefore D measures A, B, C;  therefore D is a common measure of A, B, C.  I say that it is also the greatest.  For, if D is not the greatest common measure of A, B, C, some number which is greater than D will measure the numbers A, B, C.  Let such a number measure them, and let it be E.  Since then E measures A, B, C, it will also measure A, B;  therefore it will also measure the greatest common measure of A, B. [VII. 2, Por.]  But the greatest common measure of A, B is D;  therefore E measures D, the greater the less: which is impossible.  Therefore no number which is greater than D will measure the numbers A, B, C;  therefore D is the greatest common measure of A, B, C. 
                           
                           
Next, let D not measure C;  I say first that C, D are not prime to one another.  For, since A, B, C are not prime to one another, some number will measure them.  Now that which measures A, B, C will also measure A, B,  and will measure D, the greatest common measure of A, B. [VII. 2, Por.]  But it measures C also; therefore some number will measure the numbers D, C;  therefore D, C are not prime to one another.  Let then their greatest common measure E be taken. [VII. 2]  Then, since E measures D, and D measures A, B, therefore E also measures A, B.  But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.  I say next that it is also the greatest.  For, if E is not the greatest common measure of A, B, C, some number which is greater than E will measure the numbers A, B, C.  Let such a number measure them, and let it be F.  Now, since F measures A, B, C, it also measures A, B;  therefore it will also measure the greatest common measure of A, B. [VII. 2, Por.]  But the greatest common measure of A, B is D; therefore F measures D.  And it measures C also; therefore F measures D, C;  therefore it will also measure the greatest common measure of D, C. [VII. 2, Por.]  But the greatest common measure of D, C is E;  therefore F measures E, the greater the less: which is impossible.  Therefore no number which is greater than E will measure the numbers A, B, C;  therefore E is the greatest common measure of A, B, C.  Q. E. D. 
                                             
                                             
PROPOSITION 4. 
 
 
Any number is either a part or parts of any number, the less of the greater. 
 
 
Let A, BC be two numbers, and let BC be the less;  I say that BC is either a part, or parts, of A. 
   
   
For A, BC are either prime to one another or not.  First, let A, BC be prime to one another.  Then, if BC be divided into the units in it, each unit of those in BC will be some part of A;  so that BC is parts of A. 
       
       
Next let A, BC not be prime to one another;  then BC either measures, or does not measure, A.  If now BC measures A, BC is a part of A.  But, if not, let the greatest common measure D of A, BC be taken; [VII. 2]  and let BC be divided into the numbers equal to D, namely BE, EF, FC.  Now, since D measures A, D is a part of A.  But D is equal to each of the numbers BE, EF, FC;  therefore each of the numbers BE, EF, FC is also a part of A;  so that BC is parts of A. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 5. 
 
 
If a number be a part of a number, and another be the same part of another, the sum will also be the same part of the sum that the one is of the one. 
 
 
For let the number A be a part of BC, and another, D, the same part of another EF that A is of BC;  I say that the sum of A, D is also the same part of the sum of BC, EF that A is of BC. 
   
   
For since, whatever part A is of BC, D is also the same part of EF,  therefore, as many numbers as there are in BC equal to A, so many numbers are there also in EF equal to D.  Let BC be divided into the numbers equal to A, namely BG, GC, and EF into the numbers equal to D, namely EH, HF;  then the multitude of BG, GC will be equal to the multitude of EH, HF.  And, since BG is equal to A, and EH to D, therefore BG, EH are also equal to A, D.  For the same reason GC, HF are also equal to A, D.  Therefore, as many numbers as there are in BC equal to A, so many are there also in BC, EF equal to A, D.  Therefore, whatever multiple BC is of A, the same multiple also is the sum of BC, EF of the sum of A, D.  Therefore, whatever part A is of BC, the same part also is the sum of A, D of the sum of BC, EF.  Q. E. D. 
                   
                   
PROPOSITION 6. 
 
 
If a number be parts of a number, and another be the same parts of another, the sum will also be the same parts of the sum that the one is of the one. 
 
 
For let the number AB be parts of the number C, and another, DE, the same parts of another, F, that AB is of C;  I say that the sum of AB, DE is also the same parts of the sum of C, F that AB is of C. 
   
   
For since, whatever parts AB is of C, DE is also the same parts of F,  therefore, as many parts of C as there are in AB, so many parts of F are there also in DE.  Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE;  thus the multitude of AG, GB will be equal to the multitude of DH, HE.  And since, whatever part AG is of C, the same part is DH of F also, therefore,  whatever part AG is of C, the same part also is the sum of AG, DH of the sum of C, F. [VII. 5]  For the same reason, whatever part GB is of C, the same part also is the sum of GB, HE of the sum of C, F.  Therefore, whatever parts AB is of C, the same parts also is the sum of AB, DE of the sum of C, F.  Q. E. D. 
                 
                 
PROPOSITION 7. 
 
 
If a number be that part of a number, which a number subtracted is of a number subtracted, the remainder will also be the same part of the remainder that the whole is of the whole. 
 
 
For let the number AB be that part of the number CD which AE subtracted is of CF subtracted;  I say that the remainder EB is also the same part of the remainder FD that the whole AB is of the whole CD. 
   
   
For, whatever part AE is of CF, the same part also let EB be of CG.  Now since, whatever part AE is of CF, the same part also is EB of CG,  therefore, whatever part AE is of CF, the same part also is AB of GF. [VII. 5]  But, whatever part AE is of CF, the same part also, by hypothesis, is AB of CD;  therefore, whatever part AB is of GF, the same part is it of CD also;  therefore GF is equal to CD.  Let CF be subtracted from each;  therefore the remainder GC is equal to the remainder FD.  Now since, whatever part AE is of CF, the same part also is EB of GC,  while GC is equal to FD,  therefore, whatever part AE is of CF, the same part also is EB of FD.  But, whatever part AE is of CF, the same part also is AB of CD;  therefore also the remainder EB is the same part of the remainder FD that the whole AB is of the whole CD.  Q. E. D. 
                           
                           
PROPOSITION 8. 
 
 
If a number be the same parts of a number that a number subtracted is of a number subtracted, the remainder will also be the same parts of the remainder that the whole is of the whole. 
 
 
For let the number AB be the same parts of the number CD that AE subtracted is of CF subtracted;  I say that the remainder EB is also the same parts of the remainder FD that the whole AB is of the whole CD. 
   
   
For let GH be made equal to AB.  Therefore, whatever parts GH is of CD, the same parts also is AE of CF.  Let GH be divided into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE;  thus the multitude of GK, KH will be equal to the multitude of AL, LE.  Now since, whatever part GK is of CD, the same part also is AL of CF, while. CD is greater than CF,  therefore GK is also greater than AL.  Let GM be made equal to AL.  Therefore, whatever part GK is of CD, the same part also is GM of CF;  therefore also the remainder MK is the same part of the remainder FD that the whole GK is of the whole CD. [VII. 7]  Again, since, whatever part KH is of CD, the same part also is EL of CF, while CD is greater than CF,  therefore HK is also greater than EL.  Let KN be made equal to EL.  Therefore, whatever part KH is of CD, the same part also is KN of CF;  therefore also the remainder NH is the same part of the remainder FD that the whole KH is of the whole CD. [VII. 7]  But the remainder MK was also proved to be the same part of the remainder FD that the whole GK is of the whole CD;  therefore also the sum of MK, NH is the same parts of DF that the whole HG is of the whole CD.  But the sum of MK, NH is equal to EB, and HG is equal to BA;  therefore the remainder EB is the same parts of the remainder FD that the whole AB is of the whole CD.  Q. E. D. 
                                     
                                     
PROPOSITION 9. 
 
 
If a number be a part of a number, and another be the same part of another, alternately also, whatever part or parts the first is of the third, the same part, or the same parts, will the second also be of the fourth. 
 
 
For let the number A be a part of the number BC, and another, D, the same part of another, EF, that A is of BC;  I say that, alternately also, whatever part or parts A is of D, the same part or parts is BC of EF also. 
   
   
For since, whatever part A is of BC, the same part also is D of EF,  therefore, as many numbers as there are in BC equal to A, so many also are there in EF equal to D.  Let BC be divided into the numbers equal to A, namely BG, GC, and EF into those equal to D, namely EH, HF;  thus the multitude of BG, GC will be equal to the multitude of EH, HF. 
       
       
Now, since the numbers BG, GC are equal to one another, and the numbers EH, HF are also equal to one another,  while the multitude of BG, GC is equal to the multitude of EH, HF,  therefore, whatever part or parts BG is of EH, the same part or the same parts is GC of HF also;  so that, in addition, whatever part or parts BG is of EH,  the same part also, or the same parts, is the sum BC of the sum EF. [VII. 5, 6]  But BG is equal to A, and EH to D;  therefore, whatever part or parts A is of D, the same part or the same parts is BC of EF also.  Q. E. D. 
               
               
PROPOSITION 10. 
 
 
If a number be parts of a number, and another be the same parts of another, alternately also, whatever parts or part the first is of the third, the same parts or the same part will the second also be of the fourth. 
 
 
For let the number AB be parts of the number C, and another, DE, the same parts of another, F;  I say that, alternately also, whatever parts or part AB is of DE, the same parts or the same part is C of F also. 
   
   
For since, whatever parts AB is of C, the same parts also is DE of F,  therefore, as many parts of C as there are in AB, so many parts also of F are there in DE.  Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE;  thus the multitude of AG, GB will be equal to the multitude of DH, HE.  Now since, whatever part AG is of C, the same part also is DH of F,  alternately also, whatever part or parts AG is of DH, the same part or the same parts is C of F also. [VII. 9]  For the same reason also, whatever part or parts GB is of HE, the same part or the same parts is C of F also;  so that,      in addition, whatever parts or part AB is of DE, the same parts also, or the same part, is C of F. [VII. 5, 6]  Q. E. D. 
                       
                       
PROPOSITION 11. 
 
 
If, as whole is to whole, so is a number subtracted to a number subtracted, the remainder will also be to the remainder as whole to whole. 
 
 
As the whole AB is to the whole CD, so let AE subtracted be to CF subtracted;  I say that the remainder EB is also to the remainder FD as the whole AB to the whole CD. 
   
   
Since, as AB is to CD, so is AE to CF,  whatever part or parts AB is of CD,  the same part or the same parts is AE of CF also; [VII. Def. 20]  Therefore also the remainder EB is the same part or parts of FD that AB is of CD. [VII. 7, 8]  Therefore, as EB is to FD, so is AB to CD. [VII. Def. 20]  Q. E. D. 
           
           
PROPOSITION 12. 
 
 
If there be as many numbers as we please in proportion, then, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. 
 
 
Let A, B, C, D be as many numbers as we please in proportion, so that, as A is to B, so is C to D;  I say that, as A is to B, so are A, C to B, D. 
   
   
For since, as A is to B, so is C to D,  whatever part or parts A is of B,  the same part or parts is C of D also. [VII. Def. 20]  Therefore also the sum of A, C is the same part or the same parts of the sum of B, D that A is of B. [VII. 5, 6]  Therefore, as A is to B, so are A, C to B, D. [VII. Def. 20]   
           
           
PROPOSITION 13. 
 
 
If four numbers be proportional, they will also be proportional alternately. 
 
 
Let the four numbers A, B, C, D be proportional, so that, as A is to B, so is C to D;  I say that they will also be proportional alternately, so that, as A is to C, so will B be to D. 
   
   
For since, as A is to B, so is C to D,  therefore, whatever part or parts A is of B, the same part or the same parts is C of D also. [VII. Def. 20]  Therefore, alternately, whatever part or parts A is of C, the same part or the same parts is B of D also. [VII. 10]  Therefore, as A is to C, so is B to D. [VII. Def. 20]  Q. E. D. 
         
         
PROPOSITION 14. 
 
 
If there be as many numbers as we please, and others equal to them in multitude, which taken two and two are in the same ratio, they will also be in the same ratio ex aequali. 
 
 
Let there be as many numbers as we please A, B, C, and others equal to them in multitude D, E, F, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F;  I say that, ex aequali, as A is to C, so also is D to F. 
   
   
For, since, as A is to B, so is D to E,  therefore, alternately, as A is to D, so is B to E. [VII. 13]  Again, since, as B is to C, so is E to F,  therefore, alternately, as B is to E, so is C to F. [VII. 13]  But, as B is to E, so is A to D;  therefore also, as A is to D, so is C to F.  Therefore, alternately, as A is to C, so is D to F. [id.]   
               
               
PROPOSITION 15. 
 
 
If an unit measure any number, and another number measure any other number the same number of times, alternately also, the unit will measure the third number the same number of times that the second measures the fourth. 
 
 
For let the unit A measure any number BC, and let another number D measure any other number EF the same number of times;  I say that, alternately also, the unit A measures the number D the same number of times that BC measures EF. 
   
   
For, since the unit A measures the number BC the same number of times that D measures EF,  therefore, as many units as there are in BC,  so many numbers equal to D are there in EF also.  Let BC be divided into the units in it, BG, GH, HC, and EF into the numbers EK, KL, LF equal to D.  Thus the multitude of BG, GH, HC will be equal to the multitude of EK, KL, LF.  And, since the units BG, GH, HC are equal to one another,  and the numbers EK, KL, LF are also equal to one another,  while the multitude of the units BG, GH, HC is equal to the multitude of the numbers EK, KL, LF,  therefore, as the unit BG is to the number EK,  so will the unit GH be to the number KL, and the unit HC to the number LF.  Therefore also, as one of the antecedents is to one of the consequents,  so will all the antecedents be to all the consequents; [VII. 12]  therefore, as the unit BG is to the number EK, so is BC to EF.  But the unit BG is equal to the unit A,  and the number EK to the number D.  Therefore, as the unit A is to the number D, so is BC to EF.  Therefore the unit A measures the number D the same number of times that BC measures EF.  Q. E. D. 
                                   
                                   
PROPOSITION 16. 
 
 
If two numbers by multiplying one another make certain numbers, the numbers so produced will be equal to one another. 
 
 
Let A, B be two numbers, and let A by multiplying B make C, and B by multiplying A make D;  I say that C is equal to D. 
   
   
For, since A by multiplying B has made C,  therefore B measures C according to the units in A.  But the unit E also measures the number A according to the units in it;  therefore the unit E measures A the same number of times that B measures C.  Therefore, alternately, the unit E measures the number B the same number of times that A measures C. [VII. 15]  Again, since B by multiplying A has made D,  therefore A measures D according to the units in B.  But the unit E also measures B according to the units in it;  therefore the unit E measures the number B the same number of times that A measures D.  But the unit E measured the number B the same number of times that A measures C;  therefore A measures each of the numbers C, D the same number of times.  Therefore C is equal to D.  Q. E. D. 
                         
                         
PROPOSITION 17. 
 
 
If a number by multiplying two numbers make certain numbers, the numbers so produced will have the same ratio as the numbers multiplied. 
 
 
For let the number A by multiplying the two numbers B, C make D, E;  I say that, as B is to C, so is D to E. 
   
   
For, since A by multiplying B has made D,  therefore B measures D according to the units in A.  But the unit F also measures the number A according to the units in it;  therefore the unit F measures the number A the same number of times that B measures D.  Therefore, as the unit F is to the number A, so is B to D. [VII. Def. 20]  For the same reason, as the unit F is to the number A, so also is C to E;  therefore also, as B is to D, so is C to E.  Therefore, alternately, as B is to C, so is D to E. [VII. 13]  Q. E. D. 
                 
                 
PROPOSITION 18. 
 
 
If two numbers by multiplying any number make certain numbers, the numbers so produced will have the same ratio as the multipliers. 
 
 
For let two numbers A, B by multiplying any number C make D, E;  I say that, as A is to B, so is D to E. 
   
   
For, since A by multiplying C has made D,  therefore also C by multiplying A has made D. [VII. 16]  For the same reason also C by multiplying B has made E.  Therefore the number C by multiplying the two numbers A, B has made D, E.  Therefore, as A is to B, so is D to E. [VII. 17]   
           
           
PROPOSITION 19. 
 
 
If four numbers be proportional, the number produced from the first and fourth will be equal to the number produced from the second and third; and, if the number produced from the first and fourth be equal to that produced from the second and third, the four numbers will be proportional. 
 
 
Let A, B, C, D be four numbers in proportion, so that, as A is to B, so is C to D; and let A by multiplying D make E, and let B by multiplying C make F;  I say that E is equal to F. 
   
   
For let A by multiplying C make G.  Since, then, A by multiplying C has made G, and by multiplying D has made E,  the number A by multiplying the two numbers C, D has made G, E.  Therefore, as C is to D, so is G to E. [VII. 17]  But, as C is to D, so is A to B;  therefore also, as A is to B, so is G to E.  Again, since A by multiplying C has made G,  but, further, B has also by multiplying C made F,  the two numbers A, B by multiplying a certain number C have made G, F.  Therefore, as A is to B, so is G to F. [VII. 18]  But further, as A is to B, so is G to E also; therefore also,  as G is to E, so is G to F.  Therefore G has to each of the numbers E, F the same ratio;  therefore E is equal to F. [cf. V. 9] 
                           
                           
Again, let E be equal to F;  I say that, as A is to B, so is C to D. 
   
   
For, with the same construction,  since E is equal to F  therefore, as G is to E, so is G to F. [cf. V. 7]  But, as G is to E, so is C to D, [VII. 17]  and, as G is to F, so is A to B. [VII. 18]  Therefore also, as A is to B, so is C to D.  Q. E. D. 
             
             
PROPOSITION 20. 
 
 
The least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less. 
 
 
For let CD, EF be the least numbers of those which have the same ratio with A, B;  I say that CD measures A the same number of times that EF measures B. 
   
   
Now CD is not parts of A.  For, if possible, let it be so;  therefore EF is also the same parts of B that CD is of A. [VII. 13 and Def. 20]  Therefore, as many parts of A as there are in CD, so many parts of B are there also in EF.  Let CD be divided into the parts of A, namely CG, GD, and EF into the parts of B, namely EH, HF;  thus the multitude of CG, GD will be equal to the multitude of EH, HF.  Now, since the numbers CG, GD are equal to one another, and the numbers EH, HF are also equal to one another,  while the multitude of CG, GD is equal to the multitude of EH, HF,  therefore, as CG is to EH, so is GD to HF.  Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents. [VII. 12]  Therefore, as CG is to EH, so is CD to EF.  Therefore CG, EH are in the same ratio with CD, EF, being less than they: which is impossible,  for by hypothesis CD, EF are the least numbers of those which have the same ratio with them.  Therefore CD is not parts of A; therefore it is a part of it. [VII. 4]  And EF is the same part of B that CD is of A; [VII. 13 and Def. 20]  therefore CD measures A the same number of times that EF measures B.  Q. E. D. 
                                 
                                 
PROPOSITION 21. 
 
 
Numbers prime to one another are the least of those which have the same ratio with them. 
 
 
Let A, B be numbers prime to one another;  I say that A, B are the least of those which have the same ratio with them. 
   
   
For, if not, there will be some numbers less than A, B which are in the same ratio with A, B.  Let them be C, D. 
   
   
Since, then, the least numbers of those which have the same ratio measure those which have the same ratio the same number of times, the greater the greater and the less the less,  that is, the antecedent the antecedent and the consequent the consequent, [VII. 20]  therefore C measures A the same number of times that D measures B.  Now, as many times as C measures A, so many units let there be in E.  Therefore D also measures B according to the units in E.  And, since C measures A according to the units in E,  therefore E also measures A according to the units in C. [VII. 16]  For the same reason E also measures B according to the units in D. [VII. 16]  Therefore E measures A, B which are prime to one another: which is impossible. [VII. Def. 12]  Therefore there will be no numbers less than A, B which are in the same ratio with A, B.  Therefore A, B are the least of those which have the same ratio with them.  Q. E. D. 
                       
                       
PROPOSITION 22. 
 
 
The least numbers of those which have the same ratio with them are prime to one another. 
 
 
Let A, B be the least numbers of those which have the same ratio with them;  I say that A, B are prime to one another. 
   
   
For, if they are not prime to one another, some number will measure them.  Let some number measure them, and let it be C.  And, as many times as C measures A, so many units let there be in D,  and, as many times as C measures B, so many units let there be in E.  Since C measures A according to the units in D,  therefore C by multiplying D has made A. [VII. Def. 15]  For the same reason also C by multiplying E has made B.  Thus the number C by multiplying the two numbers D, E has made A, B;  therefore, as D is to E, so is A to B; [VII. 17]  therefore D, E are in the same ratio with A, B, being less than they: which is impossible.  Therefore no number will measure the numbers A, B.  Therefore A, B are prime to one another.  Q. E. D. 
                         
                         
PROPOSITION 23. 
 
 
If two number be prime to one another, the number which measures the one of them will be prime to the remaining number. 
 
 
Let A, B be two numbers prime to one another, and let any number C measure A;  I say that C, B are also prime to one another. 
   
   
For, if C, B are not prime to one another, some number will measure C, B.  Let a number measure them, and let it be D.  Since D measures C, and C measures A, therefore D also measures A.  But it also measures B;   therefore D measures A, B which are prime to one another: which is impossible. [VII. Def. 12]  Therefore no number will measure the numbers C, B.  Therefore C, B are prime to one another.  Q. E. D. 
               
               
PROPOSITION 24. 
 
 
If two numbers be prime to any number, their product also will be prime to the same. 
 
 
For let the two numbers A, B be prime to any number C, and let A by multiplying B make D;  I say that C, D are prime to one another. 
   
   
For, if C, D are not prime to one another, some number will measure C, D.  Let a number measure them, and let it be E.  Now, since C, A are prime to one another,  and a certain number E measures C, therefore A, E are prime to one another. [VII. 23]  As many times, then, as E measures D, so many units let there be in F;  therefore F also measures D according to the units in E. [VII. 16]  Therefore E by multiplying F has made D. [VII. Def. 15]  But, further, A by multiplying B has also made D;  therefore the product of E, F is equal to the product of A, B.  But, if the product of the extremes be equal to that of the means, the four numbers are proportional; [VII. 19]  therefore, as E is to A, so is B to F.  But A, E are prime to one another,  numbers which are prime to one another are also the least of those which have the same ratio, [VII. 21]  and the least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less,  that is, the antecedent the antecedent and the consequent the consequent; [VII. 20]  therefore E measures B.  But it also measures C;  therefore E measures B, C which are prime to one another: which is impossible. [VII. Def. 12]  Therefore no number will measure the numbers C, D.  Therefore C, D are prime to one another.  Q. E. D. 
                                         
                                         
PROPOSITION 25. 
 
 
If two numbers be prime to one another, the product of one of them into itself will be prime to the remaining one. 
 
 
Let A, B be two numbers prime to one another, and let A by multiplying itself make C:  I say that B, C are prime to one another. 
   
   
For let D be made equal to A.  Since A, B are prime to one another, and A is equal to D, therefore D, B are also prime to one another.  Therefore each of the two numbers D, A is prime to B; therefore the product of D, A will also be prime to B. [VII. 24]  But the number which is the product of D, A is C.  Therefore C, B are prime to one another.  Q. E. D. 
           
           
PROPOSITION 26. 
 
 
If two numbers be prime to two numbers, both to each, their products also will be prime to one another. 
 
 
For let the two numbers A, B be prime to the two numbers C, D; both to each, and let A by multiplying B make E, and let C by multiplying D make F;  I say that E, F are prime to one another. 
   
   
For, since each of the numbers A, B is prime to C,  therefore the product of A, B will also be prime to C. [VII. 24]  But the product of A, B is E;  therefore E, C are prime to one another.  For the same reason E, D are also prime to one another.  Therefore each of the numbers C, D is prime to E.  Therefore the product of C, D will also be prime to E. [VII. 24]  But the product of C, D is F.  Therefore E, F are prime to one another.  Q. E. D. 
                   
                   
PROPOSITION 27. 
 
 
If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and, if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes]. 
 
 
Let A, B be two numbers prime to one another, let A by multiplying itself make C, and by multiplying C make D, and let B by multiplying itself make E, and by multiplying E make F;  I say that both C, E and D, F are prime to one another. 
   
   
For, since A, B are prime to one another,  and A by multiplying itself has made C,  therefore C, B are prime to one another. [VII. 25]  Since then C, B are prime to one another, and B by multiplying itself has made E,  therefore C, E are prime to one another. [id.]  Again, since A, B are prime to one another, and B by multiplying itself has made E,  therefore A, E are prime to one another. [id.]  Since then the two numbers A, C are prime to the two numbers B, E, both to each,  therefore also the product of A, C is prime to the product of B, E. [VII. 26]  And the product of A, C is D, and the product of B, E is F.  Therefore D, F are prime to one another.  Q. E. D. 
                       
                       
PROPOSITION 28. 
 
 
If two numbers be prime to one another, the sum will also be prime to each of them; and, if the sum of two numbers be prime to any one of them, the original numbers will also be prime to one another. 
 
 
For let two numbers AB, BC prime to one another be added;  I say that the sum AC is also prime to each of the numbers AB, BC. 
   
   
For, if CA, AB are not prime to one another, some number will measure CA, AB.  Let a number measure them, and let it be D.  Since then D measures CA, AB, therefore it will also measure the remainder BC.  But it also measures BA; therefore D measures AB, BC which are prime to one another: which is impossible. [VII. Def. 12]  Therefore no number will measure the numbers CA, AB;  therefore CA, AB are prime to one another.  For the same reason AC, CB are also prime to one another.  Therefore CA is prime to each of the numbers AB, BC. 
               
               
Again, let CA, AB be prime to one another;  I say that AB, BC are also prime to one another. 
   
   
For, if AB, BC are not prime to one another, some number will measure AB, BC.  Let a number measure them, and let it be D.  Now, since D measures each of the numbers AB, BC, it will also measure the whole CA.  But it also measures AB;  therefore D measures CA, AB which are prime to one another: which is impossible. [VII. Def. 12]  Therefore no number will measure the numbers AB, BC.  Therefore AB, BC are prime to one another.  Q. E. D. 
               
               
PROPOSITION 29. 
 
 
Any prime number is prime to any number which it does not measure. 
 
 
Let A be a prime number, and let it not measure B;  I say that B, A are prime to one another. 
   
   
For, if B, A are not prime to one another, some number will measure them.  Let C measure them.  Since C measures B, and A does not measure B, therefore C is not the same with A.  Now, since C measures B, A, therefore it also measures A which is prime, though it is not the same with it: which is impossible.  Therefore no number will measure B, A.  Therefore A, B are prime to one another.  Q. E. D. 
             
             
PROPOSITION 30. 
 
 
If two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers. 
 
 
For let the two numbers A, B by multiplying one another make C, and let any prime number D measure C;  I say that D measures one of the numbers A, B. 
   
   
For let it not measure A.  Now D is prime;  therefore A, D are prime to one another. [VII. 29]  And, as many times as D measures C, so many units let there be in E.  Since then D measures C according to the units in E,  therefore D by multiplying E has made C. [VII. Def. 15]  Further, A by multiplying B has also made C;  therefore the product of D, E is equal to the product of A, B.  Therefore, as D is to A, so is B to E. [VII. 19]  But D, A are prime to one another, primes are also least, [VII. 21]  and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less,  that is, the antecedent the antecedent and the consequent the consequent; [VII. 20]  therefore D measures B.  Similarly we can also show that, if D do not measure B, it will measure A.  Therefore D measures one of the numbers A, B.  Q. E. D. 
                               
                               
PROPOSITION 31. 
 
 
Any composite number is measured by some prime number. 
 
 
Let A be a composite number;  I say that A is measured by some prime number. 
   
   
For, since A is composite, some number will measure it.  Let a number measure it, and let it be B.  Now, if B is prime, what was enjoined will have been done.  But if it is composite, some number will measure it.  Let a number measure it, and let it be C.  Then, since C measures B, and B measures A,  therefore C also measures A.  And, if C is prime, what was enjoined will have been done.  But if it is composite, some number will measure it.  Thus, if the investigation be continued in this way, some prime number will be found which will measure the number before it, which will also measure A.  For, if it is not found, an infinite series of numbers will measure the number A,  each of which is less than the other: which is impossible in numbers.  Therefore some prime number will be found which will measure the one before it, which will also measure A. 
                         
                         
Therefore any composite number is measured by some prime number.   
   
   
PROPOSITION 32. 
 
 
Any number either is prime or is measured by some prime number. 
 
 
Let A be a number;  I say that A either is prime or is measured by some prime number. 
   
   
If now A is prime, that which was enjoined will have been done.  But if it is composite, some prime number will measure it. [VII. 31] 
   
   
Therefore any number either is prime or is measured by some prime number.  Q. E. D. 
   
   
PROPOSITION 33. 
 
 
Given as many numbers as we please, to find the least of those which have the same ratio with them. 
 
 
Let A, B, C be the given numbers, as many as we please;  thus it is required to find the least of those which have the same ratio with A, B, C. 
   
   
A, B, C are either prime to one another or not.  Now, if A, B, C are prime to one another,  they are the least of those which have the same ratio with them. [VII. 21]  But, if not, let D the greatest common measure of A, B, C be taken, [VII. 3]  and, as many times as D measures the numbers A, B, C respectively, so many units let there be in the numbers E, F, G respectively.  Therefore the numbers E, F, G measure the numbers A, B, C respectively according to the units in D. [VII. 16]  Therefore E, F, G measure A, B, C the same number of times;  therefore E, F, G are in the same ratio with A, B, C. [VII. Def. 20]  I say next that they are the least that are in that ratio. 
                 
                 
For, if E, F, G are not the least of those which have the same ratio with A, B, C,  there will be numbers less than E, F, G which are in the same ratio with A, B, C.  Let them be H, K, L;  therefore H measures A the same number of times that the numbers K, L measure the numbers B, C respectively.  Now, as many times as H measures A, so many units let there be in M;  therefore the numbers K, L also measure the numbers B, C respectively according to the units in M.  And, since H measures A according to the units in M,  therefore M also measures A according to the units in H. [VII. 16]  For the same reason M also measures the numbers B, C according to the units in the numbers K, L respectively;  Therefore M measures A, B, C.  Now, since H measures A according to the units in M,  therefore H by multiplying M has made A. [VII. Def. 15]  For the same reason also E by multiplying D has made A.  Therefore the product of E, D is equal to the product of H, M.  Therefore, as E is to H, so is M to D. [VII. 19]  But E is greater than H;  therefore M is also greater than D.  And it measures A, B, C: which is impossible,  for by hypothesis D is the greatest common measure of A, B, C.  Therefore there cannot be any numbers less than E, F, G which are in the same ratio with A, B, C.  Therefore E, F, G are the least of those which have the same ratio with A, B, C.  Q. E. D. 
                                           
                                           
PROPOSITION 34. 
 
 
Given two numbers, to find the least number which they measure. 
 
 
Let A, B be the two given numbers;  thus it is required to find the least number which they measure. 
   
   
Now A, B are either prime to one another or not. First, let A, B be prime to one another, and let A by multiplying B make C;therefore also B by multiplying A has made C. [VII. 16] Therefore A, B measure C.  I say next that it is also the least number they measure. 
   
   
For, if not, A, B will measure some number which is less than C.  Let them measure D.  Then, as many times as A measures D, so many units let there be in E, and, as many times as B measures D, so many units let there be in F;  therefore A by multiplying E has made D, and B by multiplying F has made D; [VII. Def. 15]  therefore the product of A, E is equal to the product of B, F.  Therefore, as A is to B, so is F E. [VII. 19]  But A, B are prime, primes are also least, [VII. 21]  and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20]   therefore B measures E, as consequent consequent.  And, since A by multiplying B, E has made C, D, therefore, as B is to E, so is C to D. [VII. 17]  But B measures E;  therefore C also measures D, the greater the less: which is impossible.  Therefore A, B do not measure any number less than C;  therefore C is the least that is measured by A, B. 
                           
                           
Next, let A, B not be prime to one another,  and let F, E, the least numbers of those which have the same ratio with A, B, be taken; [VII. 33]  therefore the product of A, E is equal to the product of B, F. [VII. 19]  And let A by multiplying E make C;  therefore also B by multiplying F has made C;  therefore A, B measure C.  I say next that it is also the least number that they measure. 
             
             
For, if not, A, B will measure some number which is less than C.  Let them measure D.  And, as many times as A measures D, so many units let there be in G,  and, as many times as B measures D, so many units let there be in H.  Therefore A by multiplying G has made D,  and B by multiplying H has made D.  Therefore the product of A, G is equal to the product of B, H;  therefore, as A is to B, so is H to G. [VII. 19]  But, as A is to B, so is F to E.  Therefore also, as F is to E, so is H to G.  But F, E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20]  therefore E measures G.  And, since A by multiplying E, G has made C, D, therefore, as E is to G, so is C to D. [VII. 17]  But E measures G;  therefore C also measures D, the greater the less: which is impossible.  Therefore A, B will not measure any number which is less than C.  Therefore C is the least that is measured by A, B.  Q. E. D. 
                                   
                                   
PROPOSITION 35. 
 
 
If two numbers measure any number, the least number measured by them will also measure the same. 
 
 
For let the two numbers A, B measure any number CD, and let E be the least that they measure;  I say that E also measures CD. 
   
   
For, if E does not measure CD, let E, measuring DF, leave CF less than itself.  Now, since A, B measure E, and E measures DF, therefore A, B will also measure DF.  But they also measure the whole CD;  therefore they will also measure the remainder CF which is less than E: which is impossible.  Therefore E cannot fail to measure CD;  therefore it measures it.  Q. E. D. 
             
             
PROPOSITION 36. 
 
 
Given three numbers, to find the least number which they measure. 
 
 
Let A, B, C be the three given numbers;  thus it is required to find the least number which they measure. 
   
   
Let D, the least number measured by the two numbers A, B, be taken. [VII. 34]  Then C either measures, or does not measure, D.  First, let it measure it.  But A, B also measure D; therefore A, B, C measure D.  I say next that it is also the least that they measure. 
         
         
For, if not, A, B, C will measure some number which is less than D.  Let them measure E.  Since A, B, C measure E, therefore also A, B measure E.  Therefore the least number measured by A, B will also measure E. [VII. 35]  But D is the least number measured by A, B;  therefore D will measure E, the greater the less: which is impossible.  Therefore A, B, C will not measure any number which is less than D;  therefore D is the least that A, B, C measure. 
               
               
Again, let C not measure D,  and let E, the least number measured by C, D, be taken. [VII. 34]  Since A, B measure D, and D measures E, therefore also A, B measure E.  But C also measures E; therefore also A, B, C measure E.  I say next that it is also the least that they measure. 
         
         
For, if not, A, B, C will measure some number which is less than E.  Let them measure F.  Since A, B, C measure F, therefore also A, B measure F;  therefore the least number measured by A, B will also measure F. [VII. 35]  But D is the least number measured by A, B; therefore D measures F.  But C also measures F; therefore D, C measure F,  so that the least number measured by D, C will also measure F.  But E is the least number measured by C, D;  therefore E measures F, the greater the less: which is impossible.  Therefore A, B, C will not measure any number which is less than E.  Therefore E is the least that is measured by A, B, C.  Q. E. D. 
                       
                       
PROPOSITION 37. 
 
 
If a number be measured by any number, the number which is measured will have a part called by the same name as the measuring number. 
 
 
For let the number A be measured by any number B;  I say that A has a part called by the same name as B. 
   
   
For, as many times as B measures A, so many units let there be in C.  Since B measures A according to the units in C,  and the unit D also measures the number C according to the units in it,  therefore the unit D measures the number C the same number of times as B measures A.  Therefore, alternately, the unit D measures the number B the same number of times as C measures A; [VII. 15]  therefore, whatever part the unit D is of the number B, the same part is C of A also.  But the unit D is a part of the number B called by the same name as it;  therefore C is also a part of A called by the same name as B,  so that A has a part C which is called by the same name as B.  Q. E. D. 
                   
                   
PROPOSITION 38. 
 
 
If a number have any part whatever, it will be measured by a number called by the same name as the part. 
 
 
For let the number A have any part whatever, B, and let C be a number called by the same name as the part B;  I say that C measures A. 
   
   
For, since B is a part of A called by the same name as C, and the unit D is also a part of C called by the same name as it,  therefore, whatever part the unit D is of the number C, the same part is B of A also;  therefore the unit D measures the number C the same number of times that B measures A.  Therefore, alternately, the unit D measures the number B the same number of times that C measures A. [VII. 15]  Therefore C measures A.  Q. E. D. 
           
           
PROPOSITION 39. 
 
 
To find the number which is the least that will have given parts. 
 
 
Let A, B, C be the given parts;  thus it is required to find the number which is the least that will have the parts A, B, C. 
   
   
Let D, E, F be numbers called by the same name as the parts A, B, C, and let G, the least number measured by D, E, F, be taken. [VII. 36] 
 
 
Therefore G has parts called by the same name as D, E, F. [VII. 37]  But A, B, C are parts called by the same name as D, E, F;  therefore G has the parts A, B, C.  I say next that it is also the least number that has. 
       
       
For, if not, there will be some number less than G which will have the parts A, B, C.  Let it be H.  Since H has the parts A, B, C, therefore H will be measured by numbers called by the same name as the parts A, B, C. [VII. 38]  But D, E, F are numbers called by the same name as the parts A, B, C; therefore H is measured by D, E, F.  And it is less than G: which is impossible.  Therefore there will be no number less than G that will have the parts A, B, C.  Q. E. D. 
             
             
BOOK VΙΙI. 
 
 
PROPOSITION I. 
 
 
If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the numbers are the least of those which have the same ratio with them. 
 
 
Let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them A, D be prime to one another;  I say that A, B, C, D are the least of those which have the same ratio with them. 
   
   
For, if not, let E, F, G, H be less than A, B, C, D, and in the same ratio with them.  Now, since A, B, C, D are in the same ratio with E, F, G, H,  and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers E, F, G, H,  therefore, ex aequali, as A is to D, so is E to H. [VII. 14]  But A, D are prime, primes are also least, [VII. 21]  and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less,  that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]  Therefore A measures E, the greater the less: which is impossible.  Therefore E, F, G, H which are less than A, B, C, D are not in the same ratio with them.  Therefore A, B, C, D are the least of those which have the same ratio with them.  Q. E. D. 
                     
                     
PROPOSITION 2. 
 
 
To find numbers in continued proportion, as many as may be prescribed, and the least that are in a given ratio. 
 
 
Let the ratio of A to B be the given ratio in least numbers;  thus it is required to find numbers in continued proportion, as many as may be prescribed, and the least that are in the ratio of A to B. 
   
   
Let four be prescribed;  let A by multiplying itself make C, and by multiplying B let it make D;  let B by multiplying itself make E;  further, let A by multiplying C, D, E make F, G, H,  and let B by multiplying E make K. 
         
         
Now, since A by multiplying itself has made C, and by multiplying B has made D,  therefore, as A is to B, so is C to D. [VII. 17]  Again, since A by multiplying B has made D,  and B by multiplying itself has made E,  therefore the numbers A, B by multiplying B have made the numbers D, E respectively.  Therefore, as A is to B, so is D to E. [VII. 18]  But, as A is to B, so is C to D;  therefore also, as C is to D, so is D to E.  And, since A by multiplying C, D has made F, G,  therefore, as C is to D, so is F to G. [VII. 17]  But, as C is to D, so was A to B;  therefore also, as A is to B, so is F to G.  Again, since A by multiplying D, E has made G, H,  therefore, as D is to E, so is G to H. [VII. 17]  But, as D is to E, so is A to B.  Therefore also, as A is to B, so is G to H.  And, since A, B by multiplying E have made H, K,  therefore, as A is to B, so is H to K. [VII. 18]  But, as A is to B, so is F to G, and G to H.  Therefore also, as F is to G, so is G to H, and H to K;  therefore C, D, E, and F, G, H, K are proportional in the ratio of A to B.  I say next that they are the least numbers that are so. 
                                           
                                           
For, since A, B are the least of those which have the same ratio with them,  and the least of those which have the same ratio are prime to one another, [VII. 22]  therefore A, B are prime to one another.  And the numbers A, B by multiplying themselves respectively have made the numbers C, E,  and by multiplying the numbers C, E respectively have made the numbers F, K;  therefore C, E and F, K are prime to one another respectively. [VII. 27]  But, if there be as many numbers as we please in continued proportion,  and the extremes of them be prime to one another,  they are the least of those which have the same ratio with them. [VIII. 1]  Therefore C, D, E and F, G, H, K are the least of those which have the same ratio with A, B  Q. E. D. 
                     
                     
PORISM.
From this it is manifest that, if three numbers in continued proportion be the least of those which have the same ratio with them, the extremes of them are squares, and, if four numbers, cubes. 
 
 
PROPOSITION 3. 
 
 
If as many numbers as we please in continued proportion be the least of those which have the same ratio with them, the extremes of them are prime to one another. 
 
 
Let as many numbers as we please, A, B, C, D, in continued proportion be the least of those which have the same ratio with them;  I say that the extremes of them A, D are prime to one another. 
   
   
For let two numbers E, F, the least that are in the ratio of A, B, C, D, be taken, [VII. 33] then three others G, H, K with the same property;  and others, more by one continually, [VIII. 2]  until the multitude taken becomes equal to the multitude of the numbers A, B, C, D.  Let them be taken, and let them be L, M, N, O.  Now, since E, F are the least of those which have the same ratio with them, they are prime to one another. [VII. 22]  And, since the numbers E, F by multiplying themselves respectively have made the numbers G, K,  and by multiplying the numbers G, K respectively have made the numbers L, O, [VIII. 2, Por.]  therefore both G, K and L, O are prime to one another. [VII. 27]  And, since A, B, C, D are the least of those which have the same ratio with them,  while L, M, N, O are the least that are in the same ratio with A, B, C, D,  and the multitude of the numbers A, B, C, D is equal to the multitude of the numbers L, M, N, O,  therefore the numbers A, B, C, D are equal to the numbers L, M, N, O respectively;  therefore A is equal to L, and D to O.  And L, O are prime to one another.  Therefore A, D are also prime to one another.  Q. E. D. 
                               
                               
PROPOSITION 4. 
 
 
Given as many ratios as we please in least numbers, to find numbers in continued proportion which are the least in the given ratios. 
 
 
Let the given ratios in least numbers be that of A to B, that of C to D, and that of E to F;  thus it is required to find numbers in continued proportion which are the least that are in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F. 
   
   
Let G, the least number measured by B, C, be taken. [VII. 34]  And, as many times as B measures G, so many times also let A measure H,  and, as many times as C measures G, so many times also let D measure K.  Now E either measures or does not measure K.  First, let it measure it.  And, as many times as E measures K, so many times let F measure L also.  Now, since A measures H the same number of times that B measures G,  therefore, as A is to B, so is H to G. [VII. Def. 20, VII. 13]  For the same reason also, as C is to D, so is G to K,  and further, as E is to F, so is K to L;  therefore H, G, K, L are continuously proportional in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F.  I say next that they are also the least that have this property. 
                       
                       
For, if H, G, K, L are not the least numbers continuously proportional in the ratios of A to B, of C to D, and of E to F, let them be N, O, M, P.  Then since, as A is to B, so is N to O,  while A, B are least, and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less,  that is, the antecedent the antecedent and the consequent the consequent;  therefore B measures O. [VII. 20]  For the same reason C also measures O;  therefore B, C measure O;  therefore the least number measured by B, C will also measure O. [VII. 35]  But G is the least number measured by B, C;  therefore G measures O, the greater the less: which is impossible.  Therefore there will be no numbers less than H, G, K, L which are continuously in the ratio of A to B, of C to D, and of E to F. 
                     
                     
Next, let E not measure K.  Let M, the least number measured by E, K, be taken.  And, as many times as K measures M, so many times let H, G measure N, O respectively,  and, as many times as E measures M, so many times let F measure P also.  Since H measures N the same number of times that G measures O,  therefore, as H is to G, so is N to O. [VII. 13 and Def. 20]  But, as H is to G, so is A to B;  therefore also, as A is to B, so is N to O.  For the same reason also, as C is to D, so is O to M.  Again, since E measures M the same number of times that F measures P,  therefore, as E is to F, so is M to P; [VII. 13 and Def. 20]  therefore N, O, M, P are continuously proportional in the ratios of A to B, of C to D, and of E to F.  I say next that they are also the least that are in the ratios A : B, C : D, E : F. 
                         
                         
For, if not, there will be some numbers less than N, O, M, P continuously proportional in the ratios A : B, C : D, E : F.  Let them be Q, R, S, T.  Now since, as Q is to R, so is A to B,  while A, B are least,  and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent, [VII. 20]  therefore B measures R.  For the same reason C also measures R;  therefore B, C measure R.  Therefore the least number measured by B, C will also measure R. [VII. 35]  But G is the least number measured by B, C;  therefore G measures R.  And, as G is to R, so is K to S: [VII. 13]  therefore K also measures S.  But E also measures S;  therefore E, K measure S.  Therefore the least number measured by E, K will also measure S. [VII. 35]  But M is the least number measured by E, K;  therefore M measures S, the greater the less: which is impossible.  Therefore there will not be any numbers less than N, O, M, P continuously proportional in the ratios of A to B, of C to D, and of E to F;  therefore N, O, M, P are the least numbers continuously proportional in the ratios A : B, C : D, E : F.  Q. E. D. 
                                         
                                         
PROPOSITION 5. 
 
 
Plane numbers have to one another the ratio compounded of the ratios of their sides. 
 
 
Let A, B be plane numbers, and let the numbers C, D be the sides of A, and E, F of B;  I say that A has to B the ratio compounded of the ratios of the sides. 
   
   
For, the ratios being given which C has to E and D to F, let the least numbers G, H, K that are continuously in the ratios C : E, D : F be taken,  so that, as C is to E, so is G to H,  and, as D is to F, so is H to K. [VIII. 4]  And let D by multiplying E make L. 
       
       
Now, since D by multiplying C has made A, and by multiplying E has made L,  therefore, as C is to E, so is A to L. [VII. 17]  But, as C is to E, so is G to H;  therefore also, as G is to H, so is A to L.  Again, since E by multiplying D has made L,  and further by multiplying F has made B,  therefore, as D is to F, so is L to B. [VII. 17]  But, as D is to F, so is H to K;  therefore also, as H is to K, so is L to B.  But it was also proved that, as G is to H, so is A to L;  therefore, ex aequali, as G is to K, so is A to B. [VII. 14]  But G has to K the ratio compounded of the ratios of the sides;  therefore A also has to B the ratio compounded of the ratios of the sides.  Q. E. D. 
                           
                           
PROPOSITION 6. 
 
 
If there be as many numbers as we please in continued proportion, and the first do not measure the second, neither will any other measure any other. 
 
 
Let there be as many numbers as we please, A, B, C, D, E, in continued proportion, and let A not measure B;  I say that neither will any other measure any other. 
   
   
Now it is manifest that A, B, C, D, E do not measure one another in order;  for A does not even measure B.  I say, then, that neither will any other measure any other. 
     
     
For, if possible, let A measure C.  And, however many A, B, C are, let as many numbers F, G, H, the least of those which have the same ratio with A, B, C, be taken. [VII. 33]  Now, since F, G, H are in the same ratio with A, B, C, and the multitude of the numbers A, B, C is equal to the multitude of the numbers F, G, H,  therefore, ex aequali, as A is to C, so is F to H. [VII. 14]  And since, as A is to B, so is F to G,  while A does not measure B,  therefore neither does F measure G; [VII. Def. 20]  therefore F is not an unit,  for the unit measures any number.  Now F, H are prime to one another. [VIII. 3]  And, as F is to H, so is A to C;  therefore neither does A measure C.  Similarly we can prove that neither will any other measure any other.  Q. E. D. 
                           
                           
PROPOSITION 7. 
 
 
If there be as many numbers as we please in continued proportion, and the first measure the last, it will measure the second also. 
 
 
Let there be as many numbers as we please, A, B, C, D, in continued proportion; and let A measure D;  I say that A also measures B. 
   
   
For, if A does not measure B, neither will any other of the numbers measure any other. [VIII. 6]  But A measures D.  Therefore A also measures B  Q. E. D. 
       
       
PROPOSITION 8. 
 
 
If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many will also fall in continued proportion between the numbers which have the same ratio with the original numbers. 
 
 
Let the numbers C, D fall between the two numbers A, B in continued proportion with them, and let E be made in the same ratio to F as A is to B;  I say that, as many numbers as have fallen between A, B in continued proportion, so many will also fall between E, F in continued proportion. 
   
   
For, as many as A, B, C, D are in multitude, let so many numbers G, H, K, L, the least of those which have the same ratio with A, C, D, B, be taken; [VII. 33]  therefore the extremes of them G, L are prime to one another. [VIII. 3]  Now, since A, C, D, B are in the same ratio with G, H, K, L,  and the multitude of the numbers A, C, D, B is equal to the multitude of the numbers G, H, K, L,  therefore, ex aequali, as A is to B, so is G to L. [VII. 14]  But, as A is to B, so is E to F;  therefore also, as G is to L, so is E to F.  But G, L are prime, primes are also least, [VII. 21]  and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less,  that is, the antecedent the antecedent and the consequent the consequent. [VII. 20]  Therefore G measures E the same number of times as L measures F.  Next, as many times as G measures E, so many times let H, K also measure M, N respectively;  therefore G, H, K, L measure E, M, N, F the same number of times.  Therefore G, H, K, L are in the same ratio with E, M, N, F. [VII. Def. 20]  But G, H, K, L are in the same ratio with A, C, D, B;  therefore A, C, D, B are also in the same ratio with E, M, N, F.  But A, C, D, B are in continued proportion;  therefore E, M, N, F are also in continued proportion.  Therefore, as many numbers as have fallen between A, B in continued proportion with them,  so many numbers have also fallen between E, F in continued proportion.  Q. E. D. 
                                         
                                         
PROPOSITION 9. 
 
 
If two numbers be prime to one another, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many will also fall between each of them and an unit in continued proportion. 
 
 
Let A, B be two numbers prime to one another, and let C, D fall between them in continued proportion, and let the unit E be set out;  I say that, as many numbers as fall between A, B in continued proportion, so many will also fall between either of the numbers A, B and the unit in continued proportion. 
   
   
For let two numbers F, G, the least that are in the ratio of A, C, D, B, be taken, three numbers H, K, L with the same property,  and others more by one continually, until their multitude is equal to the multitude of A, C, D, B. [VIII. 2]  Let them be taken, and let them be M, N, O, P.  It is now manifest that F by multiplying itself has made H and by multiplying H has made M,  while G by multiplying itself has made L and by multiplying L has made P. [VIII. 2, Por.]  And, since M, N, O, P are the least of those which have the same ratio with F, G,  and A, C, D, B are also the least of those which have the same ratio with F, G, [VIII. 1]  while the multitude of the numbers M, N, O, P is equal to the multitude of the numbers A, C, D, B,  therefore M, N, O, P are equal to A, C, D, B respectively;  therefore M is equal to A, and P to B.  Now, since F by multiplying itself has made H,  therefore F measures H according to the units in F.  But the unit E also measures F according to the units in it;  therefore the unit E measures the number F the same number of times as F measures H.  Therefore, as the unit E is to the number F, so is F to H. [VII. Def. 20]  Again, since F by multiplying H has made M,  therefore H measures M according to the units in F.  But the unit E also measures the number F according to the units in it;  therefore the unit E measures the number F the same number of times as H measures M.  Therefore, as the unit E is to the number F, so is H to M.  But it was also proved that, as the unit E is to the number F, so is F to H;  therefore also, as the unit E is to the number F, so is F to H, and H to M.  But M is equal to A;  therefore, as the unit E is to the number F, so is F to H, and H to A.  For the same reason also, as the unit E is to the number G, so is G to L and L to B.  Therefore, as many numbers as have fallen between A, B in continued proportion,  so many numbers also have fallen between each of the numbers A, B and the unit E in continued proportion.  Q. E. D. 
                                                       
                                                       
PROPOSITION 10. 
 
 
If numbers fall between each of two numbers and an unit in continued proportion, however many numbers fall between each of them and an unit in continued proportion, so many also will fall between the numbers themselves in continued proportion. 
 
 
For let the numbers D, E and F, G respectively fall between the two numbers A, B and the unit C in continued proportion;  I say that, as many numbers as have fallen between each of the numbers A, B and the unit C in continued proportion, so many numbers will also fall between A, B in continued proportion. 
   
   
For let D by multiplying F make H, and let the numbers D, F by multiplying H make K, L respectively. 
 
 
Now, since, as the unit C is to the number D, so is D to E,  therefore the unit C measures the number D the same number of times as D measures E. [VII. Def. 20]  But the unit C measures the number D according to the units in D;  therefore the number D also measures E according to the units in D;  therefore D by multiplying itself has made E.  Again, since, as C is to the number D, so is E to A,  therefore the unit C measures the number D the same number of times as E measures A.  But the unit C measures the number D according to the units in D;  therefore E also measures A according to the units in D;  therefore D by multiplying E has made A.  For the same reason also F by multiplying itself has made G, and by multiplying G has made B.  And, since D by multiplying itself has made E and by multiplying F has made H,  therefore, as D is to F, so is E to H. [VII. 17]  For the same reason also, as D is to F, so is H to G. [VII. 18]  Therefore also, as E is to H, so is H to G.  Again, since D by multiplying the numbers E, H has made A, K respectively,  therefore, as E is to H, so is A to K. [VII. 17]  But, as E is to H, so is D to F;  therefore also, as D is to F, so is A to K.  Again, since the numbers D, F by multiplying H have made K, L respectively,  therefore, as D is to F, so is K to L. [VII. 18]  But, as D is to F, so is A to K;  therefore also, as A is to K, so is K to L.  Further, since F by multiplying the numbers H, G has made L, B respectively,  therefore, as H is to G, so is L to B. [VII. 17]  But, as H is to G, so is D to F;  therefore also, as D is to F, so is L to B.  But it was also proved that, as D is to F, so is A to K and K to L;  therefore also, as A is to K, so is K to L and L to B.  Therefore A, K, L, B are in continued proportion.  Therefore, as many numbers as fall between each of the numbers A, B and the unit C in continued proportion,  so many also will fall between A, B in continued proportion.  Q. E. D. 
                                                                 
                                                                 
PROPOSITION 11. 
 
 
Between two square numbers there is one mean proportional number, and the square has to the square the ratio duplicate of that which the side has to the side. 
 
 
Let A, B be square numbers, and let C be the side of A, and D of B;  I say that between A, B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to D. 
   
   
For let C by multiplying D make E.  Now, since A is a square and C is its side,  therefore C by multiplying itself has made A.  For the same reason also D by multiplying itself has made B.  Since then C by multiplying the numbers C, D has made A, E respectively,  therefore, as C is to D, so is A to E. [VII. 17]  For the same reason also, as C is to D, so is E to B. [VII. 18]  Therefore also, as A is to E, so is E to B.  Therefore between A, B there is one mean proportional number. 
                 
                 
I say next that A also has to B the ratio duplicate of that which C has to D.  For, since A, E, B are three numbers in proportion,  therefore A has to B the ratio duplicate of that which A has to E. [V. Def. 9]  But, as A is to E, so is C to D.  Therefore A has to B the ratio duplicate of that which the side C has to D.  Q. E. D. 
           
           
PROPOSITION 12. 
 
 
Between two cube numbers there are two mean proportional numbers, and the cube has to the cube the ratio triplicate of that which the side has to the side. 
 
 
Let A, B be cube numbers, and let C be the side of A, and D of B;  I say that between A, B there are two mean proportional numbers, and A has to B the ratio triplicate of that which C has to D. 
   
   
For let C by multiplying itself make E, and by multiplying D let it make F;  let D by multiplying itself make G, and let the numbers C, D by multiplying F make H, K respectively. 
   
   
Now, since A is a cube, and C its side,  and C by multiplying itself has made E,  therefore C by multiplying itself has made E and by multiplying E has made A.  For the same reason also D by multiplying itself has made G and by multiplying G has made B.  And, since C by multiplying the numbers C, D has made E, F respectively,  therefore, as C is to D, so is E to F. [VII. 17]  For the same reason also, as C is to D, so is F to G. [VII. 18]  Again, since C by multiplying the numbers E, F has made A, H respectively,  therefore, as E is to F, so is A to H. [VII. 17]  But, as E is to F, so is C to D.  Therefore also, as C is to D, so is A to H.  Again, since the numbers C, D by multiplying F have made H, K respectively,  therefore, as C is to D, so is H to K. [VII. 18]  Again, since D by multiplying each of the numbers F, G has made K, B respectively,  therefore, as F is to G, so is K to B. [VII. 17]  But, as F is to G, so is C to D;  therefore also, as C is to D, so is A to H, H to K, and K to B.  Therefore H, K are two mean proportionals between A, B. 
                                   
                                   
I say next that A also has to B the ratio triplicate of that which C has to D.  For, since A, H, K, B are four numbers in proportion,  therefore A has to B the ratio triplicate of that which A has to H. [V. Def. 10]  But, as A is to H, so is C to D;  therefore A also has to B the ratio triplicate of that which C has to D.  Q. E. D. 
           
           
PROPOSITION 13. 
 
 
If there be as many numbers as we please in continued proportion, and each by multiplying itself make some number, the products will be proportional; and, if the original numbers by multiplying the products make certain numbers, the latter will also be proportional. 
 
 
Let there be as many numbers as we please, A, B, C, in continued proportion, so that, as A is to B, so is B to C; let A, B, C by multiplying themselves make D, E, F, and by multiplying D, E, F let them make G, H, K;  I say that D, E, F and G, H, K are in continued proportion. 
   
   
For let A by multiplying B make L, and let the numbers A, B by multiplying L make M. N respectively.  And again let B by multiplying C make O, and let the numbers B, C by multiplying O make P, Q respectively. 
   
   
Then, in manner similar to the foregoing, we can prove that D, L, E and G, M, N, H are continuously proportional in the ratio of A to B,  and further E, O, F and H, P, Q, K are continuously proportional in the ratio of B to C.  Now, as A is to B, so is B to C;  therefore D, L, E are also in the same ratio with E, O, F, and further G, M, N, H in the same ratio with H, P, Q, K.  And the multitude of D, L, E is equal to the multitude of E, O, F, and that of G, M, N, H to that of H, P, Q, K;  therefore, ex acquali, as D is to E, so is E to F,  and, as G is to H, so is H to K. [VII. 14]  Q. E. D. 
               
               
PROPOSITION 14. 
 
 
If a square measure a square, the side will also measure the side; and, if the side measure the side, the square will also measure the square. 
 
 
Let A, B be square numbers, let C, D be their sides, and let A measure B;  I say that C also measures D. 
   
   
For let C by multiplying D make E; therefore A, E, B are continuously proportional in the ratio of C to D. [VIII. 11]  And, since A, E, B are continuously proportional, and A measures B, therefore A also measures E. [VIII. 7]  And, as A is to E, so is C to D; therefore also C measures D. [VII. Def. 20] 
     
     
Again, let C measure D;  I say that A also measures B. 
   
   
For, with the same construction, we can in a similar manner prove that A, E, B are continuously proportional in the ratio of C to D.  And since, as C is to D, so is A to E, and C measures D, therefore A also measures E. [VII. Def. 20]  And A, E, B are continuously proportional;  therefore A also measures B. 
       
       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 15. 
 
 
If a cube number measure a cube number, the side will also measure the side; and, if the side measure the side, the cube will also measure the cube. 
 
 
For let the cube number A measure the cube B, and let C be the side of A and D of B;  I say that C measures D. 
   
   
For let C by multiplying itself make E, and let D by multiplying itself make G;  further, let C by multiplying D make F, and let C, D by multiplying F make H, K respectively.  Now it is manifest that E, F, G and A, H, K, B are continuously proportional in the ratio of C to D. [VIII. 11, 12]  And, since A, H, K, B are continuously proportional, and A measures B, therefore it also measures H. [VIII. 7]  And, as A is to H, so is C to D; therefore C also measures D. [VII. Def. 20]  Next, let C measure D;  I say that A will also measure B. 
             
             
For, with the same construction, we can prove in a similar manner that A, H, K, B are continuously proportional in the ratio of C to D.  And, since C measures D, and, as C is to D, so is A to H,  therefore A also measures H, [VII. Def. 20] so that A measures B also.  Q. E. D. 
       
       
PROPOSITION 16. 
 
 
If a square number do not measure a square number, neither will the side measure the side; and, if the side do not measure the side, neither will the square measure the square. 
 
 
Let A, B be square numbers, and let C, D be their sides; and let A not measure B;  I say that neither does C measure D. 
   
   
For, if C measures D, A will also measure B. [VIII. 14]  But A does not measure B; therefore neither will C measure D. 
   
   
Again, let C not measure D;  I say that neither will A measure B. 
   
   
For, if A measures B, C will also measure D. [VIII. 14]  But C does not measure D;  therefore neither will A measure B.  Q. E. D. 
       
       
PROPOSITION 17. 
 
 
If a cube number do not measure a cube number, neither will the side measure the side; and, if the side do not measure the side, neither will the cube measure the cube. 
 
 
For let the cube number A not measure the cube number B, and let C be the side of A, and D of B;  I say that C will not measure D. 
   
   
For if C measures D, A will also measure B. [VIII. 15]  But A does not measure B; therefore neither does C measure D.  Again, let C not measure D;  I say that neither will A measure B. 
       
       
For, if A measures B, C will also measure D. [VIII. 15]  But C does not measure D; therefore neither will A measure B.  Q. E. D. 
     
     
PROPOSITION 18. 
 
 
Between two similar plane numbers there is one mean proportional number; and the plane number has to the plane number the ratio duplicate of that which the corresponding side has to the corresponding side. 
 
 
Let A, B be two similar plane numbers, and let the numbers C, D be the sides of A, and E, F of B.  Now, since similar plane numbers are those which have their sides proportional, [VII. Def. 21] therefore, as C is to D, so is E to F.  I ssay then that between A, B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to E, or D to F, that is, of that which the corresponding side has to the corresponding side. 
     
     
Now since, as C is to D, so is E to F, therefore, alternately, as C is to E, so is D to F. [VII. 13]  And, since A is plane, and C, D are its sides, therefore D by multiplying C has made A.  For the same reason also E by multiplying F has made B.  Now let D by multiplying E make G.  Then, since D by multiplying C has made A, and by multiplying E has made G,  therefore, as C is to E, so is A to G. [VII. 17]  But, as C is to E, so is D to F;  therefore also, as D is to F, so is A to G.  Again, since E by multiplying D has made G, and by multiplying F has made B,  therefore, as D is to F, so is G to B. [VII. 17]  But it was also proved that, as D is to F, so is A to G;  therefore also, as A is to G, so is G to B.  Therefore A, G, B are in continued proportion.  Therefore between A, B there is one mean proportional number. 
                           
                           
I say next that A also has to B the ratio duplicate of that which the corresponding side has to the corresponding side,  that is, of that which C has to E or D to F.  For, since A, G, B are in continued proportion, A has to B the ratio duplicate of that which it has to G. [V. Def. 9]  And, as A is to G, so is C to E, and so is D to F.  Therefore A also has to B the ratio duplicate of that which C has to E or D to F.  Q. E. D. 
           
           
PROPOSITION 19. 
 
 
Between two similar solid numbers there fall two mean proportional numbers; and the solid number has to the similar solid number the ratio triplicate of that which the corresponding side has to the corresponding side. 
 
 
Let A, B be two similar solid numbers, and let C, D, E be the sides of A, and F, G, H of B.  Now, since similar solid numbers are those which have their sides proportional, [VII. Def. 21]  therefore, as C is to D, so is F to G,  and, as D is to E, so is G to H.  I say that between A, B there fall two mean proportional numbers, and A has to B the ratio triplicate of that which C has to F, D to G, and also E to H. 
         
         
For let C by multiplying D make K, and let F by multiplying G make L.  Now, since C, D are in the same ratio with F, G, and K is the product of C, D,  and L the product of F, G, K, L are similar plane numbers; [VII. Def. 21]  therefore between K, L there is one mean proportional number. [VIII. 18]  Let it be M.  Therefore M is the product of D, F, as was proved in the theorem preceding this. [VIII. 18]  Now, since D by multiplying C has made K, and by multiplying F has made M,  therefore, as C is to F, so is K to M. [VII. 17]  But, as K is to M, so is M to L.  Therefore K, M, L are continuously proportional in the ratio of C to F.  And since, as C is to D, so is F to G, alternately therefore, as C is to F, so is D to G. [VII. 13]  For the same reason also, as D is to G, so is E to H.  Therefore K, M, L are continuously proportional in the ratio of C to F, in the ratio of D to G, and also in the ratio of E to H.  Next, let E, H by multiplying M make N, O respectively.  Now, since A is a solid number, and C, D, E are its sides, therefore E by multiplying the product of C, D has made A.  But the product of C, D is K; therefore E by multiplying K has made A.  For the same reason also H by multiplying L has made B.  Now, since E by multiplying K has made A, and further also by multiplying M has made N,  therefore, as K is to M, so is A to N. [VII. 17]  But, as K is to M, so is C to F, D to G, and also E to H;  therefore also, as C is to F, D to G, and E to H, so is A to N.  Again, since E, H by multiplying M have made N, O respectively,  therefore, as E is to H, so is N to O. [VII. 18]  But, as E is to H, so is C to F and D to G;  therefore also, as C is to F, D to G, and E to H, so is A to N and N to O.  Again, since H by multiplying M has made O, and further also by multiplying L has made B,  therefore, as M is to L, so is O to B. [VII. 17]  But, as M is to L, so is C to F, D to G, and E to H.  Therefore also, as C is to F, D to G, and E to H, so not only is O to B, but also A to N and N to O.  Therefore A, N, O, B are continuously proportional in the aforesaid ratios of the sides. 
                                                           
                                                           
I say that A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side,  that is, of the ratio which the number C has to F, or D to G, and also E to H.  For, since A, N, O, B are four numbers in continued proportion, therefore A has to B the ratio triplicate of that which A has to N. [V. Def. 10]  But, as A is to N, so it was proved that C is to F, D to G, and also E to H.  Therefore A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side,  that is, of the ratio which the number C has to F, D to G, and also E to H.  Q. E. D. 
             
             
PROPOSITION 20. 
 
 
If one mean proportional number fall between two numbers, the numbers will be similar plane numbers. 
 
 
For let one mean proportional number C fall between the two numbers A, B;  I say that A, B are similar plane numbers. 
   
   
Let D, E, the least numbers of those which have the same ratio with A, C, be taken; [VII. 33]  therefore D measures A the same number of times that E measures C. [VII. 20]  Now, as many times as D measures A, so many units let there be in F;  therefore F by multiplying D has made A,  so that A is plane, and D, F are its sides.  Again, since D, E are the least of the numbers which have the same ratio with C, B,  therefore D measures C the same number of times that E measures B. [VII. 20]  As many times, then, as E measures B, so many units let there be in G;  therefore E measures B according to the units in G;  therefore G by multiplying E has made B.  Therefore B is plane, and E, G are its sides.  Therefore A, B are plane numbers.  I say next that they are also similar.  For, <*> since F by multiplying D has made A, and by multiplying E has made C,  therefore, as D is to E, so is A to C, that is, C to B. [VII. 17]  Again, <*> since E by multiplying F, G has made C, B respectively,  therefore, as F is to G, so is C to B. [VII. 17]  But, as C is to B, so is D to E;  therefore also, as D is to E, so is F to G.  And alternately, as D is to F, so is E to G. [VII. 13]  Therefore A, B are similar plane numbers; for their sides are proportional.  Q. E. D. 
                                           
                                           
PROPOSITION 21. 
 
 
If two mean proportional numbers fall between two numbers, the numbers are similar solid numbers. 
 
 
For let two mean proportional numbers C, D fall between the two numbers A, B;  I say that A, B are similar solid numbers. 
   
   
For let three numbers E, F, G, the least of those which have the same ratio with A, C, D, be taken; [VII. 33 or VIII. 2]  therefore the extremes of them E, G are prime to one another. [VIII. 3]  Now, since one mean proportional number F has fallen between E, G, therefore E, G are similar plane numbers. [VIII. 20]  Let, then, H, K be the sides of E, and L, M of G.  Therefore it is manifest from the theorem before this that E, F, G are continuously proportional in the ratio of H to L and that of K to M.  Now, since E, F, G are the least of the numbers which have the same ratio with A, C, D,  and the multitude of the numbers E, F, G is equal to the multitude of the numbers A, C, D,  therefore, ex aequali, as E is to G, so is A to D. [VII. 14]  But E, G are prime, primes are also least, [VII. 21]  and the least measure those which have the same ratio with them the same number of times, the greater the greater and the less the less,  that is, the antecedent the antecedent and the consequent the consequent; [VII. 20]  therefore E measures A the same number of times that G measures D.  Now, as many times as E measures A, so many units let there be in N.  Therefore N by multiplying E has made A.  But E is the product of H, K;  therefore N by multiplying the product of H, K has made A.  Therefore A is solid, and H, K, N are its sides.  Again, since E, F, G are the least of the numbers which have the same ratio as C, D, B,  therefore E measures C the same number of times that G measures B.  Now, as many times as E measures C, so many units let there be in O.  Therefore G measures B according to the units in O;  therefore O by multiplying G has made B.  But G is the product of L, M;  therefore O by multiplying the product of L, M has made B.  Therefore B is solid, and L, M, O are its sides;  therefore A, B are solid. 
                                                   
                                                   
I say that they are also similar.  For since N, O by multiplying E have made A, C,  therefore, as N is to O, so is A to C, that is, E to F. [VII. 18]  But, as E is to F, so is H to L and K to M;  therefore also, as H is to L, so is K to M and N to O.  And H, K, N are the sides of A, and O, L, M the sides of B.  Therefore A, B are similar solid numbers.  Q. E. D. 
               
               
PROPOSITION 22. 
 
 
If three numbers be in continued proportion, and the first be square, the third will also be square. 
 
 
Let A, B, C be three numbers in continued proportion, and let A the first be square;  I say that C the third is also square. 
   
   
For, since between A, C there is one mean proportional number, B,  therefore A, C are similar plane numbers. [VIII. 20]  But A is square;  therefore C is also square.  Q. E. D. 
         
         
PROPOSITION 23. 
 
 
If four numbers be in continued proportion, and the first be cube, the fourth will also be cube. 
 
 
Let A, B, C, D be four numbers in continued proportion, and let A be cube;  I say that D is also cube. 
   
   
For, since between A, D there are two mean proportional numbers B, C, therefore A, D are similar solid numbers. [VIII. 21]  But A is cube;  therefore D is also cube.  Q. E. D. 
       
       
PROPOSITION 24. 
 
 
If two numbers have to one another the ratio which a square number has to a square number, and the first be square, the second will also be square. 
 
 
For let the two numbers A, B have to one another the ratio which the square number C has to the square number D, and let A be square;  I say that B is also square. 
   
   
For, since C, D are square, C, D are similar plane numbers.  Therefore one mean proportional number falls between C, D. [VIII. 18]  And, as C is to D, so is A to B;  therefore one mean proportional number falls between A, B also. [VIII. 8]  And A is square;  therefore B is also square. [VIII. 22]  Q. E. D. 
             
             
PROPOSITION 25. 
 
 
If two numbers have to one another the ratio which a cube number has to a cube number, and the first be cube, the second will also be cube. 
 
 
For let the two numbers A, B have to one another the ratio which the cube number C has to the cube number D, and let A be cube;  I say that B is also cube. 
   
   
For, since C, D are cube, C, D are similar solid numbers.  Therefore two mean proportional numbers fall between C, D. [VIII. 19]  And, as many numbers as fall between C, D in continued proportion, so many will also fall between those which have the same ratio with them; [VIII. 8]  so that two mean proportional numbers fall between A, B also.  Let E, F so fall.  Since, then, the four numbers A, E, F, B are in continued proportion, and A is cube, therefore B is also cube. [VIII. 23]  Q. E. D. 
             
             
PROPOSITION 26. 
 
 
Similar plane numbers have to one another the ratio which a square number has to a square number. 
 
 
Let A, B be similar plane numbers;  I say that A has to B the ratio which a square number has to a square number. 
   
   
For, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18]  Let it so fall, and let it be C; and let D, E, F, the least numbers of those which have the same ratio with A, C, B, be taken; [VII. 33 or VIII. 2]  therefore the extremes of them D, F are square. [VIII. 2, Por.]  And since, as D is to F, so is A to B, and D, F are square,  therefore A has to B the ratio which a square number has to a square number.  Q. E. D. 
           
           
PROPOSITION 27. 
 
 
Similar solid numbers have to one another the ratio which a cube number has to a cube number. 
 
 
Let A, B be similar solid numbers;  I say that A has to B the ratio which a cube number has to a cube number.  For, since A, B are similar solid numbers, therefore two mean proportional numbers fall between A, B. [VIII. 19]  Let C, D so fall, and let E, F, G, H, the least numbers of those which have the same ratio with A, C, D, B, and equal with them in multitude, be taken; [VII. 33 or VIII. 2]  therefore the extremes of them E, H are cube. [VIII. 2, Por.]  And, as E is to H, so is A to B;  therefore A also has to B the ratio which a cube number has to a cube number.  Q. E. D. 
               
               
BOOK IX. 
 
 
PROPOSITION I. 
 
 
If two similar plane numbers by multiplying one another make some number, the product will be square. 
 
 
Let A, B be two similar plane numbers, and let A by multiplying B make C;  I say that C is square. 
   
   
For let A by multiplying itself make D.  Therefore D is square.  Since then A by multiplying itself has made D, and by multiplying B has made C,  therefore, as A is to B, so is D to C. [VII. 17]  And, since A, B are similar plane numbers,  therefore one mean proportional number falls between A, B. [VIII. 18]  But, if numbers fall between two numbers in continued proportion,  as many as fall between them, so many also fall between those which have the same ratio; [VIII. 8]  so that one mean proportional number falls between D, C also.  And D is square;  therefore C is also square. [VIII. 22]  Q. E. D. 
                       
                       
PROPOSITION 2. 
 
 
If two numbers by multiplying one another make a square number, they are similar plane numbers. 
 
 
Let A, B be two numbers, and let A by multiplying B make the square number C;  I say that A, B are similar plane numbers. 
   
   
For let A by multiplying itself make D;  therefore D is square.  Now, since A by multiplying itself has made D, and by multiplying B has made C,  therefore, as A is to B, so is D to C. [VII. 17]  And, since D is square, and C is so also,  therefore D, C are similar plane numbers.  Therefore one mean proportional number falls between D, C. [VIII. 18]  And, as D is to C, so is A to B;  therefore one mean proportional number falls between A, B also. [VIII. 8]  But, if one mean proportional number fall between two numbers, they are similar plane numbers; [VIII. 20]  therefore A, B are similar plane numbers.  Q. E. D. 
                       
                       
PROPOSITION 3. 
 
 
If a cube number by multiplying itself make some number, the product will be cube. 
 
 
For let the cube number A by multiplying itself make B;  I say that B is cube. 
   
   
For let C, the side of A, be taken, and let C by multiplying itself make D.  It is then manifest that C by multiplying D has made A.  Now, since C by multiplying itself has made D,  therefore C measures D according to the units in itself.  But further the unit also measures C according to the units in it;  therefore, as the unit is to C, so is C to D. [VII. Def. 20]  Again, since C by multiplying D has made A,  therefore D measures A according to the units in C.  But the unit also measures C according to the units in it;  therefore, as the unit is to C, so is D to A.  But, as the unit is to C, so is C to D;  therefore also, as the unit is to C, so is C to D, and D to A.  Therefore between the unit and the number A two mean proportional numbers C, D have fallen in continued proportion.  Again, since A by multiplying itself has made B,  therefore A measures B according to the units in itself.  But the unit also measures A according to the units in it;  therefore, as the unit is to A, so is A to B. [VII. Def. 20]  But between the unit and A two mean proportional numbers have fallen;  therefore two mean proportional numbers will also fall between A, B. [VIII. 8]  But, if two mean proportional numbers fall between two numbers, and the first be cube, the second will also be cube. [VIII. 23]  And A is cube;  therefore B is also cube.  Q. E. D. 
                                             
                                             
PROPOSITION 4. 
 
 
If a cube number by multiplying a cube number make some number, the product will be cube. 
 
 
For let the cube number A by multiplying the cube number B make C;  I say that C is cube. 
   
   
For let A by multiplying itself make D;  therefore D is cube. [IX. 3]  And, since A by multiplying itself has made D, and by multiplying B has made C,  therefore, as A is to B, so is D to C. [VII. 17]  And, since A, B are cube numbers, A, B are similar solid numbers.  Therefore two mean proportional numbers fall between A, B; [VIII. 19]  so that two mean proportional numbers will fall between D, C also. [VIII. 8]  And D is cube;  therefore C is also cube [VIII. 23]  Q. E. D. 
                   
                   
PROPOSITION 5. 
 
 
If a cube number by multiplying any number make a cube number, the multiplied number will also be cube. 
 
 
For let the cube number A by multiplying any number B make the cube number C;  I say that B is cube. 
   
   
For let A by multiplying itself make D;  therefore D is cube. [IX. 3]  Now, since A by multiplying itself has made D, and by multiplying B has made C,  therefore, as A is to B, so is D to C. [VII. 17]  And since D, C are cube, they are similar solid numbers.  Therefore two mean proportional numbers fall between D, C. [VIII. 19]  And, as D is to C, so is A to B;  therefore two mean proportional numbers fall between A, B also. [VIII. 8]  And A is cube;  therefore B is also cube. [VIII. 23]   
                     
                     
PROPOSITION 6. 
 
 
If a number by multiplying itself make a cube number, it will itself also be cube. 
 
 
For let the number A by multiplying itself make the cube number B;  I say that A is also cube. 
   
   
For let A by multiplying B make C.  Since, then, A by multiplying itself has made B, and by multiplying B has made C,  therefore C is cube.  And, since A by multiplying itself has made B,  therefore A measures B according to the units in itself.  But the unit also measures A according to the units in it.  Therefore, as the unit is to A, so is A to B. [VII. Def. 20]  And, since A by multiplying B has made C,  therefore B measures C according to the units in A.  But the unit also measures A according to the units in it.  Therefore, as the unit is to A, so is B to C. [VII. Def. 20]  But, as the unit is to A, so is A to B;  therefore also, as A is to B, so is B to C.  And, since B, C are cube, they are similar solid numbers.  Therefore there are two mean proportional numbers between B, C. [VIII. 19]  And, as B is to C, so is A to B.  Therefore there are two mean proportional numbers between A, B also. [VIII. 8]  And B is cube;  therefore A is also cube. [cf. VIII. 23]  Q. E. D. 
                                       
                                       
PROPOSITION 7. 
 
 
If a composite number by multiplying any number make some number, the product will be solid. 
 
 
For let the composite number A by multiplying any number B make C;  I say that C is solid. 
   
   
For, since A is composite, it will be measured by some number. [VII. Def. 13]  Let it be measured by D;  and, as many times as D measures A, so many units let there be in E.  Since then D measures A according to the units in E,  therefore E by multiplying D has made A. [VII. Def. 15]  And, since A by multiplying B has made C, and A is the product of D, E,  therefore the product of D, E by multiplying B has made C.  Therefore C is solid, and D, E, B are its sides.  Q. E. D. 
                 
                 
PROPOSITION 8. 
 
 
If as many numbers as we please beginning from an unit be in continued proportion, the third from the unit will be square, as will also those which successively leave out one; the fourth will be cube, as will also all those which leave out two; and the seventh will be at once cube and square, as will also those which leave out five. 
 
 
Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion;  I say that B, the third from the unit, is square, as are also all those which leave out one; C, the fourth, is cube, as are also all those which leave out two; and F, the seventh, is at once cube and square, as are also all those which leave out five. 
   
   
For since, as the unit is to A, so is A to B,  therefore the unit measures the number A the same number of times that A measures B. [VII. Def. 20]  But the unit measures the number A according to the units in it;  therefore A also measures B according to the units in A.  Therefore A by multiplying itself has made B;  therefore B is square.  And, since B, C, D are in continued proportion, and B is square,  therefore D is also square. [VIII. 22]  For the same reason F is also square.  Similarly we can prove that all those which leave out one are square.  I say next that C, the fourth from the unit, is cube, as are also all those which leave out two.  For since, as the unit is to A, so is B to C,  therefore the unit measures the number A the same number of times that B measures C.  But the unit measures the number A according to the units in A;  therefore B also measures C according to the units in A.  Therefore A by multiplying B has made C.  Since then A by multiplying itself has made B, and by multiplying B has made C,  therefore C is cube.  And, since C, D, E, F are in continued proportion, and C is cube,  therefore F is also cube. [VIII. 23]  But it was also proved square;  therefore the seventh from the unit is both cube and square.  Similarly we can prove that all the numbers which leave out five are also both cube and square.  Q. E. D. 
                                               
                                               
PROPOSITION 9. 
 
 
If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be square, all the rest will also be square. And, if the number after the unit be cube, all the rest will also be cube. 
 
 
Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, be square;  I say that all the rest will also be square. 
   
   
Now it has been proved that B, the third from the unit, is square, as are also all those which leave out one;  [IX. 8] I say that all the rest are also square.  For, since A, B, C are in continued proportion, and A is square,  therefore C is also square. [VIII. 22]  Again, since B, C, D are in continued proportion, and B is square, D is also square. [VIII. 22]  Similarly we can prove that all the rest are also square. 
           
           
Next, let A be cube;  I say that all the rest are also cube. 
   
   
Now it has been proved that C, the fourth from the unit, is cube, as also are all those which leave out two; [IX. 8]  I say that all the rest are also cube.  For, since, as the unit is to A, so is A to B,  therefore the unit measures A the same number of times as A measures B.  But the unit measures A according to the units in it;  therefore A also measures B according to the units in itself;  therefore A by multiplying itself has made B.  And A is cube.  But, if a cube number by multiplying itself make some number, the product is cube. [IX. 3]  Therefore B is also cube.  And, since the four numbers A, B, C, D are in continued proportion, and A is cube,  D also is cube. [VIII. 23]  For the same reason E is also cube, and similarly all the rest are cube.  Q. E. D. 
                           
                           
PROPOSITION 10. 
 
 
If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be not square, neither will any other be square except the third from the unit and all those which leave out one. And, if the number after the unit be not cube, neither will any other be cube except the fourth from the unit and all those which leave out two. 
 
 
Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, not be square;  I say that neither will any other be square except the third from the unit
   
   
For, if possible, let C be square.  But B is also square; [IX. 8]  [therefore B, C have to one another the ratio which a square number has to a square number].  And, as B is to C, so is A to B;  therefore A, B have to one another the ratio which a square number has to a square number;  [so that A, B are similar plane numbers]. [VIII. 26, converse]  And B is square;  therefore A is also square:  which is contrary to the hypothesis.  Therefore C is not square.  Similarly we can prove that neither is any other of the numbers square except the third from the unit and those which leave out one. 
                     
                     
Next, let A not be cube.  I say that neither will any other be cube except the fourth from the unit and those which leave out two. 
   
   
For, if possible, let D be cube.  Now C is also cube;  for it is fourth from the unit. [IX. 8]  And, as C is to D, so is B to C;  therefore B also has to C the ratio which a cube has to a cube.  And C is cube;  therefore B is also cube. [VIII. 25]  And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it,  therefore A also measures B according to the units in itself;  therefore A by multiplying itself has made the cube number B.  But, if a number by multiplying itself make a cube number, it is also itself cube. [IX. 6]  Therefore A is also cube: which is contrary to the hypothesis.  Therefore D is not cube.  Similarly we can prove that neither is any other of the numbers cube except the fourth from the unit and those which leave out two.  Q. E. D. 
                             
                             
PROPOSITION 11. 
 
 
If as many numbers as we please beginning from an unit be in continued proportion, the less measures the greater according to some one of the numbers which have place among the proportional numbers. 
 
 
Let there be as many numbers as we please, B, C, D, E, beginning from the unit A and in continued proportion;  I say that B, the least of the numbers B, C, D, E, measures E according to some one of the numbers C, D. 
   
   
For since, as the unit A is to B, so is D to E,  therefore the unit A measures the number B the same number of times as D measures E;  therefore, alternately, the unit A measures D the same number of times as B measures E. [VII. 15]  But the unit A measures D according to the units in it;  therefore B also measures E according to the units in D;  so that B the less measures E the greater according to some number of those which have place among the proportional numbers. 
           
           
PORISM.
And it is manifest that, whatever place the measuring number has, reckoned from the unit, the same place also has the number according to which it measures, reckoned from the number measured, in the direction of the number before it.
Q. E. D. 
 
 
PROPOSITION 12. 
 
 
If as many numbers as we please beginning from an unit be in continued proportion, by however many prime numbers the last is measured, the next to the unit will also be measured by the same. 
 
 
Let there be as many numbers as we please, A, B, C, D, beginning from an unit, and in continued proportion;  I say that, by however many prime numbers D is measured, A will also be measured by the same. 
   
   
For let D be measured by any prime number E;  I say that E measures A. 
   
   
For suppose it does not;  now E is prime, and any prime number is prime to any which it does not measure; [VII. 29]  therefore E, A are prime to one another.  And, since E measures D, let it measure it according to F,  therefore E by multiplying F has made D.  Again, since A measures D according to the units in C, [IX. 11 and Por.]  therefore A by multiplying C has made D.  But, further, E has also by multiplying F made D;  therefore the product of A, C is equal to the product of E, F.  Therefore, as A is to E, so is F to C. [VII. 19]  But A, E are prime, primes are also least, [VII. 21] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20]  therefore E measures C. 
                       
                       
Let it measure it according to G;  therefore E by multiplying G has made C.  But, further, by the theorem before this, A has also by multiplying B made C. [IX. 11 and Por.]  Therefore the product of A, B is equal to the product of E, G.  Therefore, as A is to E, so is G to B. [VII. 19]  But A, E are prime, primes are also least, [VII. 21]  and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent: [VII. 20]  therefore E measures B. 
               
               
Let it measure it according to H;  therefore E by multiplying H has made B.  But further A has also by multiplying itself made B; [IX. 8]  therefore the product of E, H is equal to the square on A.  Therefore, as E is to A, so is A to H. [VII. 19]  But A, E are prime, primes are also least, [VII. 21]  and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20]  therefore E measures A, as antecedent antecedent.  But, again, it also does not measure it: which is impossible.  Therefore E, A are not prime to one another.  Therefore they are composite to one another.  But numbers composite to one another are measured by some number. [VII. Def. 14]  And, since E is by hypothesis prime, and the prime is not measured by any number other than itself,  therefore E measures A, E, so that E measures A.  [But it also measures D;  therefore E measures A, D.]  Similarly we can prove that, by however many prime numbers D is measured, A will also be measured by the same.  Q. E. D. 
                                   
                                   
PROPOSITION 13. 
 
 
If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be prime, the greatest will not be measured by any except those which have a place among the proportional numbers. 
 
 
Let there be as many numbers as we please, A, B, C, D, beginning from an unit and in continued proportion, and let A, the number after the unit, be prime;  I say that D, the greatest of them, will not be measured by any other number except A, B, C. 
   
   
For, if possible, let it be measured by E, and let E not be the same with any of the numbers A, B, C.  It is then manifest that E is not prime.  For, if E is prime and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.  Therefore E is not prime.  Therefore it is composite.  But any composite number is measured by some prime number; [VII. 31]  therefore E is measured by some prime number.  I say next that it will not be measured by any other prime except A. 
               
               
For, if E is measured by another,  and E measures D, that other will also measure D;  so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.  Therefore A measures E.  And, since E measures D, let it measure it according to F.  I say that F is not the same with any of the numbers A, B, C. 
           
           
For, if F is the same with one of the numbers A, B, C, and measures D according to E,  therefore one of the numbers A, B, C also measures D according to E.  But one of the numbers A, B, C measures D according to some one of the numbers A, B, C; IX. 11]  therefore E is also the same with one of the numbers A, B, C: which is contrary to the hypothesis.  Therefore F is not the same as any one of the numbers A, B, C.  Similarly we can prove that F is measured by A, by proving again that F is not prime.  For, if it is, and measures D, it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible;  therefore F is not prime.  Therefore it is composite.  But any composite number is measured by some prime number; [VII. 31]  therefore F is measured by some prime number.  I say next that it will not be measured by any other prime except A. 
                       
                       
For, if any other prime number measures F, and F measures D, that other will also measure D;  so that it will also measure A [IX. 12], which is prime, though it is not the same with it: which is impossible.  Therefore A measures F.  And, since E measures D according to F,  therefore E by multiplying F has made D.  But, further, A has also by multiplying C made D; [IX. 11]  therefore the product of A, C is equal to the product of E, F.  Therefore, proportionally, as A is to E, so is F to C. [VII. 19]  But A measures E; therefore F also measures C. 
                 
                 
Let it measure it according to G.  Similarly, then, we can prove that G is not the same with any of the numbers A, B, and that it is measured by A.  And, since F measures C according to G  therefore F by multiplying G has made C.  But, further, A has also by multiplying B made C; [IX. 11]  therefore the product of A, B is equal to the product of F, G.  Therefore, proportionally, as A is to F, so is G to B. [VII. 19]  But A measures F;  therefore G also measures B. 
                 
                 
Let it measure it according to H.  Similarly then we can prove that H is not the same with A.  And, since G measures B according to H,  therefore G by multiplying H has made B.  But further A has also by multiplying itself made B; [IX. 8]  therefore the product of H, G is equal to the square on A.  Therefore, as H is to A, so is A to G. [VII. 19]  But A measures G;  therefore H also measures A, which is prime, though it is not the same with it: which is absurd.  Therefore D the greatest will not be measured by any other number except A, B, C.  Q. E. D. 
                     
                     
PROPOSITION 14. 
 
 
If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it. 
 
 
For let the number A be the least that is measured by the prime numbers B, C, D;  I say that A will not be measured by any other prime number except B, C, D. 
   
   
For, if possible, let it be measured by the prime number E, and let E not be the same with any one of the numbers B, C, D.  Now, since E measures A, let it measure it according to F;  therefore E by multiplying F has made A.  And A is measured by the prime numbers B, C, D.  But, if two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers; [VII. 30]  therefore B, C, D will measure one of the numbers E, F.  Now they will not measure E;  for E is prime and not the same with any one of the numbers B, C, D.  Therefore they will measure F, which is less than A: which is impossible,  for A is by hypothesis the least number measured by B, C, D.  Therefore no prime number will measure A except B, C, D.  Q. E. D. 
                       
                       
PROPOSITION 15. 
 
 
If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatever added together will be prime to the remaining number. 
 
 
Let A, B, C, three numbers in continued proportion, be the least of those which have the same ratio with them;  I say that any two of the numbers A, B, C whatever added together are prime to the remaining number, namely A, B to C; B, C to A; and further A, C to B. 
   
   
For let two numbers DE, EF, the least of those which have the same ratio with A, B, C, be taken. [VIII. 2]  It is then manifest that DE by multiplying itself has made A, and by multiplying EF has made B, and, further, EF by multiplying itself has made C. [VIII. 2]  Now, since DE, EF are least, they are prime to one another. [VII. 22]  But, if two numbers be prime to one another, their sum is also prime to each; [VII. 28]  therefore DF is also prime to each of the numbers DE, EF.  But further DE is also prime to EF;  therefore DF, DE are prime to EF.  But, if two numbers be prime to any number, their product is also prime to the other; [VII. 24]  so that the product of FD, DE is prime to EF;  hence the product of FD, DE is also prime to the square on EF. [VII. 25]    But the product of FD, DE is the square on DE together with the product of DE, EF; [II. 3]  therefore the square on DE together with the product of DE, EF is prime to the square on EF.  And the square on DE is A, the product of DE, EF is B, and the square on EF is C;  therefore A, B added together are prime to C.  Similarly we can prove that B, C added together are prime to A.  I say next that A, C added together are also prime to B.  For, since DF is prime to each of the numbers DE, EF, the square on DF is also prime to the product of DE, EF. [VII. 24, 25]  But the squares on DE, EF together with twice the product of DE, EF are equal to the square on DF; [II. 4]  therefore the squares on DE, EF together with twice the product of DE, EF are prime to the product of DE, EF.  Separando, the squares on DE, EF together with once the product of DE, EF are prime to the product of DE, EF.  Therefore, separando again, the squares on DE, EF are prime to the product of DE, EF.  And the square on DE is A, the product of DE, EF is B, and the square on EF is C.  Therefore A, C added together are prime to B.  Q. E. D. 
                                                 
                                                 
PROPOSITION 16. 
 
 
If two numbers be prime to one another, the second will not be to any other number as the first is to the second. 
 
 
For let the two numbers A, B be prime to one another;  I say that B is not to any other number as A is to B. 
   
   
For, if possible, as A is to B, so let B be to C.  Now A, B are prime, primes are also least, [VII. 21]  and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20]  therefore A measures B as antecedent antecedent.  But it also measures itself;  therefore A measures A, B which are prime to one another: which is absurd.  Therefore B will not be to C, as A is to B.  Q. E. D. 
               
               
PROPOSITION 17. 
 
 
If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the last will not be to any other number as the first to the second. 
 
 
For let there be as many numbers as we please, A, B, C, D, in continued proportion, and let the extremes of them, A, D, be prime to one another;  I say that D is not to any other number as A is to B. 
   
   
For, if possible, as A is to B, so let D be to E;  therefore, alternately, as A is to D, so is B to E. [VII. 13]  But A, D are prime, primes are also least, [VII. 21]  and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent. [VII. 20]  Therefore A measures B.  And, as A is to B, so is B to C.  Therefore B also measures C; so that A also measures C.  And since, as B is to C, so is C to D, and B measures C,  therefore C also measures D.  But A measured C; so that A also measures D.  But it also measures itself;  therefore A measures A, D which are prime to one another: which is impossible.  Therefore D will not be to any other number as A is to B.  Q. E. D. 
                           
                           
PROPOSITION 18. 
 
 
Given two numbers, to investigate whether it is possible to find a third proportional to them. 
 
 
Let A, B be the given two numbers, and let it be required to investigate whether it is possible to find a third proportional to them. 
 
 
Now A, B are either prime to one another or not.  And, if they are prime to one another, it has been proved that it is impossible to find a third proportional to them. [IX. 16] 
   
   
Next, let A, B not be prime to one another, and let B by multiplying itself make C.  Then A either measures C or does not measure it.  First, let it measure it according to D;  therefore A by multiplying D has made C.  But, further, B has also by multiplying itself made C;  therefore the product of A, D is equal to the square on B.  Therefore, as A is to B, so is B to D; [VII. 19]  therefore a third proportional number D has been found to A, B. 
               
               
Next, let A not measure C;  I say that it is impossible to find a third proportional number to A, B.  For, if possible, let D, such third proportional, have been found.  Therefore the product of A, D is equal to the square on B.  But the square on B is C;  therefore the product of A, D is equal to C.  Hence A by multiplying D has made C;  therefore A measures C according to D.  But, by hypothesis, it also does not measure it: which is absurd.  Therefore it is not possible to find a third proportional number to A, B when A does not measure C.  Q. E. D. 
                     
                     
PROPOSITION 19. 
 
 
Given three numbers, to investigate when it is possible to find a fourth proportional to them. 
 
 
Let A, B, C be the given three numbers, and let it be required to investigate when it is possible to find a fourth proportional to them. 
 
 
Now either they are not in continued proportion, and the extremes of them are prime to one another;  or they are in continued proportion, and the extremes of them are not prime to one another;  or they are not in continued proportion, nor are the extremes of them prime to one another;  or they are in continued proportion, and the extremes of them are prime to one another. 
       
       
If then A, B, C are in continued proportion, and the extremes of them A, C are prime to one another, it has been proved that it is impossible to find a fourth proportional number to them. [IX. 17]  Next, let A, B, C not be in continued proportion, the extremes being again prime to one another;  I say that in this case also it is impossible to find a fourth proportional to them.  For, if possible, let D have been found, so that, as A is to B, so is C to D, and let it be contrived that, as B is to C, so is D to E.  Now, since, as A is to B, so is C to D, and, as B is to C, so is D to E,  therefore, ex aequali, as A is to C, so is C to E. [VII. 14]  But A, C are prime, primes are also least, [VII. 21]  and the least numbers measure those which have the same ratio, the antecedent the antecedent and the consequent the consequent. [VII. 20]  Therefore A measures C as antecedent antecedent.  But it also measures itself;  therefore A measures A, C which are prime to one another: which is impossible.  Therefore it is not possible to find a fourth proportional to A, B, C.<*> 
                       
                       
Next, let A, B, C be again in continued proportion, but let A, C not be prime to one another.  I say that it is possible to find a fourth proportional to them. 
   
   
For let B by multiplying C make D;  therefore A either measures D or does not measure it.  First, let it measure it according to E;  therefore A by multiplying E has made D.  But, further, B has also by multiplying C made D;  therefore the product of A, E is equal to the product of B, C;  therefore, proportionally, as A is to B, so is C to E; [VII. 19]  therefore E has been found a fourth proportional to A, B, C. 
               
               
Next, let A not measure D;  I say that it is impossible to find a fourth proportional number to A, B, C. 
   
   
For, if possible, let E have been found;  therefore the product of A, E is equal to the product of B, C. [VII. 19]  But the product of B, C is D;  therefore the product of A, E is also equal to D.  Therefore A by multiplying E has made D;  therefore A measures D according to E, so that A measures D.  But it also does not measure it: which is absurd.  Therefore it is not possible to find a fourth proportional number to A, B, C when A does not measure D.  Next, let A, B, C not be in continued proportion, nor the extremes prime to one another.  And let B by multiplying C make D.  Similarly then it can be proved that, if A measures D, it is possible to find a fourth proportional to them, but, if it does not measure it, impossible.  Q. E. D. 
                       
                       
PROPOSITION 20. 
 
 
Prime numbers are more than any assigned multitude of prime numbers. 
 
 
Let A, B, C be the assigned prime numbers;  I say that there are more prime numbers than A, B, C. 
   
   
For let the least number measured by A, B, C be taken, and let it be DE;  Let the unit DF be added to DE.  Then EF is either prime or not.  First, let it be prime;  then the prime numbers A, B, C, EF have been found which are more than A, B, C. 
         
         
Next, let EF not be prime;  therefore it is measured by some prime number. [VII. 31]  Let it be measured by the prime number G.  I say that G is not the same with any of the numbers A, B, C. 
       
       
For, if possible, let it be so.  Now A, B, C measure DE;  therefore G also will measure DE.  But it also measures EF.  Therefore G, being a number, will measure the remainder, the unit DF: which is absurd.  Therefore G is not the same with any one of the numbers A, B, C.  And by hypothesis it is prime.  Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C.  Q. E. D. 
                 
                 
PROPOSITION 21. 
 
 
If as many even numbers as we please be added together, the whole is even. 
 
 
For let as many even numbers as we please, AB, BC, CD, DE, be added together;  I say that the whole AE is even. 
   
   
For, since each of the numbers AB, BC, CD, DE is even, it has a half part; [VII. Def. 6]  so that the whole AE also has a half part.  But an even number is that which is divisible into two equal parts; [id.]  therefore AE is even.  Q. E. D. 
         
         
PROPOSITION 22. 
 
 
If as many odd numbers as we please be added together, and their multitude be even, the whole will be even. 
 
 
For let as many odd numbers as we please, AB, BC, CD, DE, even in multitude, be added together;  I say that the whole AE is even. 
   
   
For, since each of the numbers AB, BC, CD, DE is odd, if an unit be subtracted from each, each of the remainders will be even; [VII. Def. 7]  so that the sum of them will be even. [IX. 21]  But the multitude of the units is also even.  Therefore the whole AE is also even. [IX. 21]  Q. E. D. 
         
         
PROPOSITION 23. 
 
 
If as many odd numbers as we please be added together, and their multitude be odd, the whole will also be odd. 
 
 
For let as many odd numbers as we please, AB, BC, CD, the multitude of which is odd, be added together;  I say that the whole AD is also odd. 
   
   
Let the unit DE be subtracted from CD;  therefore the remainder CE is even. [VII. Def. 7]  But CA is also even; [IX. 22]  therefore the whole AE is also even. [IX. 21]  And DE is an unit.  Therefore AD is odd. [VII. Def. 7]  Q. E. D. 
             
             
PROPOSITION 24. 
 
 
If from an even number an even number be subtracted, the remainder will be even. 
 
 
For from the even number AB let the even number BC be subtracted:  I say that the remainder CA is even. 
   
   
For, since AB is even, it has a half part. [VII. Def. 6]  For the same reason BC also has a half part;  so that the remainder [CA also has a half part, and] AC is therefore even.  Q. E. D. 
       
       
PROPOSITION 25. 
 
 
If from an even number an odd number be subtracted, the remainder will be odd. 
 
 
For from the even number AB let the odd number BC be subtracted;  I say that the remainder CA is odd. 
   
   
For let the unit CD be subtracted from BC;  therefore DB is even. [VII. Def. 7]  But AB is also even;  therefore the remainder AD is also even. [IX. 24]  And CD is an unit;  therefore CA is odd. [VII. Def. 7]  Q. E. D. 
             
             
PROPOSITION 26. 
 
 
If from an odd number an odd number be subtracted, the remainder will be even. 
 
 
For from the odd number AB let the odd number BC be subtracted;  I say that the remainder CA is even. 
   
   
For, since AB is odd, let the unit BD be subtracted;  therefore the remainder AD is even. [VII. Def. 7]  For the same reason CD is also even; [VII. Def. 7]  so that the remainder CA is also even. [IX. 24]  Q. E. D. 
         
         
PROPOSITION 27. 
 
 
If from an odd number an even number be subtracted, the remainder will be odd. 
 
 
For from the odd number AB let the even number BC be subtracted;  I say that the remainder CA is odd. 
   
   
Let the unit AD be subtracted;  therefore DB is even. [VII. Def. 7]  But BC is also even;  therefore the remainder CD is even. [ IX. 24 ]  Therefore CA is odd. [VII. Def. 7]  Q. E. D. 
           
           
PROPOSITION 28. 
 
 
If an odd number by multiplying an even number make some number, the product will be even. 
 
 
For let the odd number A by multiplying the even number B make C;  I say that C is even. 
   
   
For, since A by multiplying B has made C,  therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]  And B is even;  therefore C is made up of even numbers.  But, if as many even numbers as we please be added together, the whole is even. [IX. 21]  Therefore C is even.  Q. E. D. 
             
             
PROPOSITION 29. 
 
 
If an odd number by multiplying an odd number make some number, the product will be odd. 
 
 
For let the odd number A by multiplying the odd number B make C;  I say that C is odd. 
   
   
For, since A by multiplying B has made C,  therefore C is made up of as many numbers equal to B as there are units in A. [VII. Def. 15]  And each of the numbers A, B is odd;  therefore C is made up of odd numbers the multitude of which is odd.  Thus C is odd. [IX. 23]  Q. E. D. 
           
           
PROPOSITION 30. 
 
 
If an odd number measure an even number, it will also measure the half of it. 
 
 
For let the odd number A measure the even number B;  I say that it will also measure the half of it. 
   
   
For, since A measures B, let it measure it according to C;  I say that C is not odd. 
   
   
For, if possible, let it be so.  Then, since A measures B according to C,  therefore A by multiplying C has made B.  Therefore B is made up of odd numbers the multitude of which is odd.  Therefore B is odd: [IX. 23] which is absurd, for by hypothesis it is even.  Therefore C is not odd; therefore C is even.  Thus A measures B an even number of times.  For this reason then it also measures the half of it.  Q. E. D. 
                 
                 
PROPOSITION 31. 
 
 
If an odd number be prime to any number, it will also be prime to the double of it. 
 
 
For let the odd number A be prime to any number B, and let C be double of B;  I say that A is prime to C. 
   
   
For, if they are not prime to one another, some number will measure them.  Let a number measure them, and let it be D.  Now A is odd;  therefore D is also odd.  And since D which is odd measures C, and C is even,  therefore [D] will measure the half of C also. [IX. 30]  But B is half of C;  therefore D measures B.  But it also measures A;  therefore D measures A, B which are prime to one another: which is impossible.  Therefore A cannot but be prime to C.  Therefore A, C are prime to one another.  Q. E. D. 
                         
                         
PROPOSITION 32. 
 
 
Each of the numbers which are continually doubled beginning from a dyad is even-times even only. 
 
 
For let as many numbers as we please, B, C, D, have been continually doubled beginning from the dyad A;  I say that B, C, D are eventimes even only. 
   
   
Now that each of the numbers B, C, D is even-times even is manifest;  for it is doubled from a dyad.  I say that it is also even-times even only. 
     
     
For let an unit be set out.  Since then as many numbers as we please beginning from an unit are in continued proportion,  and the number A after the unit is prime,  therefore D, the greatest of the numbers A, B, C, D, will not be measured by any other number except A, B, C. [IX. 13]  And each of the numbers A, B, C is even;  therefore D is even-times even only. [VII. Def. 8]  Similarly we can prove that each of the numbers B, C is even-times even only.  Q. E. D. 
               
               
PROPOSITION 33. 
 
 
If a number have its half odd, it is even-times odd only. 
 
 
For let the number A have its half odd;  I say that A is even-times odd only. 
   
   
Now that it is even-times odd is manifest;  for the half of it, being odd, measures it an even number of times. [VII. Def. 9]  I say next that it is also even-times odd only. 
     
     
For, if A is even-times even also, it will be measured by an even number according to an even number; [VII. Def. 8]  so that the half of it will also be measured by an even number though it is odd: which is absurd.  Therefore A is even-times odd only.  Q. E. D. 
       
       
PROPOSITION 34. 
 
 
If a number neither be one of those which are continually doubled from a dyad, nor have its half odd, it is both eventimes even and even-times odd. 
 
 
For let the number A neither be one of those doubled from a dyad, nor have its half odd;  I say that A is both even-times even and even-times odd. 
   
   
Now that A is even-times even is manifest;  for it has not its half odd. [VII. Def. 8]  I say next that it is also even-times odd.  For, if we bisect A, then bisect its half, and do this continually, we shall come upon some odd number which will measure A according to an even number.  For, if not, we shall come upon a dyad, and A will be among those which are doubled from a dyad: which is contrary to the hypothesis.  Thus A is even-times odd.  But it was also proved even-times even.  Therefore A is both even-times even and even-times odd.  Q. E. D. 
                 
                 
PROPOSITION 35. 
 
 
If as many numbers as we please be in continued proportion, and there be subtracted from the second and the last numbers equal to the first, then, as the excess of the second is to the first, so will the excess of the last be to all those before it. 
 
 
Let there be as many numbers as we please in continued proportion, A, BC, D, EF, beginning from A as least, and let there be subtracted from BC and EF the numbers BG, FH, each equal to A;  I say that, as GC is to A, so is EH to A, BC, D. 
   
   
For let FK be made equal to BC, and FL equal to D.  Then, since FK is equal to BC, and of these the part FH is equal to the part BG,  therefore the remainder HK is equal to the remainder GC.  And since, as EF is to D, so is D to BC, and BC to A,  while D is equal to FL, BC to FK, and A to FH,  therefore, as EF is to FL, so is LF to FK, and FK to FH.  Separando, as EL is to LF, so is LK to FK, and KH to FH. [VII. 11, 13]  Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [VII. 12]  therefore, as KH is to FH, so are EL, LK. KH to LF, FK, HF.  But KH is equal to CG, FH to A, and LF, FK, HF to D, BC, A;  therefore, as CG is to A, so is EH to D, BC, A.  Therefore, as the excess of the second is to the first, so is the excess of the last to all those before it.  Q. E. D. 
                         
                         
PROPOSITION 36. 
 
 
If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect. 
 
 
For let as many numbers as we please, A, B, C, D, beginning from an unit be set out in double proportion, until the sum of all becomes prime, let E be equal to the sum, and let E by multiplying D make FG;  I say that FG is perfect. 
   
   
For, however many A, B, C, D are in multitude, let so many E, HK, L, M be taken in double proportion beginning from E;  therefore, ex aequali, as A is to D, so is E to M. [VII. 14]  Therefore the product of E, D is equal to the product of A, M. [VII. 19]  And the product of E, D is FG;  therefore the product of A, M is also FG.  Therefore A by multiplying M has made FG;  therefore M measures FG according to the units in A.  And A is a dyad;  therefore FG is double of M.  But M, L, HK, E are continuously double of each other;  therefore E, HK, L, M, FG are continuously proportional in double proportion.  Now let there be subtracted from the second HK and the last FG the numbers HN, FO, each equal to the first E;  therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. [IX. 35]  Therefore, as NK is to E, so is OG to M, L, KH, E.  And NK is equal to E;  therefore OG is also equal to M, L, HK, E.  But FO is also equal to E, and E is equal to A, B, C, D and the unit.  Therefore the whole FG is equal to E, HK, L, M and A, B, C, D and the unit; and it is measured by them.  I say also that FG will not be measured by any other number except A, B, C, D, E, HK, L, M and the unit.  For, if possible, let some number P measure FG, and let P not be the same with any of the numbers A, B, C, D, E, HK, L, M.  And, as many times as P measures FG, so many units let there be in Q;  therefore Q by multiplying P has made FG.  But, further, E has also by multiplying D made FG;  therefore, as E is to Q, so is P to D. [VII. 19]  And, since A, B, C, D are continuously proportional beginning from an unit,  therefore D will not be measured by any other number except A, B, C. [IX. 13]  And, by hypothesis, P is not the same with any of the numbers A, B, C;  therefore P will not measure D.  But, as P is to D, so is E to Q;  therefore neither does E measure Q. [VII. Def. 20]  And E is prime;  and any prime number is prime to any number which it does not measure. [VII. 29]  Therefore E, Q are prime to one another.  But primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20]  and, as E is to Q, so is P to D;  therefore E measures P the same number of times that Q measures D.  But D is not measured by any other number except A, B, C;  therefore Q is the same with one of the numbers A, B, C. 
                                                                           
                                                                           
Let it be the same with B.  And, however many B, C, D are in multitude, let so many E, HK, L be taken beginning from E.  Now E, HK, L are in the same ratio with B, C, D;  therefore, ex aequali, as B is to D, so is E to L. [VII. 14]  Therefore the product of B, L is equal to the product of D, E. [VII. 19]  But the product of D, E is equal to the product of Q, P;  therefore the product of Q, P is also equal to the product of B, L.  Therefore, as Q is to B, so is L to P. [VII. 19]  And Q is the same with B;  therefore L is also the same with P; which is impossible,  for by hypothesis P is not the same with any of the numbers set out.  Therefore no number will measure FG except A, B, C, D, E, HK, L, M and the unit.  And FG was proved equal to A, B, C, D, E, HK, L, M and the unit;  and a perfect number is that which is equal to its own parts; [VII. Def. 22]  therefore FG is perfect.  Q. E. D. 
                               
                               
BOOK X. 
 
 
DEFINITIONS. 
 
 
1. Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure. 
 
 
2. Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure. 
 
 
3. With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational. 
 
 
4. And let the square on the assigned straight line be called rational and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them. 
 
 
PROPOSITION 1. 
 
 
Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the lesser magnitude set out. 
 
 
Let AB, C be two unequal magnitudes of which AB is the greater:  I say that, if from AB there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C. 
   
   
For C if multiplied will sometime be greater than AB. [cf. v. Def. 4]  Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE. 
   
   
Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE.  Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.  And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.  But DF is equal to C;  therefore C is also greater than AK.  Therefore AK is less than C. 
           
           
Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C.  Q. E. D. And the theorem can be similarly proved even if the parts subtracted be halves. 
   
   
PROPOSITION 2. 
 
 
If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable. 
 
 
For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it;  I say that the magnitudes AB, CD are incommensurable. 
   
   
For, if they are commensurable, some magnitude will measure them.  Let a magnitude measure them, if possible, and let it be E;  let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself,  and let this process be repeated continually, until there is left some magnitude which is less than E.  Suppose this done, and let there be left AG less than E.  Then, since E measures AB, while AB measures DF, therefore E will also measure FD.  But it measures the whole CD also;  therefore it will also measure the remainder CF.  But CF measures BG;  therefore E also measures BG.  But it measures the whole AB also;  therefore it will also measure the remainder AG, the greater the less: which is impossible.  Therefore no magnitude will measure the magnitudes AB, CD;  therefore the magnitudes AB, CD are incommensurable. 
                           
                           
Therefore etc. [X. Def. 1] 
 
 
PROPOSITION 3. 
 
 
Given two commensurable magnitudes, to find their greatest common measure. 
 
 
Let the two given commensurable magnitudes be AB, CD of which AB is the less;  thus it is required to find the greatest common measure of AB, CD. 
   
   
Now the magnitude AB either measures CD or it does not.  If then it measures it—and it measures itself also—AB is a common measure of AB, CD.  And it is manifest that it is also the greatest;  for a greater magnitude than the magnitude AB will not measure AB. 
       
       
Next, let AB not measure CD.  Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable; [cf. X. 2]  let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE. 
     
     
Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB.  But it measures itself also;  therefore AF will also measure the whole AB.  But AB measures DE;  therefore AF will also measure ED.  But it measures CE also;  therefore it also measures the whole CD.  Therefore AF is a common measure of AB, CD.  I say next that it is also the greatest. 
                 
                 
For, if not, there will be some magnitude greater than AF which will measure AB, CD.  Let it be G.  Since then G measures AB, while AB measures ED, therefore G will also measure ED.  But it measures the whole CD also;  therefore G will also measure the remainder CE.  But CE measures FB;  therefore G will also measure FB.  But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible.  Therefore no magnitude greater than AF will measure AB, CD;  therefore AF is the greatest common measure of AB, CD. 
                   
                   
Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found.  Q. E. D. 
   
   
PORISM.
From this it is manifest that, if a magnitude measure two magnitudes, it will also measure their greatest common measure. 
 
 
PROPOSITION 4. 
 
 
Given three commensurable magnitudes, to find their greatest common measure. 
 
 
Let A, B, C be the three given commensurable magnitudes;  thus it is required to find the greatest common measure of A, B, C. 
   
   
Let the greatest common measure of the two magnitudes A, B be taken, and let it be D; [X. 3]  then D either measures C, or does not measure it.  First, let it measure it.  Since then D measures C, while it also measures A, B,  therefore D is a common measure of A, B, C.  And it is manifest that it is also the greatest;  for a greater magnitude than the magnitude D does not measure A, B. 
             
             
Next, let D not measure C.  I say first that C, D are commensurable. 
   
   
For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A, B also;  so that it will also measure the greatest common measure of A, B, namely D. [X. 3, Por.]  But it also measures C;  so that the said magnitude will measure C, D;  therefore C, D are commensurable.  Now let their greatest common measure be taken, and let it be E. [X. 3]  Since then E measures D, while D measures A, B,  therefore E will also measure A, B.  But it measures C also;  therefore E measures A, B, C;  therefore E is a common measure of A, B, C.  I say next that it is also the greatest. 
                       
                       
For, if possible, let there be some magnitude F greater than E, and let it measure A, B, C.  Now, since F measures A, B, C, it will also measure A, B, and will measure the greatest common measure of A, B. [X. 3, Por.]  But the greatest common measure of A, B is D;  therefore F measures D.  But it measures C also;  therefore F measures C, D;  therefore F will also measure the greatest common measure of C, D. [X. 3, Por.]  But that is E;  therefore F will measure E, the greater the less: which is impossible.  Therefore no magnitude greater than the magnitude E will measure A, B, C;  therefore E is the greatest common measure of A, B, C if D do not measure C, and, if it measure it, D is itself the greatest common measure. 
                     
                     
Therefore the greatest common measure of the three given commensurable magnitudes has been found. 
 
 
PORISM.
From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure. 
 
 
Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended.  Q. E. D. 
   
   
PROPOSITION 5. 
 
 
Commensurable magnitudes have to one another the ratio which a number has to a number. 
 
 
Let A, B be commensurable magnitudes;  I say that A has to B the ratio which a number has to a number. 
   
   
For, since A, B are commensurable, some magnitude will measure them.  Let it measure them, and let it be C.  And, as many times as C measures A, so many units let there be in D;  and, as many times as C measures B, so many units let there be in E. 
       
       
Since then C measures A according to the units in D, while the unit also measures D according to the units in it,  therefore the unit measures the number D the same number of times as the magnitude C measures A;  therefore, as C is to A, so is the unit to D; [VII. Def. 20]  therefore, inversely, as A is to C, so is D to the unit. [cf. V. 7, Por.]  Again, since C measures B according to the units in E, while the unit also measures E according to the units in it,  therefore the unit measures E the same number of times as C measures B;  therefore, as C is to B, so is the unit to E.  But it was also proved that, as A is to C, so is D to the unit;  therefore, ex aequali, as A is to B, so is the number D to E. [V. 22] 
                 
                 
Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E.  Q. E. D. 
   
   
PROPOSITION 6. 
 
 
If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable. 
 
 
For let the two magnitudes A, B have to one another the ratio which the number D has to the number E;  I say that the magnitudes A, B are commensurable. 
   
   
For let A be divided into as many equal parts as there are units in D, and let C be equal to one of them;  and let F be made up of as many magnitudes equal to C as there are units in E. 
   
   
Since then there are in A as many magnitudes equal to C as there are units in D, whatever part the unit is of D, the same part is C of A also;  therefore, as C is to A, so is the unit to D. [VII. Def. 20]  But the unit measures the number D;  therefore C also measures A.  And since, as C is to A, so is the unit to D, therefore,  inversely, as A is to C, so is the number D to the unit. [cf. V. 7, Por.]  Again, since there are in F as many magnitudes equal to C as there are units in E,  therefore, as C is to F, so is the unit to E. [VII. Def. 20]  But it was also proved that, as A is to C, so is D to the unit;  therefore, ex aequali, as A is to F, so is D to E. [v. 22]  But, as D is to E, so is A to B;  therefore also, as A is to B, so is it to F also. [V. 11]  Therefore A has the same ratio to each of the magnitudes B, F;  therefore B is equal to F. [V. 9]  But C measures F;  therefore it measures B also.  Further it measures A also;  therefore C measures A, B.  Therefore A is commensurable with B. 
                                     
                                     
Therefore etc. 
 
 
PORISM.
From this it is manifest that, if there be two numbers, as D, E, and a straight line, as A, it is possible to make a straight line [F] such that the given straight line is to it as the number D is to the number E.
And, if a mean proportional be also taken between A, F, as B, as A is to F, so will the square on A be to the square on B, that is, as the first is to the third, so is the figure on the first to that which is similar and similarly described on the second. [VI. 19, Por.]
But, as A is to F, so is the number D to the number E;
therefore it has been contrived that, as the number D is to the number E, so also is the figure on the straight line A to the figure on the straight line B.
Q. E. D. 
 
 
PROPOSITION 7. 
 
 
Incommensurable magnitudes have not to one another the ratio which a number has to a number. 
 
 
Let A, B be incommensurable magnitudes;  I say that A has not to B the ratio which a number has to a number. 
   
   
For, if A has to B the ratio which a number has to a number, A will be commensurable with B. [X. 6]  But it is not;  therefore A has not to B the ratio which a number has to a number. 
     
     
Therefore etc. 
 
 
PROPOSITION 8. 
 
 
If two magnitudes have not to one another the ratio which a number has to a number, the magnitudes will be incommensurable. 
 
 
For let the two magnitudes A, B not have to one another the ratio which a number has to a number;  I say that the magnitudes A, B are incommensurable. 
   
   
For, if they are commensurable, A will have to B the ratio which a number has to a number. [X. 5]  But it has not;  therefore the magnitudes A, B are incommensurable. 
     
     
Therefore etc. 
 
 
PROPOSITION 9. 
 
 
The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either. 
 
 
For let A, B be commensurable in length;  I say that the square on A has to the square on B the ratio which a square number has to a square number. 
   
   
For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [X. 5]  Let it have to it the ratio which C has to D.  Since then, as A is to B, so is C to D,  while the ratio of the square on A to the square on B is duplicate of the ratio of A to B,  for similar figures are in the duplicate ratio of their corresponding sides; [VI. 20, Por.]  and the ratio of the square on C to the square on D is duplicate of the ratio of C to D,  for between two square numbers there is one mean proportional number,  and the square number has to the square number the ratio duplicate of that which the side has to the side; [VIII. 11]  therefore also, as the square on A is to the square on B, so is the square on C to the square on D. 
                 
                 
Next, as the square on A is to the square on B, so let the square on C be to the square on D;  I say that A is commensurable in length with B. 
   
   
For since, as the square on A is to the square on B, so is the square on C to the square on D,  while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D,  therefore also, as A is to B, so is C to D.  Therefore A has to B the ratio which the number C has to the number D;  therefore A is commensurable in length with B. [X. 6] 
         
         
Next, let A be incommensurable in length with B;  I say that the square on A has not to the square on B the ratio which a square number has to a square number. 
   
   
For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B.  But it is not;  therefore the square on A has not to the square on B the ratio which a square number has to a square number. 
     
     
Again, let the square on A not have to the square on B the ratio which a square number has to a square number;  I say that A is incommensurable in length with B. 
   
   
For, if A is commensurable with B, the square on A will have to the square on B the ratio which a square number has to a square number.  But it has not;  therefore A is not commensurable in length with B. 
     
     
Therefore etc. 
 
 
PORISM.
And it is manifest from what has been proved that straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always commensurable in length also. 
 
 
LEMMA. [It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number, [VIII. 26]  and that, if two numbers have to one another the ratio which a square number has to a square number, they are similar plane numbers. [Converse of VIII. 26]  And it is manifest from these propositions that numbers which are not similar plane numbers, that is, those which have not their sides proportional, have not to one another the ratio which a square number has to a square number.  For, if they have, they will be similar plane numbers: which is contrary to the hypothesis.  Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.] 
         
         
[PROPOSITION 10. 
 
 
To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line. 
 
 
Let A be the assigned straight line;  thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A. 
   
   
Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number,  that is, which are not similar plane numbers;  and let it be contrived that, as B is to C, so is the square on A to the square on D – for we have learnt how to do this — [X. 6, Por.]  therefore the square on A is commensurable with the square on D. [X. 6]  And, since B has not to C the ratio which a square number has to a square number,  therefore neither has the square on A to the square on D the ratio which a square number has to a square number;  therefore A is incommensurable in length with D. [X. 9]  Let E be taken a mean proportional between A, D;  therefore, as A is to D, so is the square on A to the square on E. [V. Def. 9]  But A is incommensurable in length with D;  therefore the square on A is also incommensurable with the square on E; [X. 11]  therefore A is incommensurable in square with E. 
                       
                       
Therefore two straight lines D, E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A.] 
 
 
PROPOSITION 11. 
 
 
If four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable with the fourth;  and, if the first be incommensurable with the second, the third will also be incommensurable with the fourth. 
   
   
Let A, B, C, D be four magnitudes in proportion, so that, as A is to B, so is C to D, and let A be commensurable with B;  I say that C will also be commensurable with D. 
   
   
For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X. 5]  And, as A is to B, so is C to D;  therefore C also has to D the ratio which a number has to a number;  therefore C is commensurable with D. [X. 6] 
       
       
Next, let A be incommensurable with B;  I say that C will also be incommensurable with D. 
   
   
For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7]  And, as A is to B, so is C to D;  therefore neither has C to D the ratio which a number has to a number;  therefore C is incommensurable with D. [X. 8] 
       
       
Therefore etc. 
 
 
PROPOSITION 12. 
 
 
Magnitudes commensurable with the same magnitude are commensurable with one another also. 
 
 
For let each of the magnitudes A, B be commensurable with C;  I say that A is also commensurable with B. 
   
   
For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [X. 5]  Let it have the ratio which D has to E.  Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. [X. 5]  Let it have the ratio which F has to G.  And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, let the numbers H, K, L be taken continuously in the given ratios; [cf. VIII. 4]  so that, as D is to E, so is H to K,  and, as F is to G, so is K to L. 
             
             
Since, then, as A is to C, so is D to E,  while, as D is to E, so is H to K,  therefore also, as A is to C, so is H to K. [V. 11]  Again, since, as C is to B, so is F to G,  while, as F is to G, so is K to L,  therefore also, as C is to B, so is K to L. [V. 11]  But also, as A is to C, so is H to K;  therefore, ex aequali, as A is to B, so is H to L. [V. 22]  Therefore A has to B the ratio which a number has to a number;  therefore A is commensurable with B. [X. 6] 
                   
                   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 13. 
 
 
If two magnitudes be commensurable, and the one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same. 
 
 
Let A, B be two commensurable magnitudes, and let one of them, A, be incommensurable with any other magnitude C;  I say that the remaining one, B, will also be incommensurable with C. 
   
   
For, if B is commensurable with C, while A is also commensurable with B, A is also commensurable with C. [X. 12]  But it is also incommensurable with it:  which is impossible.  Therefore B is not commensurable with C;  therefore it is incommensurable with it. 
         
         
Therefore etc. 
 
 
LEMMA.
Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less. 
 
 
Let AB, C be the given two unequal straight lines, and let AB be the greater of them;  thus it is required to find by what square the square on AB is greater than the square on C. 
   
   
Let the semicircle ADB be described on AB, and let AD be fitted into it equal to C; [IV. 1] let DB be joined.  It is then manifest that the angle ADB is right, [III. 31] and that the square on AB is greater than the square on AD, that is, C, by the square on DB. [I. 47] 
   
   
Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner. 
 
 
Let AD, DB be the given two straight lines, and let it be required to find the straight line the square on which is equal to the sum of the squares on them.  Let them be placed so as to contain a right angle, that formed by AD, DB; and let AB be joined.  It is again manifest that the straight line the square on which is equal to the sum of the squares on AD, DB is AB. [I. 47]  Q. E. D. 
       
       
PROPOSITION 14. 
 
 
If four straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line commensurable with the third. And, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line incommensurable with the third. 
 
 
Let A, B, C, D be four straight lines in proportion, so that, as A is to B, so is C to D; and let the square on A be greater than the square on B by the square on E, and let the square on C be greater than the square on D by the square on F;  I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. 
   
   
For since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. [VI. 22]  But the squares on E, B are equal to the square on A, and the squares on D, F are equal to the square on C.  Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D;  therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; [V. 17]  therefore also, as E is to B, so is F to D; [VI. 22]  therefore, inversely, as B is to E, so is D to F.  But, as A is to B, so also is C to D;  therefore, ex aequali, as A is to E, so is C to F. [V. 22]  Therefore, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. [X. 11] 
                 
                 
Therefore etc.2 
 
 
PROPOSITION 15. 
 
 
If two commensurable magnitudes be added together, the whole will also be commensurable with each of them;  and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable. 
   
   
For let the two commensurable magnitudes AB, BC be added together;  I say that the whole AC is also commensurable with each of the magnitudes AB, BC. 
   
   
For, since AB, BC are commensurable, some magnitude will measure them.  Let it measure them, and let it be D.  Since then D measures AB, BC, it will also measure the whole AC.  But it measures AB, BC also;  therefore D measures AB, BC, AC;  therefore AC is commensurable with each of the magnitudes AB, BC. [X. Def. 1] 
           
           
Next, let AC be commensurable with AB;  I say that AB, BC are also commensurable. 
   
   
For, since AC, AB are commensurable, some magnitude will measure them.  Let it measure them, and let it be D.  Since then D measures CA, AB, it will also measure the remainder BC.  But it measures AB also;  therefore D will measure AB, BC;  therefore AB, BC are commensurable. [X. Def. 1] 
           
           
Therefore etc. 
 
 
PROPOSITION 16. 
 
 
If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them;  and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable. 
   
   
For let the two incommensurable magnitudes AB, BC be added together;  I say that the whole AC is also incommensurable with each of the magnitudes AB, BC. 
   
   
For, if CA, AB are not incommensurable, some magnitude will measure them.  Let it measure them, if possible, and let it be D.  Since then D measures CA, AB, therefore it will also measure the remainder BC.  But it measures AB also;  therefore D measures AB, BC.  Therefore AB, BC are commensurable;  but they were also, by hypothesis, incommensurable: which is impossible.  Therefore no magnitude will measure CA, AB;  therefore CA, AB are incommensurable. [X. Def. 1]  Similarly we can prove that AC, CB are also incommensurable.  Therefore AC is incommensurable with each of the magnitudes AB, BC. 
                     
                     
Next, let AC be incommensurable with one of the magnitudes AB, BC.  First, let it be incommensurable with AB;  I say that AB, BC are also incommensurable. 
     
     
For, if they are commensurable, some magnitude will measure them.  Let it measure them, and let it be D.  Since then D measures AB, BC. therefore it will also measure the whole AC.  But it measures AB also;  therefore D measures CA, AB.  Therefore CA, AB are commensurable;  but they were also, by hypothesis, incommensurable: which is impossible.  Therefore no magnitude will measure AB, BC;  therefore AB, BC are incommensurable. [X. Def. 1] 
                 
                 
Therefore etc. 
 
 
LEMMA.
If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application. 
 
 
For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB;  I say that AD is equal to the rectangle contained by AC, CB. 
   
   
This is indeed at once manifest;  for, since DB is a square, DC is equal to CB;  and AD is the rectangle AC, CD, that is, the rectangle AC, CB. 
     
     
Therefore etc. 
 
 
PROPOSITION 17. 
 
 
If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater. And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length. 
 
 
Let A, BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma] and let BD be commensurable in length with DC;  I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC. 
   
   
For let BC be bisected at the point E, and let EF be made equal to DE.  Therefore the remainder DC is equal to BF.  And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D,  therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [II. 5]  And the same is true of their quadruples;  therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square on EC.  But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE,  for DF is double of DE.  And the square on BC is equal to four times the square on EC,  for again BC is double of CE.  Therefore the squares on A, DF are equal to the square on BC,  so that the square on BC is greater than the square on A by the square on DF.  It is to be proved that BC is also commensurable with DF.  Since BD is commensurable in length with DC,  therefore BC is also commensurable in length with CD. [X. 15]  But CD is commensurable in length with CD, BF,  for CD is equal to BF. [X. 6]  Therefore BC is also commensurable in length with BF, CD, [X. 12]  so that BC is also commensurable in length with the remainder FD; [X. 15]  therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC. 
                                       
                                       
Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC.  It is to be proved that BD is commensurable in length with DC. 
   
   
With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.  But the square on BC is greater than the square on A by the square on a straight line commensurable with BC.  Therefore BC is commensurable in length with FD,  so that BC is also commensurable in length with the remainder, the sum of BF, DC. [X. 15]  But the sum of BF, DC is commensurable with DC, [X. 6]  so that BC is also commensurable in length with CD; [X. 12]  and therefore, separando, BD is commensurable in length with DC. [X. 15] 
             
             
Therefore etc. 
 
 
PROPOSITION 18. 
 
 
If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are incommensurable, the square on the greater will be greater than the square on the less by the square on a straight line incommensurable with the greater.  And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable. 
   
   
Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17] and let BD be incommensurable in length with DC;  I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. 
   
   
For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD.  It is to be proved that BC is incommensurable in length with DF.  Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16]  But DC is commensurable with the sum of BF, DC; [X. 6]  therefore BC is also incommensurable with the sum of BF, DC; [X. 13]  so that BC is also incommensurable in length with the remainder FD. [X. 16]  And the square on BC is greater than the square on A by the square on FD;  therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. 
               
               
Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC,  and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure.  Let this be the rectangle BD, DC.  It is to be proved that BD is incommensurable in length with DC. 
       
       
For, with the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.  But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC;  therefore BC is incommensurable in length with FD,  so that BC is also incommensurable with the remainder, the sum of BF, DC. [X. 16]  But the sum of BF, DC is commensurable in length with DC; [X. 6]  therefore BC is also incommensurable in length with DC, [X. 13]  so that, separando, BD is also incommensurable in length with DC. [X. 16] 
             
             
Therefore etc. 
 
 
LEMMA.
[Since it has been proved that straight lines commensurable in length are always commensurable in square also, while those commensurable in square are not always commensurable in length also, but can of course be either commensurable or incommensurable in length, it is manifest that, if any straight line be commensurable in length with a given rational straight line, it is called rational and commensurable with the other not only in length but in square also, since straight lines commensurable in length are always commensurable in square also. 
But, if any straight line be commensurable in square with a given rational straight line, then, if it is also commensurable in length with it, it is called in this case also rational and commensurable with it both in length and in square;  but, if again any straight line, being commensurable in square with a given rational straight line, be incommensurable in length with it, it is called in this case also rational but commensurable in square only.] 
     
     
PROPOSITION 19. 
 
 
The rectangle contained by rational straight lines commensurable in length is rational. 
 
 
For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in length;  I say that AC is rational. 
   
   
For on AB let the square AD be described;  therefore AD is rational. [X. Def. 4]  And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC.  And, as BD is to BC, so is DA to AC. [VI. 1]  Therefore DA is commensurable with AC. [X. 11]  But DA is rational;  therefore AC is also rational. [X. Def. 4] 
             
             
Therefore etc. 
 
 
PROPOSITION 20. 
 
 
If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied. 
 
 
For let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth;  I say that BC is rational and commensurable in length with BA. 
   
   
For on AB let the square AD be described;  therefore AD is rational. [X. Def. 4]  But AC is also rational;  therefore DA is commensurable with AC.  And, as DA is to AC, so is DB to BC. [VI. 1]  Therefore DB is also commensurable with BC; [X. 11]  and DB is equal to BA;  therefore AB is also commensurable with BC.  But AB is rational;  therefore BC is also rational and commensurable in length with AB. 
                   
                   
Therefore etc. 
 
 
PROPOSITION 21. 
 
 
The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called medial. 
 
 
For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only;  I say that AC is irrational, and the side of the square equal to it is irrational; and let the latter be called medial. 
   
   
For on AB let the square AD be described;  therefore AD is rational. [X. Def. 4]  And, since AB is incommensurable in length with BC,  for by hypothesis they are commensurable in square only,  while AB is equal to BD, therefore DB is also incommensurable in length with BC.  And, as DB is to BC, so is AD to AC; [VI. 1]  therefore DA is incommensurable with AC. [X. 11]  But DA is rational;  therefore AC is irrational,  so that the side of the square equal to AC is also irrational. [X. Def. 4] And let the latter be called medial.  Q. E. D. 
                     
                     
LEMMA.
If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines. 
 
 
Let FE, EG be two straight lines.  I say that, as FE is to EG, so is the square on FE to the rectangle FE, EG. 
   
   
For on FE let the square DF be described, and let GD be completed.  Since then, as FE is to EG, so is FD to DG, [VI. 1]  and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG,  therefore, as FE is to EG, so is the square on FE to the rectangle FE, EG.  Similarly also, as the rectangle GE, EF is to the square on EF, that is, as GD is to FD, so is GE to EF.  Q. E. D. 
           
           
PROPOSITION 22. 
 
 
The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied. 
 
 
Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth;  I say that CD is rational and incommensurable in length with CB. 
   
   
For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only. [X. 21]  Let the square on it be equal to GF.  But the square on it is also equal to BD;  therefore BD is equal to GF.  But it is also equiangular with it;  and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; [VI. 14]  therefore, proportionally, as BC is to EG, so is EF to CD.  Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [VI. 22]  But the square on CB is commensurable with the square on EG,  for each of these straight lines is rational;  therefore the square on EF is also commensurable with the square on CD. [X. 11]  But the square on EF is rational;  therefore the square on CD is also rational; [X. Def. 4]  therefore CD is rational.  And, since EF is incommensurable in length with EG,  for they are commensurable in square only,  and, as EF is to EG, so is the square on EF to the rectangle FE, EG,  [Lemma] therefore the square on EF is incommensurable with the rectangle FE, EG. [X. 11]  But the square on CD is commensurable with the square on EF,  for the straight lines are rational in square;  and the rectangle DC, CB is commensurable with the rectangle FE, EG,  for they are equal to the square on A;  therefore the square on CD is also incommensurable with the rectangle DC, CB. [X. 13]  But, as the square on CD is to the rectangle DC, CB, so is DC to CB;  [Lemma] therefore DC is incommensurable in length with CB. [X. 11]  Therefore CD is rational and incommensurable in length with CB.  Q. E. D. 
                                                     
                                                     
PROPOSITION 23. 
 
 
A straight line commensurable with a medial straight line is medial. 
 
 
Let A be medial, and let B be commensurable with A;  I say that B is also medial. 
   
   
For let a rational straight line CD be set out, and to CD let the rectangular area CE equal to the square on A be applied, producing ED as breadth;  therefore ED is rational and incommensurable in length with CD. [X. 22]  And let the rectangular area CF equal to the square on B be applied to CD, producing DF as breadth.  Since then A is commensurable with B, the square on A is also commensurable with the square on B.  But EC is equal to the square on A, and CF is equal to the square on B;  therefore EC is commensurable with CF.  And, as EC is to CF, so is ED to DF; [VI. 1]  therefore ED is commensurable in length with DF. [X. 11]  But ED is rational and incommensurable in length with DC;  therefore DF is also rational [X. Def. 3] and incommensurable in length with DC. [X. 13]  Therefore CD, DF are rational and commensurable in square only.  But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable in square only is medial; [X. 21]  therefore the side of the square equal to the rectangle CD, DF is medial.  And B is the side of the square equal to the rectangle CD, DF;  therefore B is medial. 
                             
                             
PORISM.
From this it is manifest that an area commensurable with a medial area is medial.
[And in the same way as was explained in the case of rationals [Lemma following X. 18] it follows, as regards medials, that a straight line commensurable in length with a medial straight line is called medial and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also.
But, if any straight line be commensurable in square with a medial straight line, then, if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only.] 
 
 
PROPOSITION 24. 
 
 
The rectangle contained by medial straight lines commensurable in length is medial. 
 
 
For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in length;  I say that AC is medial. 
   
   
For on AB let the square AD be described;  therefore AD is medial.  And, since AB is commensurable in length with BC, while AB is equal to BD, therefore DB is also commensurable in length with BC;  so that DA is also commensurable with AC. [VI. 1, X. 11]  But DA is medial;  therefore AC is also medial. [X. 23, Por.]  Q. E. D. 
             
             
PROPOSITION 25. 
 
 
The rectangle contained by medial straight lines commensurable in square only is either rational or medial. 
 
 
For let the rectangle AC be contained by the medial straight lines AB, BC which are commensurable in square only;  I say that AC is either rational or medial. 
   
   
For on AB, BC let the squares AD, BE be described;  therefore each of the squares AD, BE is medial.  Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth,  to HM let there be applied the rectangular parallelogram MK equal to AC, producing HK as breadth,  and further to KN let there be similarly applied NL equal to BE, producing KL as breadth;  therefore FH, HK, KL are in a straight line.  Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL,  therefore each of the rectangles GH, NL is also medial.  And they are applied to the rational straight line FG;  therefore each of the straight lines FH, KL is rational and incommensurable in length with FG. [X. 22]  And, since AD is commensurable with BE,  therefore GH is also commensurable with NL.  And, as GH is to NL, so is FH to KL; [VI. 1]  therefore FH is commensurable in length with KL. [X. 11]  Therefore FH, KL are rational straight lines commensurable in length;  therefore the rectangle FH, KL is rational. [X. 19]  And, since DB is equal to BA, and OB to BC,  therefore, as DB is to BC, so is AB to BO.  But, as DB is to BC, so is DA to AC, [VI. 1]  and, as AB is to BO, so is AC to CO; [id.]  therefore, as DA is to AC, so is AC to CO.  But AD is equal to GH, AC to MK and CO to NL;  therefore, as GH is to MK, so is MK to NL;  therefore also, as FH is to HK, so is HK to KL; [VI. 1, V. 11]  therefore the rectangle FH, KL is equal to the square on HK. [VI. 17]  But the rectangle FH, KL is rational;  therefore the square on HK is also rational.  Therefore HK is rational.  And, if it is commensurable in length with FG, HN is rational; [X. 19]  but, if it is incommensurable in length with FG, KH, HM are rational straight lines commensurable in square only, and therefore HN is medial. [X. 21]  Therefore HN is either rational or medial.  But HN is equal to AC;  therefore AC is either rational or medial. 
                                                                 
                                                                 
Therefore etc. 
 
 
PROPOSITION 26. 
 
 
A medial area does not exceed a medial area by a rational area. 
 
 
For, if possible, let the medial area AB exceed the medial area AC by the rational area DB, and let a rational straight line EF be set out; to EF let there be applied the rectangular parallelogram FH equal to AB, producing EH as breadth, and let the rectangle FG equal to AC be subtracted;  therefore the remainder BD is equal to the remainder KH. 
   
   
But DB is rational;  therefore KH is also rational.  Since, then, each of the rectangles AB, AC is medial, and AB is equal to FH, and AC to FG, therefore each of the rectangles FH, FG is also medial.  And they are applied to the rational straight line EF;  therefore each of the straight lines HE, EG is rational and incommensurable in length with EF. [X. 22]  And, since [DB is rational and is equal to KH, therefore] KH is [also] rational;  and it is applied to the rational straight line EF;  therefore GH is rational and commensurable in length with EF. [X. 20]  But EG is also rational, and is incommensurable in length with EF;  therefore EG is incommensurable in length with GH. [X. 13]  And, as EG is to GH, so is the square on EG to the rectangle EG, GH;  therefore the square on EG is incommensurable with the rectangle EG, GH. [X. 11]  But the squares on EG, GH are commensurable with the square on EG, for both are rational;  and twice the rectangle EG, GH is commensurable with the rectangle EG, GH, for it is double of it; [X. 6]  therefore the squares on EG, GH are incommensurable with twice the rectangle EG, GH; [X. 13]  therefore also the sum of the squares on EG, GH and twice the rectangle EG, GH,  that is, the square on EH [II. 4], is incommensurable with the squares on EG, GH. [X. 16]  But the squares on EG, GH are rational;  therefore the square on EH is irrational. [X. Def. 4]  Therefore EH is irrational.  But it is also rational: which is impossible. 
                                         
                                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 27. 
 
 
To find medial straight lines commensurable in square only which contain a rational rectangle. 
 
 
Let two rational straight lines A, B commensurable in square only be set out;  let C be taken a mean proportional between A, B, [VI. 13]  and let it be contrived that, as A is to B, so is C to D. [VI. 12] 
     
     
Then, since A, B are rational and commensurable in square only,  the rectangle A, B, that is, the square on C [VI.17], is medial. [X. 21]  Therefore C is medial. [X. 21]  And since, as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only. [X. 11]  And C is medial;  therefore D is also medial. [X. 23, addition] 
           
           
Therefore C, D are medial and commensurable in square only.  I say that they also contain a rational rectangle. 
   
   
For since, as A is to B, so is C to D,  therefore, alternately, as A is to C, so is B to D. [V. 16]  But, as A is to C, so is C to B;  therefore also, as C is to B, so is B to D;  therefore the rectangle C, D is equal to the square on B.  But the square on B is rational;  therefore the rectangle C, D is also rational. 
             
             
Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle.  Q. E. D. 
   
   
PROPOSITION 28. 
 
 
To find medial straight lines commensurable in square only which contain a medial rectangle. 
 
 
Let the rational straight lines A, B, C commensurable in square only be set out;  let D be taken a mean proportional between A, B, [VI. 13]  and let it be contrived that, as B is to C, so is D to E. [VI. 12] 
     
     
Since A, B are rational straight lines commensurable in square only,  therefore the rectangle A, B, that is, the square on D [VI. 17], is medial. [X. 21]  Therefore D is medial. [X. 21]  And since B, C are commensurable in square only, and, as B is to C, so is D to E,  therefore D, E are also commensurable in square only. [X. 11]  But D is medial;  therefore E is also medial. [X. 23, addition]  Therefore D, E are medial straight lines commensurable in square only.  I say next that they also contain a medial rectangle. 
                 
                 
For since, as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E. [V. 16]  But, as B is to D, so is D to A;  therefore also, as D is to A, so is C to E;  therefore the rectangle A, C is equal to the rectangle D, E. [VI. 16]  But the rectangle A, C is medial; [X. 21]  therefore the rectangle D, E is also medial. 
           
           
Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle.  Q. E. D. 
   
   
LEMMA I.
To find two square numbers such that their sum is also square. 
Let two numbers AB, BC be set out, and let them be either both even or both odd.  Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [IX. 24, 26] therefore the remainder AC is even.  Let AC be bisected at D.  Let AB, BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers.  Now the product of AB, BC together with the square on CD is equal to the square on BD. [II. 6]  And the product of AB, BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [IX. 1]  Therefore two square numbers, the product of AB, BC, and the square on CD, have been found which, when added together, make the square on BD. 
               
               
And it is manifest that two square numbers, the square on BD and the square on CD, have again been found such that their difference, the product of AB, BC, is a square, whenever AB, BC are similar plane numbers.  But when they are not similar plane numbers, two square numbers, the square on BD and the square on DC, have been found such that their difference, the product of AB, BC, is not square.  Q. E. D. 
     
     
LEMMA 2.
To find two square numbers such that their sum is not square. 
 
 
For let the product of AB, BC, as we said, be square, and CA even, and let CA be bisected by D.  It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See Lemma 1]  Let the unit DE be subtracted;  therefore the product of AB, BC together with the square on CE is less than the square on BD.  I say then that the square product of AB, BC together with the square on CE will not be square. 
         
         
For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the unit be divided.  First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE.  Since then the whole AC is double of the whole CD, and in them AG is double of DE, therefore the remainder GC is also double of the remainder EC;  therefore GC is bisected by E.  Therefore the product of GB, BC together with the square on CE is equal to the square on BE. [II. 6]  But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE;  therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE.  And, if the common square on CE be subtracted, it follows that AB is equal to GB: which is absurd. 
               
               
Therefore the product of AB, BC together with the square on CE is not equal to the square on BE.  I say next that neither is it less than the square on BE.  For, if possible, let it be equal to the square on BF, and let HA be double of DF.  Now it will again follow that HC is double of CF;  so that CH has also been bisected at F, and for this reason the product of HB, BC together with the square on FC is equal to the square on BF. [II. 6]  But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BF.  Thus the product of HB, BC together with the square on CF will also be equal to the product of AB, BC together with the square on CE:  which is absurd.  Therefore the product of AB, BC together with the square on CE is not less than the square on BE.  And it was proved that neither is it equal to the square on BE.  Therefore the product of AB, BC together with the square on CE is not square.  Q. E. D. 
                       
                       
PROPOSITION 29. 
 
 
To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. 
 
 
For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1]  let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.] Let FB be joined. 
   
   
Since, as the square on BA is to the square on AF, so is DC to CE,  therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE;  therefore the square on BA is commensurable with the square on AF. [X. 6]  But the square on AB is rational; [X. Def. 4]  therefore the square on AF is also rational; [id.]  therefore AF is also rational.  And, since DC has not to CE the ratio which a square number has to a square number,  neither has the square on BA to the square on AF the ratio which a square number has to a square number;  therefore AB is incommensurable in length with AF. [X. 9]  Therefore BA, AF are rational straight lines commensurable in square only.  And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]  But CD has to DE the ratio which a square number has to a square number:  therefore also the square on AB has to the square on BF the ratio which a square number has to a square number;  therefore AB is commensurable in length with BF. [X. 9]  And the square on AB is equal to the squares on AF, FB;  therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB. 
                               
                               
Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.   
   
   
PROPOSITION 30. 
 
 
To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater. 
 
 
Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not square; [Lemma 2]  let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, [X. 6, Por.] and let FB be joined. 
   
   
Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only.  And since, as DC is to CE, so is the square on BA to the square on AF,  therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47]  But CD has not to DE the ratio which a square number has to a square number;  therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number;  therefore AB is incommensurable in length with BF. [X. 9]  And the square on AB is greater than the square on AF by the square on FB incommensurable with AB. 
             
             
Therefore AB, AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB.  Q. E. D. 
   
   
PROPOSITION 31. 
 
 
To find two medial straight lines commensurable in square only, containing a rational rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. 
 
 
Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29]  And let the square on C be equal to the rectangle A, B.  Now the rectangle A, B is medial; [X. 21]  therefore the square on C is also medial;  therefore C is also medial. [X. 21]  Let the rectangle C, D be equal to the square on B.  Now the square on B is rational;  therefore the rectangle C, D is also rational.  And since, as A is to B, so is the rectangle A, B to the square on B,  while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B,  therefore, as A is to B, so is the square on C to the rectangle C, D.  But, as the square on C is to the rectangle C, D, so is C to D;  therefore also, as A is to B, so is C to D.  But A is commensurable with B in square only;  therefore C is also commensurable with D in square only. [X. 11]  And C is medial;  therefore D is also medial. [X. 23, addition]  And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A,  therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [X. 14] 
                                     
                                     
Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C. 
 
 
Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C,  when the square on A is greater than the square on B by the square on a straight line incommensurable with A. [X. 30] 
   
   
PROPOSITION 32. 
 
 
To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. 
 
 
Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on A is greater than the square on C by the square on a straight line commensurable with A, [X. 29]  and let the square on D be equal to the rectangle A, B.  Therefore the square on D is medial;  therefore D is also medial. [X. 21]  Let the rectangle D, E be equal to the rectangle B, C.  Then since, as the rectangle A, B is to the rectangle B, C, so is A to C;  while the square on D is equal to the rectangle A, B,  and the rectangle D, E is equal to the rectangle B, C,  therefore, as A is to C, so is the square on D to the rectangle D, E.  But, as the square on D is to the rectangle D, E, so is D to E;  therefore also, as A is to C, so is D to E.  But A is commensurable with C in square only;  therefore D is also commensurable with E in square only. [X. 11]  But D is medial;  therefore E is also medial. [X. 23, addition]  And, since, as A is to C, so is D to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A,  therefore also the square on D will be greater than the square on E by the square on a straight line commensurable with D.[X. 14]  I say next that the rectangle D, E is also medial. 
                                   
                                   
For, since the rectangle B, C is equal to the rectangle D, E, while the rectangle B, C is medial, [X. 21]  therefore the rectangle D, E is also medial. 
   
   
Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. 
 
 
Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [X. 30] 
 
 
LEMMA.
Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn; 
I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC. 
   
   
And first that the rectangle CB, BD is equal to the square on BA. 
 
 
For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base,  therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8]  And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4]  therefore the rectangle CB, BD is equal to the square on AB. [VI. 17] 
       
       
For the same reason the rectangle BC, CD is also equal to the square on AC. 
 
 
And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base,  the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.]  therefore, as BD is to DA, so is AD to DC;  therefore the rectangle BD, DC is equal to the square on AD. [VI. 17]  I say that the rectangle BC, AD is also equal to the rectangle BA, AC. 
         
         
For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI. 4]  Therefore the rectangle BC, AD is equal to the rectangle BA, AC. [VI. 16]  Q. E. D. 
     
     
PROPOSITION 33. 
 
 
To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial. 
 
 
Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30] let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28]  let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined. 
   
   
Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB,  while there has been applied to AB a parallelogram equal to the fourth part of the square on BC,  that is, to the square on half of it,  and deficient by a square figure, making the rectangle AE, EB,  therefore AE is incommensurable with EB. [X. 18]  And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE,  while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF;  therefore the square on AF is incommensurable with the square on FB;  therefore AF, FB are incommensurable in square.  And, since AB is rational, therefore the square on AB is also rational;  so that the sum of the squares on AF, FB is also rational. [I. 47]  And since, again, the rectangle AE, EB is equal to the square on EF,  and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD;  therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF.  But the rectangle AB, BC is medial; [X. 21]  therefore the rectangle AB, EF is also medial. [X. 23, Por.]  But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma]  therefore the rectangle AF, FB is also medial.  But it was also proved that the sum of the squares on these straight lines is rational. 
                                     
                                     
Therefore two straight lines AF, FB incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial.  Q. E. D. 
   
   
PROPOSITION 34. 
 
 
To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational. 
 
 
Let there be set out two medial straight lines AB, BC, commensurable in square only, such that the rectangle which they contain is rational, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 31, ad fin.]  let the semicircle ADB be described on AB, let BC be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28]  therefore AF is incommensurable in length with FB. [X. 18]  Let FD be drawn from F at right angles to AB, and let AD, DB be joined. 
       
       
Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [X. 11]  But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to the square on DB;  therefore the square on AD is also incommensurable with the square on DB.  And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31, I. 47]  And, since BC is double of DF, therefore the rectangle AB, BC is also double of the rectangle AB, FD.  But the rectangle AB, BC is rational;  therefore the rectangle AB, FD is also rational. [X. 6]  But the rectangle AB, FD is equal to the rectangle AD, DB; [Lemma]  so that the rectangle AD, DB is also rational. 
                 
                 
Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial, but the rectangle contained by them rational.  Q. E. D. 
   
   
PROPOSITION 35. 
 
 
To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. 
 
 
Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB; [X. 32 , ad fin.]  let the semicircle ADB be described on AB, and let the rest of the construction be as above. 
   
   
Then, since AF is incommensurable in length with FB, [X. 18 ] AD is also incommensurable in square with DB. [X. 11 ]  And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. [III. 31 , I. 47 ]  And, since the rectangle AF, FB is equal to the square on each of the straight lines BE, DF, therefore BE is equal to DF;  therefore BC is double of FD, so that the rectangle AB, BC is also double of the rectangle AB, FD.  But the rectangle AB, BC is medial; therefore the rectangle AB, FD is also medial. [X. 32, Por.]  And it is equal to the rectangle AD, DB; [Lemma after X. 32 ] therefore the rectangle AD, DB is also medial.  And, since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, [X. 13 ]  so that the square on AB is also incommensurable with the rectangle AB, BE. [X. 11 ]  But the squares on AD, DB are equal to the square on AB, [I. 47 ] and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE;  therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB. 
                   
                   
Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.  Q. E. D. 
   
   
PROPOSITION 36. 
 
 
If two rational straight lines commensurable in square only be added together, the whole is irrational; and let it be called binomial. 
 
 
For let two rational straight lines AB, BC commensurable in square only be added together;  I say that the whole AC is irrational. 
   
   
For, since AB is incommensurable in length with BC — for they are commensurable in square only — and, as AB is to BC, so is the rectangle AB, BC to the square on BC,  therefore the rectangle AB, BC is incommensurable with the square on BC. [X. 11 ]  But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [X. 6 ],  and the squares on AB, BC are commensurable with the square on BC — for AB, BC are rational straight lines commensurable in square only — [X. 15 ]  therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13 ]  And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [II. 4 ], is incommensurable with the sum of the squares on AB, BC. [X. 16 ]  But the sum of the squares on AB, BC is rational;  therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ]  And let it be called binomial.   
                   
                   
PROPOSITION 37. 
 
 
If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational; and let it be called a first bimedial straight line. 
 
 
For let two medial straight lines AB, BC commensurable in square only and containing a rational rectangle be added together;  I say that the whole AC is irrational. 
   
   
For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC; [cf. X. 36, ll. 9-20]  and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on AC [II. 4], is incommensurable with the rectangle AB, BC. [X. 16 ]  But the rectangle AB, BC is rational,  for, by hypothesis, AB, BC are straight lines containing a rational rectangle;  therefore the square on AC is irrational;  therefore AC is irrational. [X. Def. 4 ] And let it be called a first bimedial straight line.  Q. E. D. 
             
             
PROPOSITION 38. 
 
 
If two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a second bimedial straight line. 
 
 
For let two medial straight lines AB, BC commensurable in square only and containing a medial rectangle be added together;  I say that AC is irrational. 
   
   
For let a rational straight line DE be set out, and let the parallelogram DF equal to the square on AC be applied to DE, producing DG as breadth. [I. 44 ]  Then, since the square on AC is equal to the squares on AB, BC and twice the rectangle AB, BC, [II. 4 ] let EH, equal to the squares on AB, BC, be applied to DE;  therefore the remainder HF is equal to twice the rectangle AB, BC.  And, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial.  But, by hypothesis, twice the rectangle AB, BC is also medial.  And EH is equal to the squares on AB, BC, while FH is equal to twice the rectangle AB, BC;  therefore each of the rectangle EH, HF is medial.  And they are applied to the rational straight line DE;  therefore each of the straight lines DH, HG is rational and incommensurable in length with DE. [X. 22 ]  Since then AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC,  therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11 ]  But the sum of the squares on AB, BC is commensurable with the square on AB, [X. 15 ] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6 ]  Therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC. [X. 13 ]  But EH is equal to the squares on AB, BC, and HF is equal to twice the rectangle AB, BC.  Therefore EH is incommensurable with HF, so that DH is also incommensurable in length with HG. [VI. 1 , X. 11 ]  Therefore DH, HG are rational straight lines commensurable in square only;  so that DG is irrational. [X. 36 ]  But DE is rational; and the rectangle contained by an irrational and a rational straight line is irrational; [cf. X. 20 ]  therefore the area DF is irrational, and the side of the square equal to it is irrational. [X. Def. 4 ]  But AC is the side of the square equal to DF; therefore AC is irrational.  And let it be called a second bimedial straight line.  Q. E. D. 
                                           
                                           
PROPOSITION 39. 
 
 
If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational: and let it be called major. 
 
 
For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 33 ], be added together;  I say that AC is irrational. 
   
   
For, since the rectangle AB, BC is medial, twice the rectangle AB, BC is also medial. [X. 6 and 23, Por.]  But the sum of the squares on AB, BC is rational;  therefore twice the rectangle AB, BC is incommensurable with the sum of the squares on AB, BC,  so that the squares on AB, BC together with twice the rectangle AB, BC that is, the square on AC, is also incommensurable with the sum of the squares on AB, BC; [X. 16 ]  therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ]  And let it be called major.  Q. E. D. 
             
             
PROPOSITION 40. 
 
 
If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the side of a rational plus a medial area. 
 
 
For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34 ], be added together;  I say that AC is irrational. 
   
   
For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC;  so that the square on AC is also incommensurable with twice the rectangle AB, BC. [X. 16 ]  But twice the rectangle AB, BC is rational;  therefore the square on AC is irrational.  Therefore AC is irrational. [X. Def. 4 ]  And let it be called the side of a rational plus a medial area.  Q. E. D. 
             
             
PROPOSITION 41. 
 
 
If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the side of the sum of two medial areas. 
 
 
For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [X. 35 ] be added together;  I say that AC is irrational. 
   
   
Let a rational straight line DE be set out, and let there be applied to DE the rectangle DF equal to the squares on AB, BC, and the rectangle GH equal to twice the rectangle AB, BC;  therefore the whole DH is equal to the square on AC. [II. 4 ]  Now, since the sum of the squares on AB, BC is medial, and is equal to DF, therefore DF is also medial.  And it is applied to the rational straight line DE;  therefore DG is rational and incommensurable in length with DE. [X. 22 ]  For the same reason GK is also rational and incommensurable in length with GF, that is, DE.  And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, DF is incommensurable with GH;  so that DG is also incommensurable with GK. [VI. 1 , X. 11 ]  And they are rational;  therefore DG, GK are rational straight lines commensurable in square only;  therefore DK is irrational and what is called binomial. [X. 36 ]  But DE is rational;  therefore DH is irrational, and the side of the square which is equal to it is irrational. [X. Def. 4 ]  But AC is the side of the square equal to HD;  therefore AC is irrational.  And let it be called the side of the sum of two medial areas.  Q. E. D. 
                                 
                                 
LEMMA.
And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question, we will now prove after premising the following lemma. 
 
 
Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D, and let AC be supposed greater than DB;  I say that the squares on AC, CB are greater than the squares on AD, DB. 
   
   
For let AB be bisected at E.  Then, since AC is greater than DB, let DC be subtracted from each;  therefore the remainder AD is greater than the remainder CB.  But AE is equal to EB;  therefore DE is less than EC;  therefore the points C, D are not equidistant from the point of bisection.  And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5 ]  and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.]  therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE.  And of these the square on DE is less than the square on EC;  therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB,  so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.  Therefore also the remainder, the sum of the squares on AC, CB, is greater than the sum of the squares on AD, DB.  Q. E. D. 
                           
                           
PROPOSITION 42. 
 
 
A binomial straight line is divided into its terms at one point only. 
 
 
Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36 ]  I say that AB is not divided at another point into two rational straight lines commensurable in square only. 
   
   
For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only.  It is then manifest that AC is not the same with DB.  For, if possible, let it be so.  Then AD will also be the same as CB, and, as AC is to CB, so will BD be to DA;  thus AB will be divided at D also in the same way as by the division at C:  which is contrary to the hypothesis.  Therefore AC is not the same with DB.  For this reason also the points C, D are not equidistant from the point of bisection.  Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [II. 4 ]  But the squares on AC, CB differ from the squares on AD, DB by a rational area, for both are rational;  therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [X. 21 ]:  which is absurd, for a medial area does not exceed a medial by a rational area. [x. 26 ] 
                       
                       
Therefore a binomial straight line is not divided at different points; therefore it is divided at one point only.  Q. E. D. 
   
   
PROPOSITION 43. 
 
 
A first bimedial straight line is divided at one point only. 
 
 
Let AB be a first bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a rational rectangle;  I say that AB is not so divided at another point. 
   
   
For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle.  Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB, while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area — for both are rational —  therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial: which is absurd. [x. 26 ] 
     
     
Therefore a first bimedial straight line is not divided into its terms at different points; therefore it is so divided at one point only.   
   
   
PROPOSITION 44. 
 
 
A second bimedial straight line is divided at one point only. 
 
 
Let AB be a second bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a medial rectangle; [X. 38 ] it is then manifest that C is not at the point of bisection, because the segments are not commensurable in length.  I say that AB is not so divided at another point. 
   
   
For, if possible, let it be divided at D also, so that AC is not the same with DB, but AC is supposed greater;  it is then clear that the squares on AD, DB are also, as we proved above [Lemma], less than the squares on AC, CB;  and suppose that AD, DB are medial straight lines commensurable in square only and containing a medial rectangle.  Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB, and let EG equal to the squares on AC, CB be subtracted;  therefore the remainder HK is equal to twice the rectangle AC, CB. [II. 4 ]  Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma ];  therefore the remainder MK is also equal to twice the rectangle AD, DB.  Now, since the squares on AC, CB are medial, therefore EG is medial.  And it is applied to the rational straight line EF;  therefore EH is rational and incommensurable in length with EF. [X. 22 ]  For the same reason HN is also rational and incommensurable in length with EF.  And, since AC, CB are medial straight lines commensurable in square only, therefore AC is incommensurable in length with CB.  But, as AC is to CB, so is the square on AC to the rectangle AC, CB;  therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11 ]  But the squares on AC, CB are commensurable with the square on AC;  for AC, CB are commensurable in square. [x. 15 ]  And twice the rectangle AC, CB is commensurable with the rectangle AC, CB. [X. 6 ]  Therefore the squares on AC, CB are also incommensurable with twice the rectangle AC, CB. [X. 13 ]  But EG is equal to the squares on AC, CB, and HK is equal to twice the rectangle AC, CB;  therefore EG is incommensurable with HK, so that EH is also incommensurable in length with HN. [VI. 1 , X. 11 ]  And they are rational;  therefore EH, HN are rational straight lines commensurable in square only.  But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [X. 36 ]  Therefore EN is a binomial straight line divided at H.  In the same way EM, MN will also be proved to be rational straight lines commensurable in square only;  and EN will be a binomial straight line divided at different points, H and M.  And EH is not the same with MN.  For the squares on AC, CB are greater than the squares on AD, DB.  But the squares on AD, DB are greater than twice the rectangle AD, DB;  therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB,  that is, MK, so that EH is also greater than MN.  Therefore EH is not the same with MN.  Q. E. D. 
                                                                 
                                                                 
PROPOSITION 45. 
 
 
A major straight line is divided at one and the same point only. 
 
 
Let AB be a major straight line divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB rational, but the rectangle AC, CB medial; [X. 39 ]  I say that AB is not so divided at another point. 
   
   
For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle contained by them medial.  Then, since that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB,  while the squares on AC, CB exceed the squares on AD, DB by a rational area — for both are rational —  therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial: which is impossible. [X. 26 ]  Therefore a major straight line is not divided at different points;  therefore it is only divided at one and the same point.  Q. E. D. 
             
             
PROPOSITION 46. 
 
 
The side of a rational plus a medial area is divided at one point only. 
 
 
Let AB be the side of a rational plus a medial area divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, but twice the rectangle AC, CB rational; [X. 40 ]  I say that AB is not so divided at another point. 
   
   
For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB medial, but twice the rectangle AD, DB rational.  Since then that by which twice the rectangle AC, CB differs from twice the rectangle AD, DB is also that by which the squares on AD, DB differ from the squares on AC, CB, while twice the rectangle AC, CB exceeds twice the rectangle AD, DB by a rational area, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area, though they are medial:  which is impossible. [X. 26 ]  Therefore the side of a rational plus a medial area is not divided at different points;  therefore it is divided at one point only.  Q. E. D. 
           
           
PROPOSITION 47. 
 
 
The side of the sum of two medial areas is divided at one point only. 
 
 
Let AB be divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB medial, and the rectangle AC, CB medial and also incommensurable with the sum of the squares on them;  I say that AB is not divided at another point so as to fulfil the given conditions. 
   
   
For, if possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC, CB, and the rectangle HK equal to twice the rectangle AC, CB;  therefore the whole EK is equal to the square on AB. [II. 4 ]  Again, let EL, equal to the squares on AD, DB, be applied to EF;  therefore the remainder, twice the rectangle AD, DB, is equal to the remainder MK.  And since, by hypothesis, the sum of the squares on AC, CB is medial, therefore EG is also medial.  And it is applied to the rational straight line EF;  therefore HE is rational and incommensurable in length with EF. [X. 22 ]  For the same reason HN is also rational and incommensurable in length with EF.  And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN,  so that EH is also incommensurable with HN. [VI. 1 , X. 11 ]  And they are rational;  therefore EH, HN are rational straight lines commensurable in square only;  therefore EN is a binomial straight line divided at H. [X. 36 ]  Similarly we can prove that it is also divided at M.  And EH is not the same with MN;  therefore a binomial has been divided at different points:  which is absurd. [X. 42 ]  Therefore a side of the sum of two medial areas is not divided at different points;  therefore it is divided at one point only. 
                                     
                                     
DEFINITIONS II. 
 
 
1. Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line; 
 
 
2. but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial; 
 
 
3. and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial. 
 
 
4. Again, if the square on the greater term be greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a fourth binomial; 
 
 
5. if the lesser, a fifth binomial; 
 
 
6. and if neither, a sixth binomial. 
 
 
PROPOSITION 48. 
 
 
To find the first binomial straight line. 
 
 
Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to CA the ratio which a square number has to a square number; [Lemma I after X. 28]  let any rational straight line D be set out, and let EF be commensurable in length with D.  Therefore EF is also rational.  Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG. [X. 6, Por.]  But AB has to AC the ratio which a number has to a number;  therefore the square on EF also has to the square on FG the ratio which a number has to a number,  so that the square on EF is commensurable with the square on FG. [X. 6]  And EF is rational;  therefore FG is also rational.  And, since BA has not to AC the ratio which a square number has to a square number,  neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number;  therefore EF is incommensurable in length with FG. [X. 9]  Therefore EF, FG are rational straight lines commensurable in square only;  therefore EG is binomial. [X. 36]  I say that it is also a first binomial straight line. 
                             
                             
For since, as the number BA is to AC, so is the square on EF to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG.  Let then the squares on FG, H be equal to the square on EF.  Now since, as BA is to AC, so is the square on EF to the square on FG,  therefore, convertendo, as AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]  But AB has to BC the ratio which a square number has to a square number;  therefore the square on EF also has to the square on H the ratio which a square number has to a square number.  Therefore EF is commensurable in length with H; [X. 9]  therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF.  And EF, FG are rational, and EF is commensurable in length with D. 
                 
                 
Therefore EF is a first binomial straight line.  Q. E. D. 
   
   
PROPOSITION 49. 
 
 
To find the second binomial straight line. 
 
 
Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number;  let a rational straight line D be set out, and let EF be commensurable in length with D;  therefore EF is rational.  Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [X. 6, Por.]  therefore the square on EF is commensurable with the square on FG. [X. 6]  Therefore FG is also rational.  Now, since the number CA has not to AB the ratio which a square number has to a square number,  neither has the square on EF to the square on FG the ratio which a square number has to a square number.  Therefore EF is incommensurable in length with FG; [X. 9]  therefore EF, FG are rational straight lines commensurable in square only;  therefore EG is binomial. [X. 36]  It is next to be proved that it is also a second binomial straight line. 
                       
                       
For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE,  while BA is greater than AC, therefore the square on GF is greater than the square on FE.  Let the squares on EF, H be equal to the square on GF;  therefore, convertendo, as AB is to BC, so is the square on FG to the square on H. [V. 19, Por.]  But AB has to BC the ratio which a square number has to a square number;  therefore the square on FG also has to the square on H the ratio which a square number has to a square number.  Therefore FG is commensurable in length with H; [X. 9]  so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG.  And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out. 
                 
                 
Therefore EG is a second binomial straight line.  Q. E. D. 
   
   
PROPOSITION 50. 
 
 
To find the third binomial straight line. 
 
 
Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number.  Let any other number D, not square, be set out also, and let it not have to either of the numbers BA, AC the ratio which a square number has to a square number.  Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.]  therefore the square on E is commensurable with the square on FG. [X. 6]  And E is rational;  therefore FG is also rational.  And, since D has not to AB the ratio which a square number has to a square number,  neither has the square on E to the square on FG the ratio which a square number has to a square number;  therefore E is incommensurable in length with FG. [X. 9]  Next let it be contrived that, as the number BA is to AC, so is the square on FG to the square on GH; [X. 6, Por.]  therefore the square on FG is commensurable with the square on GH. [X. 6]  But FG is rational;  therefore GH is also rational.  And, since BA has not to AC the ratio which a square number has to a square number,  neither has the square on FG to the square on HG the ratio which a square number has to a square number;  therefore FG is incommensurable in length with GH. [X. 9]  Therefore FG, GH are rational straight lines commensurable in square only;  therefore FH is binomial. [X. 36]  I say next that it is also a third binomial straight line. 
                                     
                                     
For since, as D is to AB, so is the square on E to the square on FG,  and, as BA is to AC, so is the square on FG to the square on GH,  therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]  But D has not to AC the ratio which a square number has to a square number;  therefore neither has the square on E to the square on GH the ratio which a square number has to a square number;  therefore E is incommensurable in length with GH. [X. 9]  And since, as BA is to AC, so is the square on FG to the square on GH,  therefore the square on FG is greater than the square on GH.  Let then the squares on GH, K be equal to the square on FG;  therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]  But AB has to BC the ratio which a square number has to a square number;  therefore the square on FG also has to the square on K the ratio which a square number has to a square number;  therefore FG is commensurable in length with K. [X. 9]  Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.  And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E. 
                             
                             
Therefore FH is a third binomial straight line.  Q. E. D. 
   
   
PROPOSITION 51. 
 
 
To find the fourth binomial straight line. 
 
 
Let two numbers AC, CB be set out such that AB neither has to BC, nor yet to AC, the ratio which a square number has to a square number.  Let a rational straight line D be set out, and let EF be commensurable in length with D;  therefore EF is also rational.  Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG; [X. 6, Por.]  therefore the square on EF is commensurable with the square on FG; [X. 6]  therefore FG is also rational.  Now, since BA has not to AC the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number;  therefore EF is incommensurable in length with FG. [X. 9]  Therefore EF, FG are rational straight lines commensurable in square only;  so that EG is binomial.  I say next that it is also a fourth binomial straight line. 
                     
                     
For since, as BA is to AC, so is the square on EF to the square on FG,  therefore the square on EF is greater than the square on FG.  Let then the squares on FG, H be equal to the square on EF;  therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19, Por.]  But AB has not to BC the ratio which a square number has to a square number;  therefore neither has the square on EF to the square on H the ratio which a square number has to a square number.  Therefore EF is incommensurable in length with H; [X. 9]  therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF.  And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D. 
                 
                 
Therefore EG is a fourth binomial straight line.  Q. E. D. 
   
   
PROPOSITION 52. 
 
 
To find the fifth binomial straight line. 
 
 
Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number;  let any rational straight line D be set out, and let EF be commensurable with D;  therefore EF is rational.  Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [X. 6, Por.]  But CA has not to AB the ratio which a square number has to a square number;  therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number.  Therefore EF, FG are rational straight lines commensurable in square only; [X. 9]  therefore EG is binomial. [X. 36]  I say next that it is also a fifth binomial straight line. 
                 
                 
For since, as CA is to AB, so is the square on EF to the square on FG,  inversely, as BA is to AC, so is the square on FG to the square on FE;  therefore the square on GF is greater than the square on FE.  Let then the squares on EF, H be equal to the square on GF;  therefore, convertendo, as the number AB is to BC, so is the square on GF to the square on H. [V. 19, Por.]  But AB has not to BC the ratio which a square number has to a square number;  therefore neither has the square on FG to the square on H the ratio which a square number has to a square number.  Therefore FG is incommensurable in length with H; [X. 9]  so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG.  And GF, FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out. 
                   
                   
Therefore EG is a fifth binomial straight line.  Q. E. D. 
   
   
PROPOSITION 53. 
 
 
To find the sixth binomial straight line. 
 
 
Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number;  and let there also be another number D which is not square and which has not to either of the numbers BA, AC the ratio which a square number has to a square number.  Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.]  therefore the square on E is commensurable with the square on FG. [X. 6]  And E is rational; therefore FG is also rational.  Now, since D has not to AB the ratio which a square number has to a square number,  neither has the square on E to the square on FG the ratio which a square number has to a square number;  therefore E is incommensurable in length with FG. [X. 9]  Again, let it be contrived that, as BA is to AC, so is the square on FG to the square on GH. [X. 6, Por.]  Therefore the square on FG is commensurable with the square on HG. [X. 6]  Therefore the square on HG is rational; therefore HG is rational.  And, since BA has not to AC the ratio which a square number has to a square number,  neither has the square on FG to the square on GH the ratio which a square number has to a square number;  therefore FG is incommensurable in length with GH. [X. 9]  Therefore FG, GH are rational straight lines commensurable in square only;  therefore FH is binomial. [X. 36]  It is next to be proved that it is also a sixth binomial straight line. 
                                 
                                 
For since, as D is to AB, so is the square on E to the square on FG, and also, as BA is to AC, so is the square on FG to the square on GH,  therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22]  But D has not to AC the ratio which a square number has to a square number;  therefore neither has the square on E to the square on GH the ratio which a square number has to a square number;  therefore E is incommensurable in length with GH. [X. 9]  But it was also proved incommensurable with FG;  therefore each of the straight lines FG, GH is incommensurable in length with E.  And, since, as BA is to AC, so is the square on FG to the square on GH,  therefore the square on FG is greater than the square on GH.  Let then the squares on GH, K be equal to the square on FG;  therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]  But AB has not to BC the ratio which a square number has to a square number;  so that neither has the square on FG to the square on K the ratio which a square number has to a square number.  Therefore FG is incommensurable in length with K; [X. 9]  therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG.  And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out. 
                               
                               
Therefore FH is a sixth binomial straight line.  Q. E. D. 
   
   
LEMMA.
Let there be two squares AB, BC, and let them be placed so that DB is in a straight line with BE; 
therefore FB is also in a straight line with BG. 
   
   
Let the parallelogram AC be completed;  I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB. 
   
   
For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG.  But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34]  therefore each of the straight lines AH, KC is also equal to each of the straight lines AK, HC.  Therefore the parallelogram AC is equilateral.  And it is also rectangular;  therefore AC is a square. 
           
           
And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC, [VI. 1]  therefore also, as AB is to DG, so is DG to BC. [V. 11]  Therefore DG is a mean proportional between AB, BC.  I say next that DC is also a mean proportional between AC, CB. 
       
       
For since, as AD is to DK, so is KG to GC — for they are equal respectively —  and, componendo, as AK is to KD, so is KC to CG, [V. 18]  while, as AK is to KD, so is AC to CD, and, as KC is to CG, so is DC to CB, [VI. 1]  therefore also, as AC is to DC, so is DC to BC. [V. 11]  Therefore DC is a mean proportional between AC, CB.  Being what it was proposed to prove. 
           
           
PROPOSITION 54. 
 
 
If an area be contained by a rational straight line and the first binomial, the side of the area is the irrational straight line which is called binomial. 
 
 
For let the area AC be contained by the rational straight line AB and the first binomial AD;  I say that the “side” of the area AC is the irrational straight line which is called binomial. 
   
   
For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.  It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE,  and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1]  Let ED be bisected at the point F.  Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE,  therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less,  that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17]  Let then the rectangle AG, GE equal to the square on EF be applied to AE;  therefore AG is commensurable in length with EG.  Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD;  let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14]  and let them be placed so that MN is in a straight line with NO;  therefore RN is also in a straight line with NP.  And let the parallelogram SQ be completed;  therefore SQ is a square. [Lemma]  Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17]  therefore also, as AH is to EL, so is EL to KG; [VI. 1]  therefore EL is a mean proportional between AH, GK.  But AH is equal to SN, and GK to NQ;  therefore EL is a mean proportional between SN, NQ.  But MR is also a mean proportional between the same SN, NQ;  [Lemma] therefore EL is equal to MR, so that it is also equal to PO.  But AH, GK are also equal to SN, NQ;  therefore the whole AC is equal to the whole SQ, that is, to the square on MO;  therefore MO is the “side” of AC.  I say next that MO is binomial. 
                                                   
                                                   
For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG, GE. [X. 15]  But AE is also, by hypothesis, commensurable with AB;  therefore AG, GE are also commensurable with AB. [X. 12]  And AB is rational;  therefore each of the straight lines AG, GE is also rational;  therefore each of the rectangles AH, GK is rational, [X. 19] and AH is commensurable with GK.  But AH is equal to SN, and GK to NQ;  therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable.  And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF,  therefore AG is also incommensurable with EF, [X. 13]  so that AH is also incommensurable with EL. [VI. 1, X. 11]  But AH is equal to SN, and EL to MR;  therefore SN is also incommensurable with MR.  But, as SN is to MR, so is PN to NR; [VI. 1]  therefore PN is incommensurable with NR. [X. 11]  But PN is equal to MN, and NR to NO;  therefore MN is incommensurable with NO.  And the square on MN is commensurable with the square on NO, and each is rational;  therefore MN, NO are rational straight lines commensurable in square only. 
                                     
                                     
Therefore MO is binomial [X. 36] and the “side” of AC.  Q. E. D. 5 
   
   
PROPOSITION 55. 
 
 
If an area be contained by a rational straight line and the second binomial, the side of the area is the irrational straight line which is called a first bimedial. 
 
 
For let the area ABCD be contained by the rational straight line AB and the second binomial AD;  I say that the “side” of the area AC is a first bimedial straight line. 
   
   
For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term;  therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. [X. Deff. II. 2]  Let ED be bisected at F, and let there be applied to AE the rectangle AG, GE equal to the square on EF and deficient by a square figure;  therefore AG is commensurable in length with GE. [X. 17]  Through G, E, F let GH, EK, FL be drawn parallel to AB, CD, let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK,  and let them be placed so that MN is in a straight line with NO;  therefore RN is also in a straight line with NP.  Let the square SQ be completed.  It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the “side” of the area AC.  It is now to be proved that MO is a first bimedial straight line.  Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [X. 13]  And, since AG is commensurable with EG, AE is also commensurable with each of the straight lines AG, GE. [X. 15]  But AE is incommensurable in length with AB;  therefore AG, GE are also incommensurable with AB. [X. 13]  Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only;  so that each of the rectangles AH, GK is medial. [X. 21]  Hence each of the squares SN, NQ is medial.  Therefore MN, NO are also medial. 
                                   
                                   
And, since AG is commensurable in length with GE, AH is also commensurable with GK, [VI. 1. X. 11]  that is, SN is commensurable with NQ, that is, the square on MN with the square on NO.  And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF,  therefore AG is incommensurable with EF; [X. 13]  so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, [VI. 1, X. 11]  that is, MN is incommensurable in length with NO.  But MN, NO were proved to be both medial and commensurable in square;  therefore MN, NO are medial straight lines commensurable in square only.  I say next that they also contain a rational rectangle. 
                 
                 
For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, therefore EF is also commensurable with EK. [X. 12]  And each of them is rational; therefore EL, that is, MR is rational, [X. 19]  and MR is the rectangle MN, NO. 
     
     
But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line. [X. 37] 
 
 
Therefore MO is a first bimedial straight line.  Q. E. D. 6 
   
   
PROPOSITION 56. 
 
 
If an area be contained by a rational straight line and the third binomial, the side of the area is the irrational straight line called a second bimedial. 
 
 
For let the area ABCD be contained by the rational straight line AB and the third binomial AD divided into its terms at E, of which terms AE is the greater;  I say that the “side” of the area AC is the irrational straight line called a second bimedial. 
   
   
For let the same construction be made as before.  Now, since AD is a third binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and neither of the terms AE, ED is commensurable in length with AB. [X. Deff. II. 3]  Then, in manner similar to the foregoing, we shall prove that MO is the “side” of the area AC, and MN, NO are medial straight lines commensurable in square only;  so that MO is bimedial.  It is next to be proved that it is also a second bimedial straight line. 
         
         
Since DE is incommensurable in length with AB, that is, with EK, and DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]  And they are rational;  therefore FE, EK are rational straight lines commensurable in square only.  Therefore EL, that is, MR, is medial. [X. 21]  And it is contained by MN, NO;  therefore the rectangle MN, NO is medial. 
           
           
Therefore MO is a second bimedial straight line. [X. 38]  Q. E. D. 
   
   
PROPOSITION 57. 
 
 
If an area be contained by a rational straight line and the fourth binomial, the side of the area is the irrational straight line called major. 
 
 
For let the area AC be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AE be the greater;  I say that the “side” of the area AC is the irrational straight line called major. 
   
   
For, since AD is a fourth binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. [X. Deff. II. 4]  Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF;  therefore AG is incommensurable in length with GE. [X. 18]  Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before;  it is then manifest that MO is the “side” of the area AC.  It is next to be proved that MO is the irrational straight line called major. 
           
           
Since AG is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ; [VI. 1, X. 11]  therefore MN, NO are incommensurable in square.  And, since AE is commensurable with AB, AK is rational; [X. 19]  and it is equal to the squares on MN, NO;  therefore the sum of the squares on MN, NO is also rational.  And, since DE is incommensurable in length with AB, that is, with EK,  while DE is commensurable with EF, therefore EF is incommensurable in length with EK. [X. 13]  Therefore EK, EF are rational straight lines commensurable in square only;  therefore LE, that is, MR, is medial. [X. 21]  And it is contained by MN, NO;  therefore the rectangle MN, NO is medial.  And the [sum] of the squares on MN, NO is rational, and MN, NO are incommensurable in square.  But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole is irrational and is called major. [X. 39] 
                         
                         
Therefore MO is the irrational straight line called major and is the “side” of the area AC.  Q. E. D. 
   
   
PROPOSITION 58. 
 
 
If an area be contained by a rational straight line and the fifth binomial, the side of the area is the irrational straight line called the side of a rational plus a medial area. 
 
 
For let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that AE is the greater term;  I say that the “side” of the area AC is the irrational straight line called the side of a rational plus a medial area. 
   
   
For let the same construction be made as before shown;  it is then manifest that MO is the “side” of the area AC.  It is then to be proved that MO is the side of a rational plus a medial area. 
     
     
For, since AG is incommensurable with GE, [X. 18] therefore AH is also commensurable with HE, [VI. 1, X. 11] that is, the square on MN with the square on NO;  therefore MN, NO are incommensurable in square.  And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [X. Deff. II. 5]  But AE is incommensurable with ED;  therefore AB is also incommensurable in length with AE. [X. 13]  Therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21]  And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [X. 12]  And EK is rational;  therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. [X. 19]  Therefore MN, NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. 
                   
                   
Therefore MO is the side of a rational plus a medial area [X. 40] and is the “side” of the area AC.  Q. E. D. 
   
   
PROPOSITION 59. 
 
 
If an area be contained by a rational straight line and the sixth binomial, the side of the area is the irrational straight line called the side of the sum of two medial areas. 
 
 
For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term;  I say that the “side” of AC is the side of the sum of two medial areas. 
   
   
Let the same construction be made as before shown.  It is then manifest that MO is the “side” of AC, and that MN is incommensurable in square with NO.  Now, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only;  therefore AK, that is, the sum of the squares on MN, NO, is medial. [X. 21]  Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [X. 13]  therefore FE, EK are rational straight lines commensurable in square only;  therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. [X. 21]  And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. 1, X. 11]  But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO;  therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO.  And each of them is medial, and MN, NO are incommensurable in square. 
                     
                     
Therefore MO is the side of the sum of two medial areas [X. 41], and is the “side” of AC.  Q. E. D. 
   
   
[LEMMA.
If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts. 
 
 
Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater;  I say that the squares on AC, CB are greater than twice the rectangle AC, CB. 
   
   
For let AB be bisected at D.  Since then a straight line has been cut into equal parts at D, and into unequal parts at C,  therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5]  so that the rectangle AC, CB is less than double of the square on AD.  But the squares on AC, CB are double of the squares on AD, DC; [II. 9]  therefore the squares on AC, CB are greater than twice the rectangle AC, CB.  Q. E. D.] 
             
             
PROPOSITION 60. 
 
 
The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial. 
 
 
Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term; let a rational straight line DE be set out, and let DEFG equal to the square on AB be applied to DE producing DG as its breadth;  I say that DG is a first binomial straight line. 
   
   
For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC;  therefore the remainder, twice the rectangle AC, CB, is equal to MF.  Let MG be bisected at N, and let NO be drawn parallel [to ML or GF].  Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB.  Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36]  therefore the squares on AC, CB are rational and commensurable with one another,  so that the sum of the squares on AC, CB is also rational. [X. 15]  And it is equal to DL; therefore DL is rational.  And it is applied to the rational straight line DE;  therefore DM is rational and commensurable in length with DE. [X. 20]  Again, since AC, CB are rational straight lines commensurable in square only,  therefore twice the rectangle AC, CB, that is MF, is medial. [X. 21]  And it is applied to the rational straight line ML;  therefore MG is also rational and incommensurable in length with ML, that is, DE. [X. 22]  But MD is also rational and is commensurable in length with DE;  therefore DM is incommensurable in length with MG. [X. 13]  And they are rational;  therefore DM, MG are rational straight lines commensurable in square only;  therefore DG is binomial. [X. 36]  It is next to be proved that it is also a first binomial straight line. 
                                       
                                       
Since the rectangle AC, CB is a mean proportional between the squares on AC, CB, [cf. Lemma after X. 53] therefore MO is also a mean proportional between DH, KL.  Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [VI. 1]  therefore the rectangle DK, KM is equal to the square on MN. [VI. 17]  And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL,  so that DK is also commensurable with KM. [VI. 1, X. 11]  And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma]  therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]  And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM.  But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure,  and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [X. 17]  therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.  And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out. 
                       
                       
Therefore DG is a first binomial straight line. [X. Deff. II. 1]  Q. E. D. 
   
   
PROPOSITION 61. 
 
 
The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial. 
 
 
Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth;  I say that DG is a second binominal straight line. 
   
   
For let the same construction as before be made.  Then, since AB is a first bimedial divided at C,  therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37]  so that the squares on AC, CB are also medial. [X. 21]  Therefore DL is medial. [X. 15 and 23, Por.]  And it has been applied to the rational straight line DE;  therefore MD is rational and incommensurable in length with DE. [X. 22]  Again, since twice the rectangle AC, CB is rational, MF is also rational.  And it is applied to the rational straight line ML;  therefore MG is also rational and commensurable in length with ML, that is, DE; [X. 20]  therefore DM is incommensurable in length with MG. [X. 13]  And they are rational;  therefore DM, MG are rational straight lines commensurable in square only;  therefore DG is binomial. [X. 36]  It is next to be proved that it is also a second binomial straight line. 
                             
                             
For, since the squares on AC, CB are greater than twice the rectangle AC, CB,  therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]  And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL,  so that DK is also commensurable with KM. [VI. 1, X. 11]  And the rectangle DK, KM is equal to the square on MN;  therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. [X. 17]  And MG is commensurable is length with DE. 
             
             
Therefore DG is a second binomial straight line. [X. Deff. II. 2] 
 
 
PROPOSITION 62. 
 
 
The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial. 
 
 
Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth;  I say that DG is a third binomial straight line. 
   
   
Let the same construction be made as before shown.  Then, since AB is a second bimedial divided at C,  therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [X. 38]  so that the sum of the squares on AC, CB is also medial. [X. 15 and 23 Por.]  And it is equal to DL; therefore DL is also medial.  And it is applied to the rational straight line DE;  therefore MD is also rational and incommensurable in length with DE. [X. 22]  For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE;  therefore each of the straight lines DM, MG is rational and incommensurable in length with DE.  And, since AC is incommensurable in length with CB, and, as AC is to CB, so is the square on AC to the rectangle AC, CB,  therefore the square on AC is also incommensurable with the rectangle AC, CB. [X. 11]  Hence the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, [X. 12, 13]  that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [VI. 1, X. 11]  And they are rational; therefore DG is binomial. [X. 36]  It is to be proved that it is also a third binomial straight line. 
                             
                             
In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM.  And the rectangle DK, KM is equal to the square on MN;  therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.  And neither of the straight lines DM, MG is commensurable in length with DE. 
       
       
Therefore DG is a third binomial straight line. [X. Deff. II. 3]  Q. E. D. 
   
   
PROPOSITION 63. 
 
 
The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial. 
 
 
Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth;  I say that DG is a fourth binomial straight line. 
   
   
Let the same construction be made as before shown.  Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39]  Since then the sum of the squares on AC, CB is rational,  therefore DL is rational;  therefore DM is also rational and commensurable in length with DE. [X. 20]  Again, since twice the rectangle AC, CB, that is, MF, is medial, and it is applied to the rational straight line ML,  therefore MG is also rational and incommensurable in length with DE; [X. 22]  therefore DM is also incommensurable in length with MG. [X. 13]  Therefore DM, MG are rational straight lines commensurable in square only;  therefore DG is binomial. [X. 36] 
                   
                   
It is to be proved that it is also a fourth binomial straight line. 
 
 
In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN.  Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL,  so that DK is also incommensurable with KM. [VI. 1, X. 11]  But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts,  then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [X. 18]  therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM.  And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out. 
             
             
Therefore DG is a fourth binomial straight line. [X. Deff. II. 4]  Q. E. D. 
   
   
PROPOSITION 64. 
 
 
The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial. 
 
 
Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth;  I say that DG is a fifth binomial straight line. 
   
   
Let the same construction as before be made.  Since then AB is the side of a rational plus a medial area, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40]  Since then the sum of the squares on AC, CB is medial,  therefore DL is medial,  so that DM is rational and incommensurable in length with DE. [X. 22]  Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. [X. 20]  Therefore DM is incommensurable with MG; [X. 13]  therefore DM, MG are rational straight lines commensurable in square only;  therefore DG is binomial. [X. 36]  I say next that it is also a fifth binomial straight line. 
                   
                   
For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM;  therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. [X. 18]  And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE. 
     
     
Therefore DG is a fifth binomial.  Q. E. D. 
   
   
PROPOSITION 65. 
 
 
The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial. 
 
 
Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth;  I say that DG is a sixth binomial straight line. 
   
   
For let the same construction be made as before.  Then, since AB is the side of the sum of two medial areas, divided at C,  therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [X. 41]  so that, in accordance with what was before proved, each of the rectangles DL, MF is medial.  And they are applied to the rational straight line DE;  therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [X. 22]  And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB,  therefore DL is incommensurable with MF.  Therefore DM is also incommensurable with MG; [VI. 1, X. 11]  therefore DM, MG are rational straight lines commensurable in square only;  therefore DG is binomial. [X. 36]  I say next that it is also a sixth binomial straight line. 
                       
                       
Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM;  and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM.  And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out. 
     
     
Therefore DG is a sixth binomial straight line.  Q. E. D. 
   
   
PROPOSITION 66. 
 
 
A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order. 
 
 
Let AB be binomial, and let CD be commensurable in length with AB;  I say that CD is binomial and the same in order with AB. 
   
   
For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term;  therefore AE, EB are rational straight lines commensurable in square only. [X. 36]  Let it be contrived that, as AB is to CD, so is AE to CF; [VI. 12]  therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]  But AB is commensurable in length with CD;  therefore AE is also commensurable with CF, and EB with FD. [X. 11]  And AE, EB are rational;  therefore CF, FD are also rational.  And, as AE is to CF, so is EB to FD. [V. 11]  Therefore, alternately, as AE is to EB, so is CF to FD. [V. 16]  But AE, EB are commensurable in square only;  therefore CF, FD are also commensurable in square only. [X. 11]  And they are rational;  therefore CD is binomial. [X. 36]  I say next that it is the same in order with AB. 
                             
                             
For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.  If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]  And, if AE is commensurable with the rational straight line set out, CF will also be commensurable with it, [X. 12]  and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. [X. Deff. II. 1]  But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [X. 12]  and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. [X. Deff. II. 2]  But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [X. 13] and each of the straight lines AB, CD is a third binomial. [X. Deff. II. 3]  But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]  And, if AE is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. [X. Deff. II. 4]  But, if EB is so commensurable, so is FD also, and each of the straight lines AB, CD will be a fifth binomial. [X. Deff. II. 5]  But, if neither of the straight lines AE, EB is so commensurable, neither of the straight lines CF, FD is commensurable with the rational straight line set out, and each of the straight lines AB, CD will be a sixth binomial. [X. Deff. II. 6] 
                     
                     
Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order.  Q. E. D. 
   
   
PROPOSITION 67. 
 
 
A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order. 
 
 
Let AB be bimedial, and let CD be commensurable in length with AB;  I say that CD is bimedial and the same in order with AB. 
   
   
For, since AB is bimedial, let it be divided into its medials at E;  therefore AE, EB are medial straight lines commensurable in square only. [X. 37, 38]  And let it be contrived that, as AB is to CD, so is AE to CF;  therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]  But AB is commensurable in length with CD;  therefore AE, EB are also commensurable with CF, FD respectively. [X. 11]  But AE, EB are medial;  therefore CF, FD are also medial. [X. 23]  And since, as AE is to EB, so is CF to FD, [V. 11] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. [X. 11]  But they were also proved medial;  therefore CD is bimedial.  I say next that it is also the same in order with AB. 
                       
                       
For since, as AE is to EB, so is CF to FD,  therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD;  therefore, alternately, as the square on AE is to the square on CF, so is the rectangle AE, EB to the rectangle CF, FD. [V. 16]  But the square on AE is commensurable with the square on CF;  therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.  If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [X. 37]  but if medial, medial, [X. 23, Por.] and each of the straight lines AB, CD is a second bimedial. [X. 38] 
             
             
And for this reason CD will be the same in order with AB.  Q. E. D. 
   
   
PROPOSITION 68. 
 
 
A straight line commensurable with a major straight line is itself also major. 
 
 
Let AB be major, and let CD be commensurable with AB;  I say that CD is major. 
   
   
Let AB be divided at E;  therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [X. 39]  Let the same construction be made as before.  Then since, as AB is to CD, so is AE to CF, and EB to FD,  therefore also, as AE is to CF, so is EB to FD. [V. 11]  But AB is commensurable with CD;  therefore AE, EB are also commensurable with CF, FD respectively. [X. 11]  And since, as AE is to CF, so is EB to FD,  alternately also, as AE is to EB, so is CF to FD; [V. 16]  therefore also, componendo, as AB is to BE, so is CD to DF; [V. 18]  therefore also, as the square on AB is to the square on BE, so is the square on CD to the square on DF. [VI. 20]  Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF.  Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD;  therefore also, alternately, as the square on AB is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. [V. 16]  But the square on AB is commensurable with the square on CD;  therefore the squares on AE, EB are also commensurable with the squares on CF, FD.  And the squares on AE, EB together are rational; therefore the squares on CF, FD together are rational.  Similarly also twice the rectangle AE, EB is commensurable with twice the rectangle CF, FD.  And twice the rectangle AE, EB is medial;  therefore twice the rectangle CF, FD is also medial. [X. 23, Por.]  Therefore CF, FD are straight lines incommensurable in square which make, at the same time, the sum of the squares on them rational, but the rectangle contained by them medial;  therefore the whole CD is the irrational straight line called major. [X. 39] 
                                           
                                           
Therefore a straight line commensurable with the major straight line is major.  Q. E. D. 
   
   
PROPOSITION 69. 
 
 
A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area. 
 
 
Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB;  it is to be proved that CD is also the side of a rational plus a medial area. 
   
   
Let AB be divided into its straight lines at E;  therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [X. 40]  Let the same construction be made as before.  We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD;  so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational. 
         
         
Therefore CD is the side of a rational plus a medial area.  Q. E. D. 
   
   
PROPOSITION 70. 
 
 
A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas. 
 
 
Let AB be the side of the sum of two medial areas, and CD commensurable with AB;  it is to be proved that CD is also the side of the sum of two medial areas. 
   
   
For, since AB is the side of the sum of two medial areas, let it be divided into its straight lines at E;  therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and furthermore the sum of the squares on AE, EB incommensurable with the rectangle AE, EB. [X. 41]  Let the same construction be made as before.  We can then prove similarly that CF, FD are also incommensurable in square, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD;  so that the sum of the squares on CF, FD is also medial, the rectangle CF, FD is medial, and moreover the sum of the squares on CF, FD is incommensurable with the rectangle CF, FD. 
         
         
Therefore CD is the side of the sum of two medial areas.  Q. E. D. 
   
   
PROPOSITION 71. 
 
 
If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area. 
 
 
Let AB be rational, and CD medial;  I say that the “side” of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area. 
   
   
For AB is either greater or less than CD.  First, let it be greater;  let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB,  producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth.  Then, since AB is rational and is equal to EG, therefore EG is also rational.  And it has been applied to EF, producing EH as breadth;  therefore EH is rational and commensurable in length with EF. [X. 20]  Again, since CD is medial and is equal to HI, therefore HI is also medial.  And it is applied to the rational straight line EF, producing HK as breadth;  therefore HK is rational and incommensurable in length with EF [X. 22]  And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD,  so that EG is also incommensurable with HI.  But, as EG is to HI, so is EH to HK; [VI. 1]  therefore EH is also incommensurable in length with HK. [X. 11]  And both are rational;  therefore EH, HK are rational straight lines commensurable in square only;  therefore EK is a binomial straight line, divided at H. [X. 36]  And, since AB is greater than CD, while AB is equal to EG and CD to HI, therefore EG is also greater than HI;  therefore EH is also greater than HK.  The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it.  First, let the square on it be greater by the square on a straight line commensurable with itself.  Now the greater straight line HE is commensurable in length with the rational straight line EF set out;  therefore EK is a first binomial. [X. Deff. II. 1]  But EF is rational;  and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [X. 54]  Therefore the “side” of EI is binomial;  so that the “side” of AD is also binomial.  Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH.  Now the greater straight line EH is commensurable in length with the rational straight line EF set out;  therefore EK is a fourth binomial. [X. Deff. II. 4]  But EF is rational;  and, if an area be contained by a rational straight line and the fourth binomial, the “side” of the area is the irrational straight line called major. [X. 57]  Therefore the “side” of the area EI is major;  so that the “side” of the area AD is also major. 
                                                                   
                                                                   
Next, let AB be less than CD;  therefore EG is also less than HI, so that EH is also less than HK.  Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.  First, let the square on it be greater by the square on a straight line commensurable in length with itself.  Now the lesser straight line EH is commensurable in length with the rational straight line EF set out;  therefore EK is a second binomial. [X. Deff. II. 2]  But EF is rational,  and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a first bimedial; [X. 55]  therefore the “side” of the area EI is a first bimedial,  so that the “side” of AD is also a first bimedial.  Next, let the square on HK be greater than the square on HE by the square on a straight line incommensurable with HK.  Now the lesser straight line EH is commensurable with the rational straight line EF set out;  therefore EK is a fifth binomial. [X. Deff. II. 5]  But EF is rational;  and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a side of a rational plus a medial area. [X. 58]  Therefore the “side” of the area EI is a side of a rational plus a medial area,  so that the “side” of the area AD is also a side of a rational plus a medial area. 
                                 
                                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 72. 
 
 
If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas. 
 
 
For let two medial areas AB, CD incommensurable with one another be added together;  I say that the “side” of the area AD is either a second bimedial or a side of the sum of two medial areas. 
   
   
For AB is either greater or less than CD.  First, if it so chance, let AB be greater than CD.  Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth.  Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial.  And they are applied to the rational straight line FE, producing EH, HK as breadth;  therefore each of the straight lines EH, HK is rational and incommensurable in length with EF. [X. 22]  And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI,  therefore EG is also incommensurable with HI.  But, as EG is to HI, so is EH to HK; [VI. 1]  therefore EH is incommensurable in length with HK. [X. 11]  Therefore EH, HK are rational straight lines commensurable in square only;  therefore EK is binomial. [X. 36]  But the square on EH is greater than the square on HK either by the square on a straight line commensurable with EH or by the square on a straight line incommensurable with it.  First, let the square on it be greater by the square on a straight line commensurable in length with itself.  Now neither of the straight lines EH, HK is commensurable in length with the rational straight line EF set out;  therefore EK is a third binomial. [X. Deff. II. 3]  But EF is rational;  and, if an area be contained by a rational straight line and the third binomial, the “side” of the area is a second bimedial; [X. 56]  therefore the “side” of EI, that is, of AD, is a second bimedial.  Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH.  Now each of the straight lines EH, HK is incommensurable in length with EF;  therefore EK is a sixth binomial. [X. Deff. II. 6]  But, if an area be contained by a rational straight line and the sixth binomial, the “side” of the area is the side of the sum of two medial areas; [X. 59]  so that the “side” of the area AD is also the side of the sum of two medial areas. 
                                               
                                               
 
 
 
Therefore etc.  Q. E. D. 
   
   
                   
                   
                   
PROPOSITION 73. 
 
 
If from a rational straight line there be subtracted a rational straight line commensurable with the whole in square only, the remainder is irrational;  and let it be called an apotome. 
   
   
For from the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subtracted;  I say that the remainder AC is the irrational straight line called apotome. 
   
   
For, since AB is incommensurable in length with BC,  and, as AB is to BC, so is the square on AB to the rectangle AB, BC,  therefore the square on AB is incommensurable with the rectangle AB, BC. [X. 11]  But the squares on AB, BC are commensurable with the square on AB, [X. 15]  and twice the rectangle AB, BC is commensurable with the rectangle AB, BC. [X. 6]  And, inasmuch as the squares on AB, BC are equal to twice the rectangle AB, BC together with the square on CA, [II. 7]  therefore the squares on AB, BC are also incommensurable with the remainder, the square on AC. [X. 13, 16]  But the squares on AB, BC are rational;  therefore AC is irrational. [X. Def. 4]  And let it be called an apotome.  Q. E. D. 
                     
                     
PROPOSITION 74. 
 
 
If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a first apotome of a medial straight line. 
 
 
For from the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in square only and with AB makes the rectangle AB, BC rational;  I say that the remainder AC is irrational; and let it be called a first apotome of a medial straight line. 
   
   
For, since AB, BC are medial, the squares on AB, BC are also medial.  But twice the rectangle AB, BC is rational;  therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC;  therefore twice the rectangle AB, BC is also incommensurable with the remainder, the square on AC, [Cf. II. 7]  since, if the whole is incommensurable with one of the magnitudes, the original magnitudes will also be incommensurable. [X. 16]  But twice the rectangle AB, BC is rational;  therefore the square on AC is irrational;  therefore AC is irrational. [X. Def. 4]  And let it be called a first apotome of a medial straight line. 
                 
                 
PROPOSITION 75 
 
 
If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a second apotome of a medial straight line. 
 
 
For from the medial straight line AB let there be subtracted the medial straight line CB which is commensurable with the whole AB in square only and such that the rectangle AB, BC, which it contains with the whole AB, is medial; [X. 28]  I say that the remainder AC is irrational; and let it be called a second apotome of a medial straight line. 
   
   
For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth;  therefore the remainder FE is equal to the square on AC. [II. 7]  Now, since the squares on AB, BC are medial and commensurable, therefore DE is also medial. [X. 15 and 23, Por.]  And it is applied to the rational straight line DI, producing DG as breadth;  therefore DG is rational and incommensurable in length with DI. [X. 22]  Again, since the rectangle AB, BC is medial, therefore twice the rectangle AB, BC is also medial. [X. 23, Por.]  And it is equal to DH;  therefore DH is also medial.  And it has been applied to the rational straight line DI, producing DF as breadth;  therefore DF is rational and incommensurable in length with DI. [X. 22]  And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC;  therefore the square on AB is also incommensurable with the rectangle AB, BC. [X. 11]  But the squares on AB, BC are commensurable with the square on AB, [X. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC; [X. 6]  therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13]  But DE is equal to the squares on AB, BC, and DH to twice the rectangle AB, BC;  therefore DE is incommensurable with DH.  But, as DE is to DH, so is GD to DF; [VI. 1]  therefore GD is incommensurable with DF. [X. 11]  And both are rational;  therefore GD, DF are rational straight lines commensurable in square only;  therefore FG is an apotome. [X. 73]  But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from X. 20] and its ’side’ is irrational.  And AC is the ’side’ of FE;  therefore AC is irrational.  And let it be called a second apotome of a medial straight line.  Q. E. D. 
                                                   
                                                   
PROPOSITION 76 
 
 
If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the squares on them added together rational, but the rectangle contained by them medial, the remainder is irrational;  and let it be called minor. 
   
   
For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with the whole and fulfils the given conditions. [X. 33]  I say that the remainder AC is the irrational straight line called minor. 
   
   
For, since the sum of the squares on AB, BC is rational, while twice the rectangle AB, BC is medial,  therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC;  and, convertendo, the squares on AB, BC are incommensurable with the remainder, the square on AC. [II. 7, X. 16]  But the squares on AB, BC are rational;  therefore the square on AC is irrational;  therefore AC is irrational.  And let it be called minor.   
               
               
PROPOSITION 77 
 
 
If from a straight line there be subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial, but twice the rectangle contained by them rational, the remainder is irrational: and let it be called that which produces with a rational area a medial whole. 
 
 
For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [X. 34]  I say that the remainder AC is the irrational straight line aforesaid. 
   
   
For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC;  therefore the remainder also, the square on AC, is incommensurable with twice the rectangle AB, BC. [II. 7, X. 16]  And twice the rectangle AB, BC is rational;  therefore the square on AC is irrational;  therefore AC is irrational.  And let it be called that which produces with a rational area a medial whole.  Q. E. D. 
             
             
PROPOSITION 78 
 
 
If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them, the remainder is irrational; and let it be called that which produces with a medial area a medial whole. 
 
 
For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions; [X. 35]  I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole. 
   
   
For let a rational straight line DI be set out, to DI let there be applied DE equal to the squares on AB, BC, producing DG as breadth, and let DH equal to twice the rectangle AB, BC be subtracted.  Therefore the remainder FE is equal to the square on AC, [II. 7]  so that AC is the “side” of FE.  Now, since the sum of the squares on AB, BC is medial and is equal to DE, therefore DE is medial.  And it is applied to the rational straight line DI, producing DG as breadth;  therefore DG is rational and incommensurable in length with DI. [X. 22]  Again, since twice the rectangle AB, BC is medial and is equal to DH, therefore DH is medial.  And it is applied to the rational straight line DI, producing DF as breadth;  therefore DF is also rational and incommensurable in length with DI. [X. 22]  And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC,  therefore DE is also incommensurable with DH.  But, as DE is to DH, so also is DG to DF; [VI. 1]  therefore DG is incommensurable with DF. [X. 11]  And both are rational;  therefore GD, DF are rational straight lines commensurable in square only.  Therefore FG is an apotome. [X. 73]  And FH is rational;  but the rectangle contained by a rational straight line and an apotome is irrational, [deduction from X. 20] and its “side” is irrational.  And AC is the “side” of FE;  therefore AC is irrational.  And let it be called that which produces with a medial area a medial whole.  Q. E. D. 
                                           
                                           
PROPOSITION 79 
 
 
To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only. 
 
 
Let AB be an apotome, and BC an annex to it; therefore AC, CB are rational straight lines commensurable in square only. [X. 73]  I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only. 
   
   
For, if possible, let BD be so annexed;  therefore AD, DB are also rational straight lines commensurable in square only. [X. 73]  Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB,  for both exceed by the same, the square on AB, [II. 7]  therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.  But the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational;  therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible,  for both are medial [X. 21], and a medial area does not exceed a medial by a rational area. [X. 26]  Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only. 
                 
                 
Therefore only one rational straight line can be annexed to an apotome which is commensurable with the whole in square only.  Q. E. D. 
   
   
PROPOSITION 80. 
 
 
To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle. 
 
 
For let AB be a first apotome of a medial straight line, and let BC be an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is rational; [X. 74]  I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a rational area. 
   
   
For, if possible, let DB also be so annexed;  therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational. [X. 74]  Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB,  for they exceed by the same, the square on AB, [II. 7]  therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB.  But twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area, for both are rational.  Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area. which is impossible, for both are medial [X. 15 and 23, Por.],  and a medial area does not exceed a medial by a rational area. [X. 26] 
               
               
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 81. 
 
 
To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle. 
 
 
Let AB be a second apotome of a medial straight line and BC an annex to AB; therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is medial. [X. 75]  I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a medial rectangle. 
   
   
For, if possible, let BD also be so annexed;  therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is medial. [X. 75]  Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth,  and let HG equal to twice the rectangle AC, CB be subtracted, producing HM as breadth;  therefore the remainder EL is equal to the square on AB, [II. 7]  so that AB is the “side” of EL.  Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.  But EL is also equal to the square on AB;  therefore the remainder HI is equal to twice the rectangle AD, DB. [II. 7]  Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial.  And they are equal to EG;  therefore EG is also medial. [X. 15 and 23, Por.]  And it is applied to the rational straight line EF, producing EM as breadth;  therefore EM is rational and incommensurable in length with EF. [X. 22]  Again, since the rectangle AC, CB is medial, twice the rectangle AC, CB is also medial. [X. 23, Por.]  And it is equal to HG;  therefore HG is also medial.  And it is applied to the rational straight line EF, producing HM as breadth;  therefore HM is also rational and incommensurable in length with EF. [X. 22]  And, since AC, CB are commensurable in square only,  therefore AC is incommensurable in length with CB.  But, as AC is to CB, so is the square on AC to the rectangle AC, CB;  therefore the square on AC is incommensurable with the rectangle AC, CB. [X. 11]  But the squares on AC, CB are commensurable with the square on AC, while twice the rectangle AC, CB is commensurable with the rectangle AC, CB; [X. 6]  therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB. [X. 13]  And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB;  therefore EG is incommensurable with HG.  But, as EG is to HG, so is EM to HM; [VI. 1]  therefore EM is incommensurable in length with MH. [X. 11]  And both are rational;  therefore EM, MH are rational straight lines commensurable in square only;  therefore EH is an apotome, and HM an annex to it. [X. 73]  Similarly we can prove that HN is also an annex to it;  therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only: which is impossible. [X. 79] 
                                                                   
                                                                   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 82. 
 
 
To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial. 
 
 
Let AB be the minor straight line, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76]  I say that no other straight line can be annexed to AB fulfilling the same conditions. 
   
   
For, if possible, let BD be so annexed;  therefore AD, DB are also straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 76]  Now, since the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on AC, CB by a rational area,  for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible,  for both are medial. [X. 26] 
         
         
Therefore to a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes the squares on them added together rational, but twice the rectangle contained by them medial.  Q. E. D. 
   
   
PROPOSITION 83. 
 
 
To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational. 
 
 
Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which fulfil the given conditions. [X. 77]  I say that no other straight line can be annexed to AB which fulfils the same conditions. 
   
   
For, if possible, let BD be so annexed;  therefore AD, DB are also straight lines incommensurable in square which fulfil the given conditions. [X. 77]  Since then, as in the preceding cases, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area,  for both are rational, therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area: which is impossible,  for both are medial. [X. 26] 
         
         
Therefore no other straight line can be annexed to AB which is incommensurable in square with the whole and which with the whole fulfils the aforesaid conditions; therefore only one straight line can be so annexed.  Q. E. D. 
   
   
PROPOSITION 84. 
 
 
To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them. 
 
 
Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it; therefore AC, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 78]  I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions. 
   
   
For, if possible, let BD be so annexed, so that AD, DB are also straight lines incommensurable in square which make the squares on AD, DB added together medial, twice the rectangle AD, DB medial, and also the squares on AD, DB incommensurable with twice the rectangle AD, DB. [X. 78]  Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle AC, CB be applied to EF, producing HM as breadth;  therefore the remainder, the square on AB [II. 7], is equal to EL;  therefore AB is the “side” of EL.  Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth.  But the square on AB is also equal to EL;  therefore the remainder, twice the rectangle AD, DB [II. 7], is equal to HI.  Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial.  And it is applied to the rational straight line EF, producing EM as breadth;  therefore EM is rational and incommensurable in length with EF. [X. 22]  Again, since twice the rectangle AC, CB is medial and is equal to HG, therefore HG is also medial.  And it is applied to the rational straight line EF, producing HM as breadth;  therefore HM is rational and incommensurable in length with EF. [X. 22]  And, since the squares on AC, CB are incommensurable with twice the rectangle AC, CB, EG is also incommensurable with HG;  therefore EM is also incommensurable in length with MH. [VI. 1, X. 11]  And both are rational;  therefore EM, MH are rational straight lines commensurable in square only;  therefore EH is an apotome, and HM an annex to it. [X. 73]  Similarly we can prove that EH is again an apotome and HN an annex to it.  Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only:  which was proved impossible. [X. 79]  Therefore no other straight line can be so annexed to AB. 
                                           
                                           
Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the squares on them incommensurable with twice the rectangle contained by them.  Q. E. D. 
   
   
DEFINITIONS III. 
 
 
1. Given a rational straight line and an apotome, if the square on the whole be greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole be commensurable in length with the rational straight line set out, let the apotome be called a first apotome. 
 
 
2. But if the annex be commensurable in length with the rational straight line set out, and the square on the whole be greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome. 
 
 
3. But if neither be commensurable in length with the rational straight line set out, and the square on the whole be greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome. 
 
 
4. Again, if the square on the whole be greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole be commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome; 
 
 
5. if the annex be so commensurable, a fifth; 
 
 
6. and, if neither, a sixth. 
 
 
PROPOSITION 85. 
 
 
To find the first apotome. 
 
 
Let a rational straight line A be set out, and let BG be commensurable in length with A;  therefore BG is also rational.  Let two square numbers DE, EF be set out, and let their difference FD not be square;  therefore neither has ED to DF the ratio which a square number has to a square number.  Let it be contrived that, as ED is to DF, so is the square on BG to the square on GC; [X. 6, Por.]  therefore the square on BG is commensurable with the square on GC. [X. 6]  But the square on BG is rational;  therefore the square on GC is also rational;  therefore GC is also rational.  And, since ED has not to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number;  therefore BG is incommensurable in length with GC. [X. 9]  And both are rational;  therefore BG, GC are rational straight lines commensurable in square only;  therefore BC is an apotome. [X. 73]  I say next that it is also a first apotome. 
                             
                             
For let the square on H be that by which the square on BG is greater than the square on GC.  Now since. as ED is to FD, so is the square on BG to the square on GC,  therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H.  But DE has to EF the ratio which a square number has to a square number, for each is square;  therefore the square on GB also has to the square on H the ratio which a square number has to a square number;  therefore BG is commensurable in length with H. [X. 9]  And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG.  And the whole BG is commensurable in length with the rational straight line A set out.  Therefore BC is a first apotome. [X. Deff. III. 1] 
                 
                 
Therefore the first apotome BC has been found.  (Being) that which it was required to find. 
   
   
PROPOSITION 86. 
 
 
To find the second apotome. 
 
 
Let a rational straight line A be set out, and GC commensurable in length with A;  therefore GC is rational.  Let two square numbers DE, EF be set out, and let their difference DF not be square.  Now let it be contrived that, as FD is to DE, so is the square on CG to the square on GB. [X. 6, Por.]  Therefore the square on CG is commensurable with the square on GB. [X. 6]  But the square on CG is rational;  therefore the square on GB is also rational;  therefore BG is rational.  And, since the square on GC has not to the square on GB the ratio which a square number has to a square number, CG is incommensurable in length with GB. [X. 9]  And both are rational;  therefore CG, GB are rational straight lines commensurable in square only;  therefore BC is an apotome. [X. 73]  I say next that it is also a second apotome. 
                         
                         
For let the square on H be that by which the square on BG is greater than the square on GC.  Since then, as the square on BG is to the square on GC, so is the number ED to the number DF,  therefore, convertendo, as the square on BG is to the square on H, so is DE to EF. [V. 19, Por.]  And each of the numbers DE, EF is square;  therefore the square on BG has to the square on H the ratio which a square number has to a square number;  therefore BG is commensurable in length with H. [X. 9]  And the square on BG is greater than the square on GC by the square on H;  therefore the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG.  And CG, the annex, is commensurable with the rational straight line A set out.  Therefore BC is a second apotome. [X. Deff. III. 2] 
                   
                   
Therefore the second apotome BC has been found.  Q. E. D. 
   
   
PROPOSITION 87. 
 
 
To find the third apotome. 
 
 
Let a rational straight line A be set out, let three numbers E, BC, CD be set out which have not to one another the ratio which a square number has to a square number,  but let CB have to BD the ratio which a square number has to a square number.  Let it be contrived that, as E is to BC, so is the square on A to the square on FG,  and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.]  Since then, as E is to BC, so is the square on A to the square on FG,  therefore the square on A is commensurable with the square on FG. [X. 6]  But the square on A is rational;  therefore the square on FG is also rational;  therefore FG is rational.  And, since E has not to BC the ratio which a square number has to a square number,  therefore neither has the square on A to the square on FG the ratio which a square number has to a square number;  therefore A is incommensurable in length with FG. [X. 9]  Again, since, as BC is to CD, so is the square on FG to the square on GH,  therefore the square on FG is commensurable with the square on GH. [X. 6]  But the square on FG is rational;  therefore the square on GH is also rational;  therefore GH is rational.  And, since BC has not to CD the ratio which a square number has to a square number,  therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number;  therefore FG is incommensurable in length with GH. [X. 9]  And both are rational;  therefore FG, GH are rational straight lines commensurable in square only;  therefore FH is an apotome. [X. 73]  I say next that it is also a third apotome. 
                                               
                                               
For since, as E is to BC, so is the square on A to the square on FG  and, as BC is to CD, so is the square on FG to the square on HG,  therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22]  But E has not to CD the ratio which a square number has to a square number;  therefore neither has the square on A to the square on GH the ratio which a square number has to a square number;  therefore A is incommensurable in length with GH. [X. 9]  Therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out.  Now let the square on K be that by which the square on FG is greater than the square on GH.  Since then, as BC is to CD, so is the square on FG to the square on GH,  therefore, convertendo, as BC is to BD, so is the square on FG to the square on K. [V. 19, Por.]  But BC has to BD the ratio which a square number has to a square number;  therefore the square on FG also has to the square on K the ratio which a square number has to a square number.  Therefore FG is commensurable in length with K, [X. 9] and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.  And neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out;  therefore FH is a third apotome. [X. Deff. III. 3] 
                             
                             
Therefore the third apotome FH has been found.  Q. E. D. 
   
   
PROPOSITION 88. 
 
 
To find the fourth apotome. 
 
 
Let a rational straight line A be set out, and BG commensurable in length with it;  therefore BG is also rational.  Let two numbers DF, FE be set out such that the whole DE has not to either of the numbers DF, EF the ratio which a square number has to a square number.  Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC; [X. 6, Por.]  therefore the square on BG is commensurable with the square on GC. [X. 6]  But the square on BG is rational;  therefore the square on GC is also rational;  therefore GC is rational.  Now, since DE has not to EF the ratio which a square number has to a square number,  therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number;  therefore BG is incommensurable in length with GC. [X. 9]  And both are rational;  therefore BG, GC are rational straight lines commensurable in square only;  therefore BC is an apotome. [X. 73]   
                             
                             
Now let the square on H be that by which the square on BG is greater than the square on GC.  Since then, as DE is to EF, so is the square on BG to the square on GC,  therefore also, convertendo, as ED is to DF, so is the square on GB to the square on H. [v. 19, Por.]  But ED has not to DF the ratio which a square number has to a square number;  therefore neither has the square on GB to the square on H the ratio which a square number has to a square number;  therefore BG is incommensurable in length with H. [X. 9]  And the square on BG is greater than the square on GC by the square on H;  therefore the square on BG is greater than the square on GC by the square on a straight line incommensurable with BG.  And the whole BG is commensurable in length with the rational straight line A set out.  Therefore BC is a fourth apotome. [X. Deff. III. 4] 
                   
                   
Therefore the fourth apotome has been found.  Q. E. D. 
   
   
PROPOSITION 89. 
 
 
To find the fifth apotome. 
 
 
Let a rational straight line A be set out, and let CG be commensurable in length with A;  therefore CG is rational.  Let two numbers DF, FE be set out such that DE again has not to either of the numbers DF, FE the ratio which a square number has to a square number;  and let it be contrived that, as FE is to ED, so is the square on CG to the square on GB.  Therefore the square on GB is also rational; [X. 6]  therefore BG is also rational.  Now since, as DE is to EF, so is the square on BG to the square on GC, while DE has not to EF the ratio which a square number has to a square number,  therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number;  therefore BG is incommensurable in length with GC. [X. 9]  And both are rational;  therefore BG, GC are rational straight lines commensurable in square only;  therefore BC is an apotome. [X. 73]  I say next that it is also a fifth apotome. 
                         
                         
For let the square on H be that by which the square on BG is greater than the square on GC.  Since then, as the square on BG is to the square on GC, so is DE to EF,  therefore, convertendo, as ED is to DF, so is the square on BG to the square on H. [V. 19, Por.]  But ED has not to DF the ratio which a square number has to a square number;  therefore neither has the square on BG to the square on H the ratio which a square number has to a square number;  therefore BG is incommensurable in length with H. [X. 9]  And the square on BG is greater than the square on GC by the square on H;  therefore the square on GB is greater than the square on GC by the square on a straight line incommensurable in length with GB.  And the annex CG is commensurable in length with the rational straight line A set out;  therefore BC is a fifth apotome. [X. Deff. III. 5] 
                   
                   
Therefore the fifth apotome BC has been found.  Q. E. D. 
   
   
PROPOSITION 90. 
 
 
To find the sixth apotome. 
 
 
Let a rational straight line A be set out, and three numbers E, BC, CD not having to one another the ratio which a square number has to a square number; and further let CB also not have to BD the ratio which a square number has to a square number.  Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.] 
   
   
Now since, as E is to BC, so is the square on A to the square on FG,  therefore the square on A is commensurable with the square on FG. [X. 6]  But the square on A is rational;  therefore the square on FG is also rational;  therefore FG is also rational.  And, since E has not to BC the ratio which a square number has to a square number,  therefore neither has the square on A to the square on FG the ratio which a square number has to a square number;  therefore A is incommensurable in length with FG. [X. 9]  Again, since, as BC is to CD, so is the square on FG to the square on GH,  therefore the square on FG is commensurable with the square on GH. [X. 6]  But the square on FG is rational;  therefore the square on GH is also rational;  therefore GH is also rational.  And, since BC has not to CD the ratio which a square number has to a square number,  therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number;  therefore FG is incommensurable in length with GH. [X. 9]  And both are rational;  therefore FG, GH are rational straight lines commensurable in square only;  therefore FH is an apotome. [X. 73]  I say next that it is also a sixth apotome. 
                                       
                                       
For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH,  therefore, ex aequali, as E is to CD, so is the square on A to the square on GH. [v. 22]  But E has not to CD the ratio which a square number has to a square number;  therefore neither has the square on A to the square on GH the ratio which a square number has to a square number;  therefore A is incommensurable in length with GH; [X. 9]  therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A.  Now let the square on K be that by which the square on FG is greater than the square on GH.  Since then, as BC is to CD, so is the square on FG to the square on GH,  therefore, convertendo, as CB is to BD, so is the square on FG to the square on K. [v. 19, Por.]  But CB has not to BD the ratio which a square number has to a square number;  therefore neither has the square on FG to the square on K the ratio which a square number has to a square number;  therefore FG is incommensurable in length with K. [X. 9]  And the square on FG is greater than the square on GH by the square on K;  therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG.  And neither of the straight lines FG, GH is commensurable with the rational straight line A set out.  Therefore FH is a sixth apotome. [X. Deff. III. 6] 
                               
                               
Therefore the sixth apotome FH has been found.  Q. E. D. 
   
   
PROPOSITION 91. 
 
 
If an area be contained by a rational straight line and a first apotome, the side of the area is an apotome. 
 
 
For let the area AB be contained by the rational straight line AC and the first apotome AD;  I say that the “side” of the area AB is an apotome. 
   
   
For, since AD is a first apotome, let DG be its annex;  therefore AG, GD are rational straight lines commensurable in square only. [X. 73]  And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1]  if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17]  Let DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG;  therefore AF is commensurable with FG.  And through the points E, F, G let EH, FI, GK be drawn parallel to AC. 
             
             
Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]  But AG is commensurable with AC;  therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12]  And AC is rational;  therefore each of the straight lines AF, FG is also rational,  so that each of the rectangles AI, FK is also rational. [X. 19]  Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]  But DG is rational and incommensurable in length with AC;  therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13]  therefore each of the rectangles DH, EK is medial. [X. 21] 
                   
                   
Now let the square LM be made equal to AI, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK;  therefore the squares LM, NO are about the same diameter. [VI. 26]  Let PR be their diameter, and let the figure be drawn.  Since then the rectangle contained by AF, FG is equal to the square on EG,  therefore, as AF is to EG, so is EG to FG. [VI. 17]  But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to KF; [VI. 1]  therefore EK is a mean proportional between AI, KF. [V. 11]  But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after X. 53] and AI is equal to the square LM, and KF to NO;  therefore MN is also equal to EK.  But EK is equal to DH, and MN to LO;  therefore DK is equal to the gnomon UVW and NO.  But AK is also equal to the squares LM, NO;  therefore the remainder AB is equal to ST.  But ST is the square on LN;  therefore the square on LN is equal to AB;  therefore LN is the “side” of AB.  I say next that LN is an apotome. 
                                 
                                 
For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO,  therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational;  therefore each of the straight lines LP, PN is also rational.  Again, since DH is medial and is equal to LO, therefore LO is also medial.  Since then LO is medial, while NO is rational, therefore LO is incommensurable with NO.  But, as LO is to NO, so is LP to PN; [VI. 1]  therefore LP is incommensurable in length with PN. [X. 11]  And both are rational;  therefore LP, PN are rational straight lines commensurable in square only;  therefore LN is an apotome. [X. 73]  And it is the “side” of the area AB;  therefore the “side” of the area AB is an apotome. 
                       
                       
Therefore etc. 
 
 
PROPOSITION 92. 
 
 
If an area be contained by a rational straight line and a second apotome, the side of the area is a first apotome of a medial straight line. 
 
 
For let the area AB be contained by the rational straight line AC and the second apotome AD;  I say that the “side” of the area AB is a first apotome of a medial straight line. 
   
   
For let DG be the annex to AD;  therefore AG, GD are rational straight lines commensurable in square only, [X. 73]  and the annex DG is commensurable with the rational straight line AC set out,  while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2]  Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG,  therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, it divides it into commensurable parts. [X. 17]  Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG;  therefore AF is commensurable in length with FG.  Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]  But AG is rational and incommensurable in length with AC;  therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13]  therefore each of the rectangles AI, FK is medial. [X. 21]  Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15]  But DG is commensurable in length with AC.  Therefore each of the rectangles DH, EK is rational. [X. 19] 
                             
                             
Let then the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM, namely the angle LPM;  therefore the squares LM, NO are about the same diameter. [VI. 26]  Let PR be their diameter, and let the figure be drawn.  Since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial;  therefore LP, PN are also medial straight lines commensurable in square only.  And, since the rectangle AF, FG is equal to the square on EG,  therefore, as AF is to EG, so is EG to FG, [VI. 17]  while, as AF is to EG, so is AI to EK,  and, as EG is to FG, so is EK to FK; [VI. 1]  therefore EK is a mean proportional between AI, FK. [V. 11]  But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO;  therefore MN is also equal to EK.  But DH is equal to EK, and LO equal to MN;  therefore the whole DK is equal to the gnomon UVW and NO.  Since then the whole AK is equal to LM, NO, and, in these, DK is equal to the gnomon UVW and NO,  therefore the remainder AB is equal to TS.  But TS is the square on LN;  therefore the square on LN is equal to the area AB;  therefore LN is the “side” of the area AB.  I say that LN is a first apotome of a medial straight line. 
                                       
                                       
For, since EK is rational and is equal to LO, therefore LO, that is, the rectangle LP, PN, is rational.  But NO was proved medial;  therefore LO is incommensurable with NO.  But, as LO is to NO, so is LP to PN; [VI. 1]  therefore LP, PN are incommensurable in length. [X. 11]  Therefore LP, PN are medial straight lines commensurable in square only which contain a rational rectangle;  therefore LN is a first apotome of a medial straight line. [X. 74]  And it is the “side” of the area AB. 
               
               
Therefore the “side” of the area AB is a first apotome of a medial straight line.  Q. E. D. 
   
   
PROPOSITION 93. 
 
 
If an area be contained by a rational straight line and a third apotome, the side of the area is a second apotome of a medial straight line. 
 
 
For let the area AB be contained by the rational straight line AC and the third apotome AD;  I say that the “side” of the area AB is a second apotome of a medial straight line. 
   
   
For let DG be the annex to AD;  therefore AG, GD are rational straight lines commensurable in square only, and neither of the straight lines AG, GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. [X. Deff. III. 3]  Since then the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into commensurable parts. [X. 17]  Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG.  Let EH, FI, GK be drawn through the points E, F, G parallel to AC.  Therefore AF, FG are commensurable;  therefore AI is also commensurable with FK. [VI. 1, X. 11]  And, since AF, FG are commensurable in length,  therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]  But AG is rational and incommensurable in length with AC;  so that AF, FG are so also. [X. 13]  Therefore each of the rectangles AI, FK is medial. [X. 21]  Again, since DE is commensurable in length with EG,  therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]  But GD is rational and incommensurable in length with AC;  therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13]  therefore each of the rectangles DH, EK is medial. [X. 21]  And, since AG, GD are commensurable in square only,  therefore AG is incommensurable in length with GD.  But AG is commensurable in length with AF, and DG with EG;  therefore AF is incommensurable in length with EG. [X. 13]  But, as AF is to EG, so is AI to EK; [VI. 1]  therefore AI is incommensurable with EK. [X. 11] 
                                             
                                             
Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM;  therefore LM, NO are about the same diameter. [VI. 26]  Let PR be their diameter, and let the figure be drawn.  Now, since the rectangle AF, FG is equal to the square on EG,  therefore, as AF is to EG, so is EG to FG. [VI. 17]  But, as AF is to EG, so is AI to EK,  and, as EG is to FG, so is EK to FK; [VI. 1]  therefore also, as AI is to EK, so is EK to FK; [V. 11]  therefore EK is a mean proportional between AI, FK.  But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO;  therefore EK is also equal to MN.  But MN is equal to LO, and EK equal to DH;  therefore the whole DK is also equal to the gnomon UVW and NO.  But AK is also equal to LM, NO;  therefore the remainder AB is equal to ST, that is, to the square on LN;  therefore LN is the “side” of the area AB.  I say that LN is a second apotome of a medial straight line. 
                                 
                                 
For, since AI, FK were proved medial, and are equal to the squares on LP, PN, therefore each of the squares on LP, PN is also medial;  therefore each of the straight lines LP, PN is medial.  And, since AI is commensurable with FK, [VI. 1, X. 11]  therefore the square on LP is also commensurable with the square on PN.  Again, since AI was proved incommensurable with EK,  therefore LM is also incommensurable with MN,  that is, the square on LP with the rectangle LP, PN;  so that LP is also incommensurable in length with PN; [VI. 1, X. 11]  therefore LP, PN are medial straight lines commensurable in square only.  I say next that they also contain a medial rectangle. 
                   
                   
For, since EK was proved medial, and is equal to the rectangle LP, PN,  therefore the rectangle LP, PN is also medial,  so that LP, PN are medial straight lines commensurable in square only which contain a medial rectangle.  Therefore LN is a second apotome of a medial straight line; [X. 75]  and it is the “side” of the area AB. 
         
         
Therefore the “side” of the area AB is a second apotome of a medial straight line.  Q. E. D. 
   
   
PROPOSITION 94. 
 
 
If an area be contained by a rational straight line and a fourth apotome, the side of the area is minor. 
 
 
For let the area AB be contained by the rational straight line AC and the fourth apotome AD;  I say that the “side” of the area AB is minor. 
   
   
For let DG be the annex to AD;  therefore AG, GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [X. Deff. III. 4]  Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]  Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG;  therefore AF is incommensurable in length with FG.  Let EH, FI, GK be drawn through E, F, G parallel to AC, BD.  Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19]  Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. [X. 21]  Again, since AF is incommensurable in length with FG, therefore AI is also incommensurable with FK. [VI. 1, X. 11]  Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and about the same angle, the angle LPM.  Therefore the squares LM, NO are about the same diameter. [VI. 26]  Let PR be their diameter, and let the figure be drawn.  Since then the rectangle AF, FG is equal to the square on EG,  therefore, proportionally, as AF is to EG, so is EG to FG. [VI. 17]  But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1]  therefore EK is a mean proportional between AI, FK. [V. 11]  But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO;  therefore EK is also equal to MN.  But DH is equal to EK, and LO is equal to MN;  therefore the whole DK is equal to the gnomon UVW and NO.  Since, then, the whole AK is equal to the squares LM, NO, and, in these, DK is equal to the gnomon UVW and the square NO,  therefore the remainder AB is equal to ST, that is, to the square on LN;  therefore LN is the “side” of the area AB.  I say that LN is the irrational straight line called minor. 
                                               
                                               
For, since AK is rational and is equal to the squares on LP, PN,  therefore the sum of the squares on LP, PN is rational.  Again, since DK is medial, and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial.  And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN.  Therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial.  Therefore LN is the irrational straight line called minor; [X. 76]  and it is the “side” of the area AB. 
             
             
Therefore the “side” of the area AB is minor.  Q. E. D. 
   
   
PROPOSITION 95. 
 
 
If an area be contained by a rational straight line and a fifth apotome, the side of the area is a straight line which produces with a rational area a medial whole. 
 
 
For let the area AB be contained by the rational straight line AC and the fifth apotome AD;  I say that the “side” of the area AB is a straight line which produces with a rational area a medial whole. 
   
   
For let DG be the annex to AD;  therefore AG, GD are rational straight lines commensurable in square only, the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. [X. Deff. III. 5]  Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]  Let then DG be bisected at the point E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG;  therefore AF is incommensurable in length with FG.  Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21]  Again, since DG is rational and commensurable in length with AC, DK is rational. [X. 19] 
             
             
Now let the square LM be constructed equal to AI, and let the square NO equal to FK and about the same angle, the angle LPM, be subtracted;  therefore the squares LM, NO are about the same diameter. [VI. 26]  Let PR be their diameter, and let the figure be drawn.  Similarly then we can prove that LN is the “side” of the area AB.  I say that LN is the straight line which produces with a rational area a medial whole. 
         
         
For, since AK was proved medial and is equal to the squares on LP, PN,  therefore the sum of the squares on LP, PN is medial.  Again, since DK is rational and is equal to twice the rectangle LP, PN, the latter is itself also rational.  And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN;  therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational.  Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; [X. 77]  and it is the “side” of the area AB. 
             
             
Therefore the “side” of the area AB is a straight line which produces with a rational area a medial whole.  Q. E. D. 
   
   
PROPOSITION 96. 
 
 
If an area be contained by a rational straight line and a sixth apotome, the side of the area is a straight line which produces with a medial area a medial whole. 
 
 
For let the area AB be contained by the rational straight line AC and the sixth apotome AD;  I say that the “side” of the area AB is a straight line which produces with a medial area a medial whole. 
   
   
For let DG be the annex to AD;  therefore AG, GD are rational straight lines commensurable in square only,  neither of them is commensurable in length with the rational straight line AC set out,  and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6]  Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG,  therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]  Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure,  and let it be the rectangle AF, FG;  therefore AF is incommensurable in length with FG.  But, as AF is to FG, so is AI to FK, [VI. 1]  therefore AI is incommensurable with FK. [X. 11]  And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21]  Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21]  Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.  But, as AG is to GD, so is AK to KD; [VI. 1]  therefore AK is incommensurable with KD. [X. 11] 
                               
                               
Now let the square LM be constructed equal to AI, and let NO equal to FK, and about the same angle, be subtracted;  therefore the squares LM, NO are about the same diameter. [VI. 26]  Let PR be their diameter, and let the figure be drawn.  Then in manner similar to the above we can prove that LN is the “side” of the area AB.  I say that LN is a straight line which produces with a medial area a medial whole. 
         
         
For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial.  Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial.  And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN.  And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN;  therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them.  Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78]  and it is the “side” of the area AB. 
             
             
Therefore the “side” of the area is a straight line which produces with a medial area a medial whole.  Q. E. D. 
   
   
PROPOSITION 97. 
 
 
The square on an apotome applied to a rational straight line produces as breadth a first apotome. 
 
 
Let AB be an apotome, and CD rational, and to CD let there be applied CE equal to the square on AB and producing CF as breadth;  I say that CF is a first apotome. 
   
   
For let BG be the annex to AB;  therefore AG, GB are rational straight lines commensurable in square only. [X. 73]  To CD let there be applied CH equal to the square on AG, and KL equal to the square on BG.  Therefore the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB;  therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]  Let FM be bisected at the point N, and let NO be drawn through N parallel to CD;  therefore each of the rectangles FO, LN is equal to the rectangle AG, GB.  Now, since the squares on AG, GB are rational, and DM is equal to the squares on AG, GB, therefore DM is rational.  And it has been applied to the rational straight line CD, producing CM as breadth;  therefore CM is rational and commensurable in length with CD. [X. 20]  Again, since twice the rectangle AG, GB is medial, and FL is equal to twice the rectangle AG, GB, therefore FL is medial.  And it is applied to the rational straight line CD, producing FM as breadth;  therefore FM is rational and incommensurable in length with CD. [X. 22]  And, since the squares on AG, GB are rational, while twice the rectangle AG, GB is medial,  therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB.  And CL is equal to the squares on AG, GB, and FL to twice the rectangle AG, GB;  therefore DM is incommensurable with FL.  But, as DM is to FL, so is CM to FM; [VI. 1]  therefore CM is incommensurable in length with FM. [X. 11]  And both are rational;  therefore CM, MF are rational straight lines commensurable in square only;  therefore CF is an apotome. [X. 73]  I say next that it is also a first apotome. 
                                             
                                             
For, since the rectangle AG, GB is a mean proportional between the squares on AG, GB,  and CH is equal to the square on AG, KL equal to the square on BG, and NL equal to the rectangle AG, GB,  therefore NL is also a mean proportional between CH, KL;  therefore, as CH is to NL, so is NL to KL.  But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1]  therefore the rectangle CK, KM is equal to the square on NM [VI. 17],  that is, to the fourth part of the square on FM.  And, since the square on AG is commensurable with the square on GB, CH is also commensurable with KL.  But, as CH is to KL, so is CK to KM; [VI. 1]  therefore CK is commensurable with KM. [X. 11]  Since then CM, MF are two unequal straight lines, and to CM there has been applied the rectangle CK, KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]  And CM is commensurable in length with the rational straight line CD set out;  therefore CF is a first apotome. [X. Deff. III. 1] 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 98. 
 
 
The square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome. 
 
 
Let AB be a first apotome of a medial straight line and CD a rational straight line, and to CD let there be applied CE equal to the square on AB, producing CF as breadth;  I say that CF is a second apotome. 
   
   
For let BG be the annex to AB;  therefore AG, GB are medial straight lines commensurable in square only which contain a rational rectangle. [X. 74]  To CD let there be applied CH equal to the square on AG, producing CK as breadth, and KL equal to the square on GB, producing KM as breadth;  therefore the whole CL is equal to the squares on AG, GB; therefore CL is also medial. [X. 15 and 23, Por.]  And it is applied to the rational straight line CD, producing CM as breadth;  therefore CM is rational and incommensurable in length with CD. [X. 22]  Now, since CL is equal to the squares on AG, GB, and, in these, the square on AB is equal to CE, therefore the remainder, twice the rectangle AG, GB, is equal to FL. [II. 7]  But twice the rectangle AG, GB is rational; therefore FL is rational.  And it is applied to the rational straight line FE, producing FM as breadth;  therefore FM is also rational and commensurable in length with CD. [X. 20]  Now, since the sum of the squares on AG, GB, that is, CL, is medial, while twice the rectangle AG, GB, that is, FL, is rational, therefore CL is incommensurable with FL.  But, as CL is to FL, so is CM to FM; [VI. 1]  therefore CM is incommensurable in length with FM. [X. 11]  And both are rational;  therefore CM, MF are rational straight lines commensurable in square only;  therefore CF is an apotome. [X. 73]  I say next that it is also a second apotome. 
                                 
                                 
For let FM be bisected at N, and let NO be drawn through N parallel to CD;  therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.  Now, since the rectangle AG, GB is a mean proportional between the squares on AG, GB,  and the square on AG is equal to CH, the rectangle AG, GB to NL, and the square on BG to KL,  therefore NL is also a mean proportional between CH, KL;  therefore, as CH is to NL, so is NL to KL.  But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to MK; [VI. 1]  therefore, as CK is to NM, so is NM, so is KM; [V. 11]  therefore the rectangle CK, KM is equal to the square on NM [VI. 17],  that is, to the fourth part of the square on FM.    Since the CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to the greater, CM, and divides it into commensurable parts,  therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. [X. 17]  And the annex FM is commensurable in length with the rational straight line CD set out;  therefore CF is a second apotome. [X. Deff. III. 2] 
                             
                             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 99. 
 
 
The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome. 
 
 
Let AB be a second apotome of a medial straight line, and CD rational, and to CD let there be applied CE equal to the square on AB, producing CF as breadth;  I say that CF is a third apotome. 
   
   
For let BG be the annex to AB;  therefore AG, GB are medial straight lines commensurable in square only which contain a medial rectangle. [X. 75]  Let CH equal to the square on AG be applied to CD, producing CK as breadth, and let KL equal to the square on BG be applied to KH, producing KM as breadth;  therefore the whole CL is equal to the squares on AG, GB;  therefore CL is also medial. [X. 15 and 23, Por.]  And it is applied to the rational straight line CD, producing CM as breadth;  therefore CM is rational and incommensurable in length with CD. [X. 22]  Now, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB,  therefore the remainder LF is equal to twice the rectangle AG, GB. [II. 7]  Let then FM be bisected at the point N, and let NO be drawn parallel to CD;  therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.  But the rectangle AG, GB is medial;  therefore FL is also medial.  And it is applied to the rational straight line EF, producing FM as breadth;  therefore FM is also rational and incommensurable in length with CD. [X. 22]  And, since AG, GB are commensurable in square only,  therefore AG is incommensurable in length with GB;  therefore the square on AG is also incommensurable with the rectangle AG, GB. [VI. 1, X. 11]  But the squares on AG, GB are commensurable with the square on AG, and twice the rectangle AG, GB with the rectangle AG, GB;  therefore the squares on AG, GB are incommensurable with twice the rectangle AG, GB. [X. 13]  But CL is equal to the squares on AG, GB, and FL is equal to twice the rectangle AG, GB;  therefore CL is also incommensurable with FL.  But, as CL is to FL, so is CM to FM; [VI. 1]  therefore CM is incommensurable in length with FM. [X. 11]  And both are rational;  therefore CM, MF are rational straight lines commensurable in square only;  therefore CF is an apotome. [X. 73]  I say next that it is also a third apotome. 
                                                       
                                                       
For, since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. [VI. 1, X. 11]  And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB,  and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB,  therefore NL is also a mean proportional between CH, KL;  therefore, as CH is to NL, so is NL to KL.  But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1]  therefore, as CK is to MN, so is MN to KM; [V. 11]  therefore the rectangle CK, KM is equal to [the square on MN, that is, to] the fourth part of the square on FM.  Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into commensurable parts,  therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. [X. 17]  And neither of the straight lines CM, MF is commensurable in length with the rational straight line CD set out;  therefore CF is a third apotome. [X. Deff. III. 3] 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 100. 
 
 
The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome. 
 
 
Let AB be a minor and CD a rational straight line, and to the rational straight line CD let CE be applied equal to the square on AB and producing CF as breadth;  I say that CF is a fourth apotome. 
   
   
For let BG be the annex to AB;  therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on AG, GB rational, but twice the rectangle AG, GB medial. [X. 76]  To CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG, producing KM as breadth;  therefore the whole CL is equal to the squares on AG, GB.  And the sum of the squares on AG, GB is rational;  therefore CL is also rational.  And it is applied to the rational straight line CD, producing CM as breadth;  therefore CM is also rational and commensurable in length with CD. [X. 20]  And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB,  therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]  Let then FM be bisected at the point N, and let NO be drawn through N parallel to either of the straight lines CD, ML;  therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.  And, since twice the rectangle AG, GB is medial and is equal to FL, therefore FL is also medial.  And it is applied to the rational straight line FE, producing FM as breadth;  therefore FM is rational and incommensurable in length with CD. [X. 22]  And, since the sum of the squares on AG, GB is rational, while twice the rectangle AG, GB is medial,  the squares on AG, GB are incommensurable with twice the rectangle AG, GB.  But CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB;  therefore CL is incommensurable with FL.  But, as CL is to FL, so is CM to MF; [VI. 1]  therefore CM is incommensurable in length with MF. [X. 11]  And both are rational;  therefore CM, MF are rational straight lines commensurable in square only;  therefore CF is an apotome. [X. 73]  I say that it is also a fourth apotome. 
                                                 
                                                 
For, since AG, GB are incommensurable in square,  therefore the square on AG is also incommensurable with the square on GB.  And CH is equal to the square on AG, and KL equal to the square on GB;  therefore CH is incommensurable with KL.  But, as CH is to KL, so is CK to KM; [VI. 1]  therefore CK is incommensurable in length with KM. [X. 11]  And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and the square on AG is equal to CH, the square on GB to KL, and the rectangle AG, GB to NL,  therefore NL is a mean proportional between CH, KL;  therefore, as CH is to NL, so is NL to KL.  But, as CH is to NL, so is CK to NM, and, as NL is to KL, so is NM to KM; [VI. 1]  therefore, as CK is to MN, so is MN to KM; [V. 11]  therefore the rectangle CK, KM is equal to the square on MN [VI. 17], that is, to the fourth part of the square on FM.  Since then CM, MF are two unequal straight lines, and the rectangle CK, KM equal to the fourth part of the square on MF and deficient by a square figure has been applied to CM and divides it into incommensurable parts,  therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]  And the whole CM is commensurable in length with the rational straight line CD set out;  therefore CF is a fourth apotome. [X. Deff. III. 4] 
                               
                               
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 101. 
 
 
The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome. 
 
 
Let AB be the straight line which produces with a rational area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth;  I say that CF is a fifth apotome. 
   
   
For let BG be the annex to AB;  therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [X. 77]  To CD let there be applied CH equal to the square on AG, and KL equal to the square on GB;  therefore the whole CL is equal to the squares on AG, GB.  But the sum of the squares on AG, GB together is medial;  therefore CL is medial.  And it is applied to the rational straight line CD, producing CM as breadth;  therefore CM is rational and incommensurable with CD. [X. 22]  And, since the whole CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB,  therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]  Let then FM be bisected at N, and through N let NO be drawn parallel to either of the straight lines CD, ML;  therefore each of the rectangles FO, NL is equal to the rectangle AG, GB:  And, since twice the rectangle AG, GB is rational and equal to FL, therefore FL is rational.  And it is applied to the rational straight line EF, producing FM as breadth;  therefore FM is rational and commensurable in length with CD. [X. 20]  Now, since CL is medial, and FL rational, therefore CL is incommensurable with FL.  But, as CL is to FL, so is CM to MF; [VI. 1]  therefore CM is incommensurable in length with MF. [X. 11]  And both are rational;  therefore CM, MF are rational straight lines commensurable in square only;  therefore CF is an apotome. [X. 73]  I say next that it is also a fifth apotome. 
                                           
                                           
For we can prove similarly that the rectangle CK, KM is equal to the square on NM, that is, to the fourth part of the square on FM.  And, since the square on AG is incommensurable with the square on GB, while the square on AG is equal to CH, and the square on GB to KL, therefore CH is incommensurable with KL.  But, as CH is to KL, so is CK to KM; [VI. 1]  therefore CK is incommensurable in length with KM. [X. 11]  Since then CM, MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts,  therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]  And the annex FM is commensurable with the rational straight line CD set out;  therefore CF is a fifth apotome. [X. Deff. III. 5]  Q. E. D. 
                 
                 
PROPOSITION 102. 
 
 
The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome. 
 
 
Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth;  I say that CF is a sixth apotome. 
   
   
For let BG be the annex to AB;  therefore AG, GB are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle AG, GB medial, and the squares on AG, GB incommensurable with twice the rectangle AG, GB. [X. 78]  Now to CD let there be applied CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG;  therefore the whole CL is equal to the squares on AG, GB;  therefore CL is also medial.  And it is applied to the rational straight line CD, producing CM as breadth;  therefore CM is rational and incommensurable in length with CD. [X. 22]  Since now CL is equal to the squares on AG, GB, and, in these, CE is equal to the square on AB, therefore the remainder FL is equal to twice the rectangle AG, GB. [II. 7]  And twice the rectangle AG, GB is medial;  therefore FL is also medial.  And it is applied to the rational straight line FE, producing FM as breadth;  therefore FM is rational and incommensurable in length with CD. [X. 22]  And, since the squares on AG, GB are incommensurable with twice the rectangle AG, GB, and CL is equal to the squares on AG, GB, and FL equal to twice the rectangle AG, GB, therefore CL is incommensurable with FL.  But, as CL is to FL, so is CM to MF; [VI. 1]  therefore CM is incommensurable in length with MF. [X. 11]  And both are rational.  Therefore CM, MF are rational straight lines commensurable in square only;  therefore CF is an apotome. [X. 73]  I say next that it is also a sixth apotome. 
                                     
                                     
For, since FL is equal to twice the rectangle AG, GB, let FM be bisected at N, and let NO be drawn through N parallel to CD;  therefore each of the rectangles FO, NL is equal to the rectangle AG, GB.  And, since AG, GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB.  But CH is equal to the square on AG, and KL is equal to the square on GB;  therefore CH is incommensurable with KL.  But, as CH is to KL, so is CK to KM; [VI. 1]  therefore CK is incommensurable with KM. [X. 11]  And, since the rectangle AG, GB is a mean proportional between the squares on AG, GB, and CH is equal to the square on AG, KL equal to the square on GB, and NL equal to the rectangle AG, GB,  therefore NL is also a mean proportional between CH, KL;  therefore, as CH is to NL, so is NL to KL.  And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. [X. 18]  And neither of them is commensurable with the rational straight line CD set out;  therefore CF is a sixth apotome. [X. Deff. III. 6]  Q. E. D. 
                           
                           
PROPOSITION 103. 
 
 
A straight line commensurable in length with an apotome is an apotome and the same in order. 
 
 
Let AB be an apotome, and let CD be commensurable in length with AB;  I say that CD is also an apotome and the same in order with AB. 
   
   
For, since AB is an apotome, let BE be the annex to it;  therefore AE, EB are rational straight lines commensurable in square only. [X. 73]  Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD; [VI. 12]  therefore also, as one is to one, so are all to all; [V. 12]  therefore also, as the whole AE is to the whole CF, so is AB to CD.  But AB is commensurable in length with CD.  Therefore AE is also commensurable with CF, and BE with DF. [X. 11]  And AE, EB are rational straight lines commensurable in square only;  therefore CF, FD are also rational straight lines commensurable in square only. [X. 13]     
                     
                     
Now since, as AE is to CF, so is BE to DF, alternately therefore, as AE is to EB, so is CF to FD. [V. 16]  And the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.  If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]  And, if AE is commensurable in length with the rational straight line set out, CF is so also, [X. 12] if BE, then DF also, [id.] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13]  But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line incommensurable with CF. [X. 14]  And, if AE is commensurable in length with the rational straight line set out, CF is so also, if BE, then DF also, [X. 12] and, if neither of the straight lines AE, EB, then neither of the straight lines CF, FD. [X. 13] 
           
           
Therefore CD is an apotome and the same in order with AB.  Q. E. D. 
   
   
PROPOSITION 104. 
 
 
A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order. 
 
 
Let AB be an apotome of a medial straight line, and let CD be commensurable in length with AB;  I say that CD is also an apotome of a medial straight line and the same in order with AB. 
   
   
For, since AB is an apotome of a medial straight line, let EB be the annex to it.  Therefore AE, EB are medial straight lines commensurable in square only. [X. 74, 75]  Let it be contrived that, as AB is to CD, so is BE to DF; [VI. 12]  therefore AE is also commensurable with CF, and BE with DF. [V. 12, X. 11]  But AE, EB are medial straight lines commensurable in square only;  therefore CF, FD are also medial straight lines [X. 23] commensurable in square only; [X. 13]  therefore CD is an apotome of a medial straight line. [X. 74, 75]  I say next that it is also the same in order with AB. 
               
               
Since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD.  But the square on AE is commensurable with the square on CF;  therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD. [V. 16, X. 11]  Therefore, if the rectangle AE, EB is rational, the rectangle CF, FD will also be rational, [X. Def. 4] and if the rectangle AE, EB is medial, the rectangle CF, FD is also medial. [X. 23, Por.] 
       
       
Therefore CD is an apotome of a medial straight line and the same in order with AB. [X. 74, 75]  Q. E. D. 
   
   
PROPOSITION 105. 
 
 
A straight line commensurable with a minor straight line is minor. 
 
 
Let AB be a minor straight line, and CD commensurable with AB;  I say that CD is also minor. 
   
   
Let the same construction be made as before;  then, since AE, EB are incommensurable in square, [X. 76] therefore CF, FD are also incommensurable in square. [X. 13]  Now since, as AE is to EB, so is CF to FD, [V. 12, V. 16] therefore also, as the square on AE is to the square on EB, so is the square on CF to the square on FD. [VI. 22]  Therefore, componendo, as the squares on AE, EB are to the square on EB, so are the squares on CF, FD to the square on FD. [V. 18]  But the square on BE is commensurable with the square on DF;  therefore the sum of the squares on AE, EB is also commensurable with the sum of the squares on CF, FD. [V. 16, X. 11]  But the sum of the squares on AE, EB is rational; [X. 76]  therefore the sum of the squares on CF, FD is also rational. [X. Def. 4]  Again, since, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD,  while the square on AE is commensurable with the square on CF, therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.  But the rectangle AE, EB is medial; [X. 76]  therefore the rectangle CF, FD is also medial; [X. 23, Por.]  therefore CF, FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. 
                         
                         
Therefore CD is minor. [X. 76]  Q. E. D. 
   
   
PROPOSITION 106. 
 
 
A straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole. 
 
 
Let AB be a straight line which produces with a rational area a medial whole, and CD commensurable with AB;  I say that CD is also a straight line which produces with a rational area a medial whole. 
   
   
For let BE be the annex to AB;  therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on AE, EB medial, but the rectangle contained by them rational. [X. 77]  Let the same construction be made.  Then we can prove, in manner similar to the foregoing, that CF, FD are in the same ratio as AE, EB, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD;  so that CF, FD are also straight lines incommensurable in square which make the sum of the squares on CF, FD medial, but the rectangle contained by them rational. 
         
         
Therefore CD is a straight line which produces with a rational area a medial whole. [X. 77]  Q. E. D. 
   
   
PROPOSITION 107. 
 
 
A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole. 
 
 
Let AB be a straight line which produces with a medial area a medial whole, and let CD be commensurable with AB;  I say that CD is also a straight line which produces with a medial area a medial whole. 
   
   
For let BE be the annex to AB, and let the same construction be made;  therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them. [X. 78]  Now, as was proved, AE, EB are commensurable with CF, FD, the sum of the squares on AE, EB with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD;  therefore CF, FD are also straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further the sum of the squares on them incommensurable with the rectangle contained by them. 
       
       
Therefore CD is a straight line which produces with a medial area a medial whole. [X. 78]   
   
   
PROPOSITION 108. 
 
 
If from a rational area a medial area be subtracted, the side of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line. 
 
 
For from the rational area BC let the medial area BD be subtracted;  I say that the “side” of the remainder EC becomes one of two irrational straight lines, either an apotome or a minor straight line. 
   
   
For let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to BC, and let GK equal to DB be subtracted;  therefore the remainder EC is equal to LH.  Since then BC is rational, and BD medial, while BC is equal to GH, and BD to GK,  therefore GH is rational, and GK medial.  And they are applied to the rational straight line FG;  therefore FH is rational and commensurable in length with FG, [X. 20] while FK is rational and incommensurable in length with FG; [X. 22]  therefore FH is incommensurable in length with FK. [X. 13]  Therefore FH, FK are rational straight lines commensurable in square only;  therefore KH is an apotome [X. 73], and KF the annex to it.  Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable. 
                   
                   
First, let the square on it be greater by the square on a straight line commensurable with it.  Now the whole HF is commensurable in length with the rational straight line FG set out;  therefore KH is a first apotome. [X. Deff. III. 1]  But the “side” of the rectangle contained by a rational straight line and a first apotome is an apotome. [X. 91]  Therefore the “side” of LH, that is, of EC, is an apotome. 
         
         
But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the whole FH is commensurable in length with the rational straight line FG set out, KH is a fourth apotome. [X. Deff. III. 4]  But the “side” of the rectangle contained by a rational straight line and a fourth apotome is minor. [X. 94]  Q. E. D. 
     
     
PROPOSITION 109. 
 
 
If from a medial area a rational area be subtracted, there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole. 
 
 
For from the medial area BC let the rational area BD be subtracted.  I say that the “side” of the remainder EC becomes one of two irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole. 
   
   
For let a rational straight line FG be set out, and let the areas be similarly applied.  It follows then that FH is rational and incommensurable in length with FG, while KF is rational and commensurable in length with FG;  therefore FH, FK are rational straight lines commensurable in square only; [X. 13]  therefore KH is an apotome, and FK the annex to it. [X. 73]  Now the square on HF is greater than the square on FK either by the square on a straight line commensurable with HF or by the square on a straight line incommensurable with it. 
         
         
If then the square on HF is greater than the square on FK by the square on a straight line commensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a second apotome. [X. Deff. III. 2]  But FG is rational;  so that the “side” of LH, that is, of EC, is a first apotome of a medial straight line. [X. 92] 
     
     
But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, KH is a fifth apotome; [X. Deff. III. 5]  so that the “side” of EC is a straight line which produces with a rational area a medial whole. [X. 95]   
     
     
PROPOSITION 110. 
 
 
If from a medial area there be subtracted a medial area incommensurable with the whole, the two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole. 
 
 
For, as in the foregoing figures, let there be subtracted from the medial area BC the medial area BD incommensurable with the whole;  I say that the “side” of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole. 
   
   
For, since each of the rectangles BC, BD is medial, and BC is incommensurable with BD, it follows that each of the straight lines FH, FK will be rational and incommensurable in length with FG. [X. 22]  And, since BC is incommensurable with BD, that is, GH with GK, HF is also incommensurable with FK; [VI. 1, X. 11]  therefore FH, FK are rational straight lines commensurable in square only;  therefore KH is an apotome. [X. 73] 
       
       
If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH, FK is commensurable in length with the rational straight line FG set out, KH is a third apotome. [X. Deff. III. 3]  But KL is rational, and the rectangle contained by a rational straight line and a third apotome is irrational, and the “side” of it is irrational, and is called a second apotome of a medial straight line; [X. 93]  so that the “side” of LH, that is, of EC, is a second apotome of a medial straight line. 
     
     
But, if the square on FH is greater than the square on FK by the square on a straight line incommensurable with FH, while neither of the straight lines HF, FK is commensurable in length with FG, KH is a sixth apotome. [X. Deff. III. 6]  But the “side” of the rectangle contained by a rational straight line and a sixth apotome is a straight line which produces with a medial area a medial whole. [X. 96]  Therefore the “side” of LH, that is, of EC, is a straight line which produces with a medial area a medial whole.  Q. E. D. 
       
       
PROPOSITION 111. 
 
 
The apotome is not the same with the binomial straight line. 
 
 
Let AB be an apotome;  I say that AB is not the same with the binomial straight line. 
   
   
For, if possible, let it be so;  let a rational straight line DC be set out, and to CD let there be applied the rectangle CE equal to the square on AB and producing DE as breadth.  Then, since AB is an apotome, DE is a first apotome. [X. 97]  Let EF be the annex to it;  therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF,  and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1]  Again, since AB is binomial, therefore DE is a first binomial straight line. [X. 60]  Let it be divided into its terms at G, and let DG be the greater term;  therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1]  Therefore DF is also commensurable in length with DG; [X. 12]  therefore the remainder GF is also commensurable in length with DF. [X. 15]    But DF is incommensurable in length with EF;  therefore FG is also incommensurable in length with EF. [X. 13]  Therefore GF, FE are rational straight lines commensurable in square only;  therefore EG is an apotome. [X. 73]  But it is also rational: which is impossible. 
                                 
                                 
Therefore the apotome is not the same with the binomial straight line.  Q. E. D. 
   
   
 
 
 
PROPOSITION 112. 
 
 
The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line. 
 
 
Let A be a rational straight line, let BC be a binomial, and let DC be its greater term; let the rectangle BC, EF be equal to the square on A;  I say that EF is an apotome the terms of which are commensurable with CD, DB, and in the same ratio, and further EF will have the same order as BC. 
   
   
For again let the rectangle BD, G be equal to the square on A.  Since then the rectangle BC, EF is equal to the rectangle BD, G,  therefore, as CB is to BD, so is G to EF. [VI. 16]  But CB is greater than BD;  therefore G is also greater than EF. [V. 16, V. 14]  Let EH be equal to G;  therefore, as CB is to BD, so is HE to EF;  therefore, separando, as CD is to BD, so is HF to FE. [V. 17]  Let it be contrived that, as HF is to FE, so is FK to KE;  therefore also the whole HK is to the whole KF as FK is to KE;  for, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents. [V. 12]  But, as FK is to KE, so is CD to DB; [V. 11]  therefore also, as HK is to KF, so is CD to DB. [id.]  But the square on CD is commensurable with the square on DB; [X. 36]  therefore the square on HK is also commensurable with the square on KF. [VI. 22, X. 11]  And, as the square on HK is to the square on KF, so is HK to KE, since the three straight lines HK, KF, KE are proportional. [V. Def. 9]  Therefore HK is commensurable in length with KE,  so that HE is also commensurable in length with EK. [X. 15]  Now, since the square on A is equal to the rectangle EH, BD, while the square on A is rational, therefore the rectangle EH, BD is also rational.  And it is applied to the rational straight line BD;  therefore EH is rational and commensurable in length with BD; [X. 20]  so that EK, being commensurable with it, is also rational and commensurable in length with BD.  Since, then, as CD is to DB, so is FK to KE, while CD, DB are straight lines commensurable in square only,  therefore FK, KE are also commensurable in square only. [X. 11]  But KE is rational;  therefore FK is also rational.  Therefore FK, KE are rational straight lines commensurable in square only;  therefore EF is an apotome. [X. 73] 
                                                       
                                                       
Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it. 
 
 
If then the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. [X. 14]  And, if CD is commensurable in length with the rational straight line set out, so also is FK; [X. 11, 12]  if BD is so commensurable, so also is KE; [X. 12]  but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so. 
       
       
But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK. [X. 14]  And, if CD is commensurable with the rational straight line set out, so also is FK;  if BD is so commensurable, so also is KE;  but, if neither of the straight lines CD, DB is so commensurable, neither of the straight lines FK, KE is so;  so that FE is an apotome, the terms of which FK, KE are commensurable with the terms CD, DB of the binomial straight line and in the same ratio, and it has the same order as BC.  Q. E. D. 
           
           
PROPOSITION 113. 
 
 
The square on a rational straight line, if applied to an apotome, produces as, breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome. 
 
 
Let A be a rational straight line and BD an apotome, and let the rectangle BD, KH be equal to the square on A, so that the square on the rational straight line A when applied to the apotome BD produces KH as breadth;  I say that KH is a binomial straight line the terms of which are commensurable with the terms of BD and in the same ratio; and further KH has the same order as BD. 
   
   
For let DC be the annex to BD;  therefore BC, CD are rational straight lines commensurable in square only. [X. 73]  Let the rectangle BC, G be also equal to the square on A.  But the square on A is rational;  therefore the rectangle BC, G is also rational.  And it has been applied to the rational straight line BC;  therefore G is rational and commensurable in length with BC. [X. 20]  Since now the rectangle BC, G is equal to the rectangle BD, KH, therefore, proportionally, as CB is to BD, so is KH to G. [VI. 16]  But BC is greater than BD;  therefore KH is also greater than G. [V. 16, V. 14]  Let KE be made equal to G;  therefore KE is commensurable in length with BC.  And since, as CB is to BD, so is HK to KE, therefore, convertendo, as BC is to CD, so is KH to HE. [V. 19, Por.]  Let it be contrived that, as KH is to HE, so is HF to FE;  therefore also the remainder KF is to FH as KH is to HE, that is, as BC is to CD. [V. 19]  But BC, CD are commensurable in square only;  therefore KF, FH are also commensurable in square only. [X. 11]  And since, as KH is to HE, so is KF to FH, while, as KH is to HE, so is HF to FE,  therefore also, as KF is to FH, so is HF to FE, [V. 11]  so that also, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9]  therefore also, as KF is to FE, so is the square on KF to the square on FH.  But the square on KF is commensurable with the square on FH,  for KF, FH are commensurable in square;  therefore KF is also commensurable in length with FE, [X. 11]  so that KF is also commensurable in length with KE. [X. 15]  But KE is rational and commensurable in length with BC;  therefore KF is also rational and commensurable in length with BC. [X. 12]  And, since, as BC is to CD, so is KF to FH, alternately, as BC is to KF, so is DC to FH. [V. 16]  But BC is commensurable with KF;  therefore FH is also commensurable in length with CD. [X. 11]  But BC, CD are rational straight lines commensurable in square only;  therefore KF, FH are also rational straight lines [X. Def. 3] commensurable in square only;  therefore KH is binomial. [X. 36] 
                                                                 
                                                                 
If now the square on BC is greater than the square on CD by the square on a straight line commensurable with BC, the square on KF will also be greater than the square on FH by the square on a straight line commensurable with KF. [X 14]  And, if BC is commensurable in length with the rational straight line set out, so also is KF;  if CD is commensurable in length with the rational straight line set out, so also is FH, but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH. 
     
     
But, if the square on BC is greater than the square on CD by the square on a straight line incommensurable with BC, the square on KF is also greater than the square on FH by the square on a straight line incommensurable with KF. [X. 14]  And, if BC is commensurable with the rational straight line set out, so also is KF;  if CD is so commensurable, in length with the rational straight line set out, so also is FH; but, if neither of the straight lines BC, CD, then neither of the straight lines KF, FH. 
     
     
Therefore KH is a binomial straight line, the terms of which KF, FH are commensurable with the terms BC, CD of the apotome and in the same ratio, and further KH has the same order as BD.  Q. E. D. 
   
   
PROPOSITION 114. 
 
 
If an area be contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, the side of the area is rational. 
 
 
For let an area, the rectangle AB, CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter; let the terms CE, ED of the binomial straight line be commensurable with the terms AF, FB of the apotome and in the same ratio; and let the “side” of the rectangle AB, CD be G;  I say that G is rational. 
   
   
For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth.  Therefore KL is an apotome.  Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112]  But CE, ED are also commensurable with AF, FB and in the same ratio;  therefore, as AF is to FB, so is KM to ML.  Therefore, alternately, as AF is to KM, so is BF to LM;  therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19]  But AF is commensurable with KM; [X. 12]  therefore AB is also commensurable with KL. [X. 11]  And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1]  therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11]  But the rectangle CD, KL is equal to the square on H;  therefore the rectangle CD, AB is commensurable with the square on H.  But the square on G is equal to the rectangle CD, AB;  therefore the square on G is commensurable with the square on H.  But the square on H is rational;  therefore the square on G is also rational;  therefore G is rational.  And it is the “side” of the rectangle CD, AB. 
                                     
                                     
Therefore etc. 
 
 
PORISM.
And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines. 
Q. E. D. 
   
   
PROPOSITION 115. 
 
 
From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding. 
 
 
Let A be a medial straight line;  I say that from A there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding. 
   
   
Let a rational straight line B be set out, and let the square on C be equal to the rectangle B, A;  therefore C is irrational; [X. Def. 4]  for that which is contained by an irrational and a rational straight line is irrational. [deduction from X. 20]  And it is not the same with any of the preceding;  for the square on none of the preceding, if applied to a rational straight line produces as breadth a medial straight line.  Again, let the square on D be equal to the rectangle B, C;  therefore the square on D is irrational. [deduction from X. 20]  Therefore D is irrational; [X. Def. 4]  and it is not the same with any of the preceding,  for the square on none of the preceding, if applied to a rational straight line, produces C as breadth.  Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding.  Q. E. D. 
                       
                       
BOOK XI. 
 
 
DEFINITIONS. 
 
 
1. A solid is that which has length, breadth, and depth. 
 
 
2. An extremity of a solid is a surface. 
 
 
3. A straight line is at right angles to a plane, when it makes right angles with all the straight lines which meet it and are in the plane. 
 
 
4. A plane is at right angles to a plane when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane. 
 
 
5. The inclination of a straight line to a plane is, assuming a perpendicular drawn from the extremity of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the extremity of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up. 
 
 
6. The inclination of a plane to a plane is the acute angle contained by the straight lines drawn at right angles to the common section at the same point, one in each of the planes. 
 
 
7. A plane is said to be similarly inclined to a plane as another is to another when the said angles of the inclinations are equal to one another. 
 
 
8. Parallel planes are those which do not meet. 
 
 
9. Similar solid figures are those contained by similar planes equal in multitude. 
 
 
10. Equal and similar solid figures are those contained by similar planes equal in multitude and in magnitude. 
 
 
11. A solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines.
Otherwise: A solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point. 
 
 
12. A pyramid is a solid figure, contained by planes, which is constructed from one plane to one point. 
 
 
13. A prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar and parallel, while the rest are parallelograms. 
 
 
14. When, the diameter of a semicircle remaining fixed, the semicircle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a sphere. 
 
 
15. The axis of the sphere is the straight line which remains fixed and about which the semicircle is turned. 
 
 
16. The centre of the sphere is the same as that of the semicircle. 
 
 
17. A diameter of the sphere is any straight line drawn through the centre and terminated in both directions by the surface of the sphere. 
 
 
18. When, one side of those about the right angle in a right-angled triangle remaining fixed, the triangle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone.
And, if the straight line which remains fixed be equal to the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuse-angled; and if greater, acute-angled. 
 
 
19. The axis of the cone is the straight line which remains fixed and about which the triangle is turned. 
 
 
20. And the base is the circle described by the straight line which is carried round. 
 
 
21. When, one side of those about the right angle in a rectangular parallelogram remaining fixed, the parallelogram is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cylinder. 
 
 
22. The axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned. 
 
 
23. And the bases are the circles described by the two sides opposite to one another which are carried round. 
 
 
24. Similar cones and cylinders are those in which the axes and the diameters of the bases are proportional. 
 
 
25. A cube is a solid figure contained by six equal squares. 
 
 
26. An octahedron is a solid figure contained by eight equal and equilateral triangles. 
 
 
27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. 
 
 
28. A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons. 
 
 
PROPOSITION 1. 
 
 
A part of a straight line cannot be in the plane of reference and a part in a plane more elevated. 
 
 
For, if possible, let a part AB of the straight line ABC be in the plane of reference, and a part BC in a plane more elevated. 
 
 
There will then be in the plane of reference some straight line continuous with AB in a straight line.  Let it be BD;  therefore AB is a common segment of the two straight lines ABC, ABD: which is impossible,  inasmuch as, if we describe a circle with centre B and distance AB, the diameters will cut off unequal circumferences of the circle. 
       
       
Therefore a part of a straight line cannot be in the plane of reference, and a part in a plane more elevated.  Q. E. D. 
   
   
PROPOSITION 2. 
 
 
If two straight lines cut one another, they are in one plane, and every triangle is in one plane. 
 
 
For let the two straight lines AB, CD cut one another at the point E;  I say that AB, CD are in one plane, and every triangle is in one plane. 
   
   
For let points F, G be taken at random on EC, EB, let CB, FG be joined, and let FH, GK be drawn across;  I say first that the triangle ECB is in one plane.  For, if part of the triangle ECB, either FHC or GBK, is in the plane of reference, and the rest in another,  a part also of one of the straight lines EC, EB will be in the plane of reference, and a part in another.  But, if the part FCBG of the triangle ECB be in the plane of reference, and the rest in another,  a part also of both the straight lines EC, EB will be in the plane of reference and a part in another: which was proved absurd. [XI. 1]  Therefore the triangle ECB is in one plane.  But, in whatever plane the triangle ECB is, in that plane also is each of the straight lines EC, EB, and, in whatever plane each of the straight lines EC, EB is, in that plane are AB, CD also. [XI. 1]  Therefore the straight lines AB, CD are in one plane, and every triangle is in one plane.  Q. E. D. 
                   
                   
PROPOSITION 3. 
 
 
If two planes cut one another, their common section is a straight line. 
 
 
For let the two planes AB, BC cut one another, and let the line DB be their common section;  I say that the line DB is a straight line. 
   
   
For, if not, from D to B let the straight line DEB be joined in the plane AB, and in the plane BC the straight line DFB.  Then the two straight lines DEB, DFB will have the same extremities, and will clearly enclose an area: which is absurd.  Therefore DEB, DFB are not straight lines.  Similarly we can prove that neither will there be any other straight line joined from D to B except DB the common section of the planes AB, BC. 
       
       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 4. 
 
 
If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them. 
 
 
For let a straight line EF be set up at right angles to the two straight lines AB, CD, which cut one another at the point E, from E;  I say that EF is also at right angles to the plane through AB, CD. 
   
   
For let AE, EB, CE, ED be cut off equal to one another, and let any straight line GEH be drawn across through E, at random;  let AD, CB be joined, and further let FA, FG, FD, FC, FH, FB be joined from the point F taken at random Now, since the two straight lines AE, ED are equal to the two straight lines CE, EB, and contain equal angles, [I. 15]  therefore the base AD is equal to the base CB, and the triangle AED will be equal to the triangle CEB; [I. 4]   so that the angle DAE is also equal to the angle EBC.  But the angle AEG is also equal to the angle BEH; [I. 15]  therefore AGE, BEH are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, AE to EB;  therefore they will also have the remaining sides equal to the remaining sides. [I. 26]  Therefore GE is equal to EH, and AG to BH.  And, since AE is equal to EB, while FE is common and at right angles,  therefore the base FA is equal to the base FB. [I. 4]  For the same reason FC is also equal to FD.  And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively;  and the base FD was proved equal to the base FC;  therefore the angle FAD is also equal to the angle FBC. [I. 8]  And since, again, AG was proved equal to BH, and further FA also equal to FB, the two sides FA, AG are equal to the two sides FB, BH.  And the angle FAG was proved equal to the angle FBH;  therefore the base FG is equal to the base FH. [I. 4]  Now since, again, GE was proved equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF;  and the base FG is equal to the base FH;  therefore the angle GEF is equal to the angle HEF. [I. 8]  Therefore each of the angles GEF, HEF is right.  Therefore FE is at right angles to GH drawn at random through E.  Similarly we can prove that FE will also make right angles with all the straight lines which meet it and are in the plane of reference.  But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane;  [XI. Def. 3] therefore FE is at right angles to the plane of reference.  But the plane of reference is the plane through the straight lines AB, CD.  Therefore FE is at right angles to the plane through AB, CD. 
                                                       
                                                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 5. 
 
 
If a straight line be set up at right angles to three straight lines which meet one another, at their common point of section, the three straight lines are in one plane. 
 
 
For let a straight line AB be set up at right angles to the three straight lines BC, BD, BE, at their point of meeting at B;  I say that BC, BD, BE are in one plane. 
   
   
For suppose they are not, but, if possible, let BD, BE be in the plane of reference and BC in one more elevated;  let the plane through AB, BC be produced;  it will thus make, as common section in the plane of reference, a straight line. [XI. 3]  Let it make BF.  Therefore the three straight lines AB, BC, BF are in one plane, namely that drawn through AB, BC.  Now, since AB is at right angles to each of the straight lines BD, BE,  therefore AB is also at right angles to the plane through BD, BE. [XI. 4]  But the plane through BD, BE is the plane of reference;  therefore AB is at right angles to the plane of reference.  Thus AB will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3]  But BF which is in the plane of reference meets it;  therefore the angle ABF is right.  But, by hypothesis, the angle ABC is also right;  therefore the angle ABF is equal to the angle ABC.  And they are in one plane: which is impossible.  Therefore the straight line BC is not in a more elevated plane;  therefore the three straight lines BC, BD, BE are in one plane. 
                                 
                                 
Therefore, if a straight line be set up at right angles to three straight lines, at their point of meeting, the three straight lines are in one plane.  Q. E. D. 
   
   
PROPOSITION 6. 
 
 
If two straight lines be at right angles to the same plane, the straight lines will be parallel. 
 
 
For let the two straight lines AB, CD be at right angles to the plane of reference;  I say that AB is parallel to CD. 
   
   
For let them meet the plane of reference at the points B, D, let the straight line BD be joined, let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined. 
 
 
Now, since AB is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3]  But each of the straight lines BD, BE is in the plane of reference and meets AB;  therefore each of the angles ABD, ABE is right.  For the same reason each of the angles CDB, CDE is also right.  And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB;  and they include right angles;  therefore the base AD is equal to the base BE. [I. 4]  And, since AB is equal to DE, while AD is also equal to BE, the two sides AB, BE are equal to the two sides ED, DA;  and AE is their common base;  therefore the angle ABE is equal to the angle EDA. [I. 8]  But the angle ABE is right;  therefore the angle EDA is also right;  therefore ED is at right angles to DA.  But it is also at right angles to each of the straight lines BD, DC;  therefore ED is set up at right angles to the three straight lines BD, DA, DC at their point of meeting;  therefore the three straight lines BD, DA, DC are in one plane. [XI. 5]  But, in whatever plane DB, DA are, in that plane is AB also,  for every triangle is in one plane; [XI. 2]  therefore the straight lines AB, BD, DC are in one plane.  And each of the angles ABD, BDC is right;  therefore AB is parallel to CD. [I. 28] 
                                         
                                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 7. 
 
 
If two straight lines be parallel and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallel straight lines. 
 
 
Let AB, CD be two parallel straight lines, and let points E, F be taken at random on them respectively;  I say that the straight line joining the points E, F is in the same plane with the parallel straight lines. 
   
   
For suppose it is not, but, if possible, let it be in a more elevated plane as EGF, and let a plane be drawn through EGF;  it will then make, as section in the plane of reference, a straight line. [XI. 3]  Let it make it, as EF;  therefore the two straight lines EGF, EF will enclose an area: which is impossible.  Therefore the straight line joined from E to F is not in a plane more elevated;  therefore the straight line joined from E to F is in the plane through the parallel straight lines AB, CD. 
           
           
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 8. 
 
 
If two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane. 
 
 
Let AB, CD be two parallel straight lines, and let one of them, AB, be at right angles to the plane of reference;  I say that the remaining one, CD, will also be at right angles to the same plane. 
   
   
For let AB, CD meet the plane of reference at the points B, D, and let BD be joined;  therefore AB, CD, BD are in one plane. [XI. 7]  Let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined.  Now, since AB is at right angles to the plane of reference,  therefore AB is also at right angles to all the straight lines which meet it and are in the plane of reference;  [XI. Def. 3] therefore each of the angles ABD, ABE is right.  And, since the straight line BD has fallen on the parallels AB, CD,  therefore the angles ABD, CDB are equal to two right angles. [I. 29]  But the angle ABD is right;  therefore the angle CDB is also right;  therefore CD is at right angles to BD.  And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB;  and the angle ABD is equal to the angle EDB,  for each is right;  therefore the base AD is equal to the base BE.  And, since AB is equal to DE, and BE to AD, the two sides AB, BE are equal to the two sides ED, DA respectively,  and AE is their common base;  therefore the angle ABE is equal to the angle EDA.  But the angle ABE is right;  therefore the angle EDA is also right;  therefore ED is at right angles to AD.  But it is also at right angles to DB;  therefore ED is also at right angles to the plane through BD, DA. [XI. 4]  Therefore ED will also make right angles with all the straight lines which meet it and are in the plane through BD, DA.  But DC is in the plane through BD, DA, inasmuch as AB, BD are in the plane through BD, DA, [XI. 2] and DC is also in the plane in which AB, BD are.  Therefore ED is at right angles to DC,  so that CD is also at right angles to DE.  But CD is also at right angles to BD.  Therefore CD is set up at right angles to the two straight lines DE, DB which cut one another, from the point of section at D;  so that CD is also at right angles to the plane through DE, DB. [XI. 4]  But the plane through DE, DB is the plane of reference;  therefore CD is at right angles to the plane of reference. 
                                                               
                                                               
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 9. 
 
 
Straight lines which are parallel to the same straight line and are not in the same plane with it are also parallel to one another. 
 
 
For let each of the straight lines AB, CD be parallel to EF, not being in the same plane with it;  I say that AB is parallel to CD. 
   
   
For let a point G be taken at random on EF, and from it let there be drawn GH, in the plane through EF, AB, at right angles to EF, and GK in the plane through FE, CD again at right angles to EF.  Now, since EF is at right angles to each of the straight lines GH, GK,  therefore EF is also at right angles to the plane through GH, GK. [XI. 4]  And EF is parallel to AB;  therefore AB is also at right angles to the plane through HG, GK. [XI. 8]  For the same reason CD is also at right angles to the plane through HG, GK;  therefore each of the straight lines AB, CD is at right angles to the plane through HG, GK.  But if two straight lines be at right angles to the same plane, the straight lines are parallel; [XI. 6]  therefore AB is parallel to CD.   
                   
                   
PROPOSITION 10. 
 
 
If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane, they will contain equal angles. 
 
 
For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not in the same plane;  I say that the angle ABC is equal to the angle DEF. 
   
   
For let BA, BC, ED, EF be cut off equal to one another, and let AD, CF, BE, AC, DF be joined. 
 
 
Now, since BA is equal and parallel to ED,  therefore AD is also equal and parallel to BE. [I. 33]  For the same reason CF is also equal and parallel to BE.  Therefore each of the straight lines AD, CF is equal and parallel to BE.  But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another; [XI. 9]  therefore AD is parallel and equal to CF.  And AC, DF join them;  therefore AC is also equal and parallel to DF. [I. 33]  Now, since the two sides AB, BC are equal to the two sides DE, EF, and the base AC is equal to the base DF,  therefore the angle ABC is equal to the angle DEF. [I. 8] 
                   
                   
Therefore etc.   
   
   
PROPOSITION 11. 
 
 
From a given elevated point to draw a straight line perpendicular to a given plane. 
 
 
Let A be the given elevated point, and the plane of reference the given plane;  thus it is required to draw from the point A a straight line perpendicular to the plane of reference. 
   
   
Let any straight line BC be drawn, at random, in the plane of reference, and let AD be drawn from the point A perpendicular to BC. [I. 12]  If then AD is also perpendicular to the plane of reference, that which was enjoined will have been done.  But, if not, let DE be drawn from the point D at right angles to BC and in the plane of reference, [I. 11] let AF be drawn from A perpendicular to DE, [I. 12] and let GH be drawn through the point F parallel to BC. [I. 31] 
     
     
Now, since BC is at right angles to each of the straight lines DA, DE,  therefore BC is also at right angles to the plane through ED, DA. [XI. 4]  And GH is parallel to it;  but, if two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane; [XI. 8]  therefore GH is also at right angles to the plane through ED, DA.  Therefore GH is also at right angles to all the straight lines which meet it and are in the plane through ED, DA. [XI. Def. 3]  But AF meets it and is in the plane through ED, DA;  therefore GH is at right angles to FA,  so that FA is also at right angles to GH.  But AF is also at right angles to DE;  therefore AF is at right angles to each of the straight lines GH, DE.  But, if a straight line be set up at right angles to two straight lines which cut one another, at the point of section, it will also be at right angles to the plane through them; [XI. 4]  therefore FA is at right angles to the plane through ED, GH.  But the plane through ED, GH is the plane of reference;  therefore AF is at right angles to the plane of reference. 
                             
                             
Therefore from the given elevated point A the straight line AF has been drawn perpendicular to the plane of reference.  Q. E. F. 
   
   
PROPOSITION 12. 
 
 
To set up a straight line at right angles to a given plane from a given point in it. 
 
 
Let the plane of reference be the given plane, and A the point in it;  thus it is required to set up from the point A a straight line at right angles to the plane of reference. 
   
   
Let any elevated point B be conceived, from B let BC be drawn perpendicular to the plane of reference, [XI. 11] and through the point A let AD be drawn parallel to BC. [I. 31] 
 
 
Then, since AD, CB are two parallel straight lines, while one of them, BC, is at right angles to the plane of reference,  therefore the remaining one, AD, is also at right angles to the plane of reference. [XI. 8] 
   
   
Therefore AD has been set up at right angles to the given plane from the point A in it.   
   
   
PROPOSITION 13. 
 
 
From the same point two straight lines cannot be set up at right angles to the same plane on the same side.  For, if possible, from the same point A let the two straight lines AB, AC be set up at right angles to the plane of reference and on the same side, and let a plane be drawn through BA, AC;  it will then make, as section through A in the plane of reference, a straight line. [XI. 3]  Let it make DAE;  therefore the straight lines AB, AC, DAE are in one plane.  And, since CA is at right angles to the plane of reference,  it will also make right angles with all the straight lines which meet it and are in the plane of reference. [XI. Def. 3]  But DAE meets it and is in the plane of reference;  therefore the angle CAE is right.  For the same reason the angle BAE is also right;  therefore the angle CAE is equal to the angle BAE.  And they are in one plane: which is impossible. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 14. 
 
 
Planes to which the same straight line is at right angles will be parallel. 
 
 
For let any straight line AB be at right angles to each of the planes CD, EF;  I say that the planes are parallel. 
   
   
For, if not, they will meet when produced.  Let them meet; they will then make, as common section, a straight line. [XI. 3]  Let them make GH; let a point K be taken at random on GH, and let AK, BK be joined. 
     
     
Now, since AB is at right angles to the plane EF,  therefore AB is also at right angles to BK which is a straight line in the plane EF produced; [XI. Def. 3]  therefore the angle ABK is right.  For the same reason the angle BAK is also right.  Thus, in the triangle ABK, the two angles ABK, BAK are equal to two right angles: which is impossible. [I. 17]  Therefore the planes CD, EF will not meet when produced;  therefore the planes CD, EF are parallel. [XI. Def. 8] 
             
             
Therefore planes to which the same straight line is at right angles are parallel.  Q. E. D. 
   
   
PROPOSITION 15. 
 
 
If two straight lines meeting one another be parallel to two straight lines meeting one another, not being in the same plane, the planes through them are parallel. 
 
 
For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not being in the same plane;  I say that the planes produced through AB, BC and DE, EF will not meet one another. 
   
   
For let BG be drawn from the point B perpendicular to the plane through DE, EF, [XI. 11]  and let it meet the plane at the point G;  through G let GH be drawn parallel to ED, and GK parallel to EF. [I. 31] 
     
     
Now, since BG is at right angles to the plane through DE, EF,  therefore it will also make right angles with all the straight lines which meet it and are in the plane through DE, EF. [XI. Def. 3]  But each of the straight lines GH, GK meets it and is in the plane through DE, EF;  therefore each of the angles BGH, BGK is right.  And, since BA is parallel to GH, [XI. 9] therefore the angles GBA, BGH are equal to two right angles. [I. 29]  But the angle BGH is right;  therefore the angle GBA is also right;  therefore GB is at right angles to BA.  For the same reason GB is also at right angles to BC.  Since then the straight line GB is set up at right angles to the two straight lines BA, BC which cut one another,  therefore GB is also at right angles to the plane through BA, BC. [XI. 4]          But planes to which the same straight line is at right angles are parallel; [XI. 14]  therefore the plane through AB, BC is parallel to the plane through DE, EF. 
                                 
                                 
Therefore, if two straight lines meeting one another be parallel to two straight lines meeting one another, not in the same plane, the planes through them are parallel.  Q. E. D. 
   
   
PROPOSITION 16. 
 
 
If two parallel planes be cut by any plane, their common sections are parallel. 
 
 
For let the two parallel planes AB, CD be cut by the plane EFGH, and let EF, GH be their common sections;  I say that EF is parallel to GH. 
   
   
For, if not, EF, GH will, when produced, meet either in the direction of F, H or of E, G.  Let them be produced, as in the direction of F, H, and let them, first, meet at K.  Now, since EFK is in the plane AB,  therefore all the points on EFK are also in the plane AB. [XI. 1]  But K is one of the points on the straight line EFK;  therefore K is in the plane AB.  For the same reason K is also in the plane CD;  therefore the planes AB, CD will meet when produced.  But they do not meet, because they are, by hypothesis, parallel;  therefore the straight lines EF, GH will not meet when produced in the direction of F, H.  Similarly we can prove that neither will the straight lines EF, GH meet when produced in the direction of E, G.  But straight lines which do not meet in either direction are parallel. [I. Def. 23]  Therefore EF is parallel to GH. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 17. 
 
 
If two straight lines be cut by parallel planes, they will be cut in the same ratios. 
 
 
For let the two straight lines AB, CD be cut by the parallel planes GH, KL, MN at the points A, E, B and C, F, D;  I say that, as the straight line AE is to EB, so is CF to FD. 
   
   
For let AC, BD, AD be joined, let AD meet the plane KL at the point O, and let EO, OF be joined. 
 
 
Now, since the two parallel planes KL, MN are cut by the plane EBDO, their common sections EO, BD are parallel. [XI. 16]  For the same reason, since the two parallel planes GH, KL are cut by the plane AOFC, their common sections AC, OF are parallel. [id.]  And, since the straight line EO has been drawn parallel to BD, one of the sides of the triangle ABD, therefore, proportionally, as AE is to EB, so is AO to OD. [VI. 2]  Again, since the straight line OF has been drawn parallel to AC, one of the sides of the triangle ADC, proportionally, as AO is to OD, so is CF to FD. [id.]  But it was also proved that, as AO is to OD, so is AE to EB;  therefore also, as AE is to EB, so is CF to FD. [V. 11] 
           
           
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 18. 
 
 
If a straight line be at right angles to any plane, all the planes through it will also be at right angles to the same plane. 
 
 
For let any straight line AB be at right angles to the plane of reference;  I say that all the planes through AB are also at right angles to the plane of reference. 
   
   
For let the plane DE be drawn through AB, let CE be the common section of the plane DE and the plane of reference, let a point F be taken at random on CE, and from F let FG be drawn in the plane DE at right angles to CE. [I. 11] 
 
 
Now, since AB is at right angles to the plane of reference, AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3]  so that it is also at right angles to CE;  therefore the angle ABF is right.  But the angle GFB is also right;  therefore AB is parallel to FG. [I. 28]  But AB is at right angles to the plane of reference;  therefore FG is also at right angles to the plane of reference. [XI. 8]  Now a plane is at right angles to a plane, when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane. [XI. Def. 4]  And FG, drawn in one of the planes DE at right angles to CE, the common section of the planes, was proved to be at right angles to the plane of reference;  therefore the plane DE is at right angles to the plane of reference.  Similarly also it can be proved that all the planes through AB are at right angles to the plane of reference. 
                     
                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 19. 
 
 
If two planes which cut one another be at right angles to any plane, their common section will also be at right angles to the same plane. 
 
 
For let the two planes AB, BC be at right angles to the plane of reference, and let BD be their common section;  I say that BD is at right angles to the plane of reference. 
   
   
For suppose it is not, and from the point D let DE be drawn in the plane AB at right angles to the straight line AD, and DF in the plane BC at right angles to CD. 
 
 
Now, since the plane AB is at right angles to the plane of reference, and DE has been drawn in the plane AB at right angles to AD, their common section,  therefore DE is at right angles to the plane of reference. [XI. Def. 4]  Similarly we can prove that DF is also at right angles to the plane of reference.  Therefore from the same point D two straight lines have been set up at right angles to the plane of reference on the same side: which is impossible. [XI. 13]  Therefore no straight line except the common section DB of the planes AB, BC can be set up from the point D at right angles to the plane of reference. 
         
         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 20. 
 
 
If a solid angle be contained by three plane angles, any two, taken together in any manner, are greater than the remaining one. 
 
 
For let the solid angle at A be contained by the three plane angles BAC, CAD, DAB;  I say that any two of the angles BAC, CAD, DAB, taken together in any manner, are greater than the remaining one. 
   
   
If now the angles BAC, CAD, DAB are equal to one another, it is manifest that any two are greater than the remaining one.  But, if not, let BAC be greater, and on the straight line AB, and at the point A on it, let the angle BAE be constructed, in the plane through BA, AC, equal to the angle DAB;  let AE be made equal to AD, and let BEC, drawn across through the point E, cut the straight lines AB, AC at the points B, C; let DB, DC be joined.  Now, since DA is equal to AE, and AB is common, two sides are equal to two sides;  and the angle DAB is equal to the angle BAE;  therefore the base DB is equal to the base BE. [I. 4]  And, since the two sides BD, DC are greater than BC, [I. 20] and of these DB was proved equal to BE,  therefore the remainder DC is greater than the remainder EC. 
               
               
Now, since DA is equal to AE, and AC is common, and the base DC is greater than the base EC,  therefore the angle DAC is greater than the angle EAC. [I. 25]  But the angle DAB was also proved equal to the angle BAE;  therefore the angles DAB, DAC are greater than the angle BAC.  Similarly we can prove that the remaining angles also, taken together two and two, are greater than the remaining one. 
         
         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 21. 
 
 
Any solid angle is contained by plane angles less than four right angles. 
 
 
Let the angle at A be a solid angle contained by the plane angles BAC, CAD, DAB;  I say that the angles BAC, CAD, DAB are less than four right angles. 
   
   
For let points B, C, D be taken at random on the straight lines AB, AC, AD respectively, and let BC, CD, DB be joined.  Now, since the solid angle at B is contained by the three plane angles CBA, ABD, CBD, any two are greater than the remaining one; [XI. 20]  therefore the angles CBA, ABD are greater than the angle CBD.  For the same reason the angles BCA, ACD are also greater than the angle BCD, and the angles CDA, ADB are greater than the angle CDB;  therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles CBD, BCD, CDB.  But the three angles CBD, BDC, BCD are equal to two right angles; [I. 32]  therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles.  And, since the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles,  therefore the nine angles of the three triangles, the angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, BAD are equal to six right angles;  and of them the six angles ABC, BCA, ACD, CDA, ADB, DBA are greater than two right angles;  therefore the remaining three angles BAC, CAD, DAB containing the solid angle are less than four right angles. 
                     
                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 22. 
 
 
If there be three plane angles of which two, taken together in any manner, are greater than the remaining one, and they are contained by equal straight lines, it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines. 
 
 
Let there be three plane angles ABC, DEF, GHK, of which two, taken together in any manner, are greater than the remaining one, namely the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles GHK, ABC greater than the angle DEF; let the straight lines AB, BC, DE, EF, GH, HK be equal, and let AC, DF, GK be joined;  I say that it is possible to construct a triangle out of straight lines equal to AC, DF, GK, that is, that any two of the straight lines AC, DF, GK are greater than the remaining one. 
   
   
Now, if the angles ABC, DEF, GHK are equal to one another, it is manifest that, AC, DF, GK being equal also, it is possible to construct a triangle out of straight lines equal to AC, DF, GK.  But, if not, let them be unequal, and on the straight line HK, and at the point H on it, let the angle KHL be constructed equal to the angle ABC;  let HL be made equal to one of the straight lines AB, BC, DE, EF, GH, HK, and let KL, GL be joined.  Now, since the two sides AB, BC are equal to the two sides KH, HL, and the angle at B is equal to the angle KHL,  therefore the base AC is equal to the base KL. [I. 4]  And, since the angles ABC, GHK are greater than the angle DEF, while the angle ABC is equal to the angle KHL,  therefore the angle GHL is greater than the angle DEF.  And, since the two sides GH, HL are equal to the two sides DE, EF, and the angle GHL is greater than the angle DEF,  therefore the base GL is greater than the base DF. [I. 24]  But GK, KL are greater than GL.  Therefore GK, KL are much greater than DF.  But KL is equal to AC;  therefore AC, GK are greater than the remaining straight line DF.  Similarly we can prove that AC, DF are greater than GK, and further DF, GK are greater than AC. 
                           
                           
Therefore it is possible to construct a triangle out of straight lines equal to AC, DF, GK.  Q. E. D. 
   
   
PROPOSITION 23. 
 
 
To construct a solid angle out of three plane angles two of which, taken together in any manner, are greater than the remaining one: thus the three angles must be less than four right angles. 
 
 
Let the angles ABC, DEF, GHK be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles;  thus it is required to construct a solid angle out of angles equal to the angles ABC, DEF, GHK. 
   
   
Let AB, BC, DE, EF, GH, HK be cut off equal to one another, and let AC, DF, GK be joined;  it is therefore possible to construct a triangle out of straight lines equal to AC, DF, GK. [XI. 22]  Let LMN be so constructed that AC is equal to LM, DF to MN, and further GK to NL, let the circle LMN be described about the triangle LMN, let its centre be taken, and let it be O; let LO, MO, NO be joined;  I say that AB is greater than LO. 
       
       
For, if not, AB is either equal to LO, or less.  First, let it be equal.  Then, since AB is equal to LO, while AB is equal to BC, and OL to OM, the two sides AB, BC are equal to the two sides LO, OM respectively;  and, by hypothesis, the base AC is equal to the base LM;  therefore the angle ABC is equal to the angle LOM. [I. 8]  For the same reason the angle DEF is also equal to the angle MON, and further the angle GHK to the angle NOL;  therefore the three angles ABC, DEF, GHK are equal to the three angles LOM, MON, NOL.  But the three angles LOM, MON, NOL are equal to four right angles;  therefore the angles ABC, DEF, GHK are equal to four right angles.  But they are also, by hypothesis, less than four right angles: which is absurd.  Therefore AB is not equal to LO. 
                     
                     
I say next that neither is AB less than LO. 
 
 
For, if possible, let it be so, and let OP be made equal to AB, and OQ equal to BC, and let PQ be joined.  Then, since AB is equal to BC, OP is also equal to OQ, so that the remainder LP is equal to QM.  Therefore LM is parallel to PQ, [VI. 2] and LMO is equiangular with PQO; [I. 29]  therefore, as OL is to LM, so is OP to PQ; [VI. 4]  and alternately, as LO is to OP, so is LM to PQ. [V. 16]  But LO is greater than OP; therefore LM is also greater than PQ.  But LM was made equal to AC; therefore AC is also greater than PQ.  Since, then, the two sides AB, BC are equal to the two sides PO, OQ, and the base AC is greater than the base PQ,  therefore the angle ABC is greater than the angle POQ. [I. 25]  Similarly we can prove that the angle DEF is also greater than the angle MON, and the angle GHK greater than the angle NOL.  Therefore the three angles ABC, DEF, GHK are greater than the three angles LOM, MON, NOL.  But, by hypothesis, the angles ABC, DEF, GHK are less than four right angles;  therefore the angles LOM, MON, NOL are much less than four right angles.  But they are also equal to four right angles: which is absurd.  Therefore AB is not less than LO.  And it was proved that neither is it equal;  therefore AB is greater than LO. 
                                 
                                 
Let then OR be set up from the point O at right angles to the plane of the circle LMN, [XI. 12] and let the square on OR be equal to that area by which the square on AB is greater than the square on LO; let RL, RM, RN be joined. 
 
 
Then, since RO is at right angles to the plane of the circle LMN,  therefore RO is also at right angles to each of the straight lines LO, MO, NO.  And, since LO is equal to OM, while OR is common and at right angles,  therefore the base RL is equal to the base RM. [I. 4]  For the same reason RN is also equal to each of the straight lines RL, RM;  therefore the three straight lines RL, RM, RN are equal to one another.  Next, since by hypothesis the square on OR is equal to that area by which the square on AB is greater than the square on LO,  therefore the square on AB is equal to the squares on LO, OR.  But the square on LR is equal to the squares on LO, OR,  for the angle LOR is right; [I. 47]  therefore the square on AB is equal to the square on RL;  therefore AB is equal to RL.  But each of the straight lines BC, DE, EF, GH, HK is equal to AB, while each of the straight lines RM, RN is equal to RL;  therefore each of the straight lines AB, BC, DE, EF, GH, HK is equal to each of the straight lines RL, RM, RN.  And, since the two sides LR, RM are equal to the two sides AB, BC, and the base LM is by hypothesis equal to the base AC,  therefore the angle LRM is equal to the angle ABC. [I. 8]  For the same reason the angle MRN is also equal to the angle DEF, and the angle LRN to the angle GHK. 
                                 
                                 
Therefore, out of the three plane angles LRM, MRN, LRN, which are equal to the three given angles ABC, DEF, GHK, the solid angle at R has been constructed, which is contained by the angles LRM, MRN, LRN.  Q. E. F. 
   
   
LEMMA.
But how it is possible to take the square on OR equal to that area by which the square on AB is greater than the square on LO, we can show as follows. 
Let the straight lines AB, LO be set out, and let AB be the greater;  let the semicircle ABC be described on AB, and into the semicircle ABC let AC be fitted equal to the straight line LO, not being greater than the diameter AB; [IV. 1]  let CB be joined  Since then the angle ACB is an angle in the semicircle ACB, therefore the angle ACB is right. [III. 31]  Therefore the square on AB is equal to the squares on AC, CB. [I. 47]  Hence the square on AB is greater than the square on AC by the square on CB.  But AC is equal to LO.  Therefore the square on AB is greater than the square on LO by the square on CB.  If then we cut off OR equal to BC, the square on AB will be greater than the square on LO by the square on OR.  Q. E. F. 
                     
                     
PROPOSITION 24. 
 
 
If a solid be contained by parallel planes, the opposite planes in it are equal and parallelogrammic. 
 
 
For let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, AE;  I say that the opposite planes in it are equal and parallelogrammic. 
   
   
For, since the two parallel planes BG, CE are cut by the plane AC, their common sections are parallel. [XI. 16]  Therefore AB is parallel to DC.  Again, since the two parallel planes BF, AE are cut by the plane AC, their common sections are parallel. [XI. 16]  Therefore BC is parallel to AD.  But AB was also proved parallel to DC;  therefore AC is a parallelogram.  Similarly we can prove that each of the planes DF, FG, GB, BF, AE is a parallelogram. 
             
             
Let AH, DF be joined.  Then, since AB is parallel to DC, and BH to CF, the two straight lines AB, BH which meet one another are parallel to the two straight lines DC, CF which meet one another, not in the same plane;  therefore they will contain equal angles; [XI. 10]  therefore the angle ABH is equal to the angle DCF.  And, since the two sides AB, BH are equal to the two sides DC, CF, [I. 34] and the angle ABH is equal to the angle DCF,  therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DCF. [I. 4]  And the parallelogram BG is double of the triangle ABH, and the parallelogram CE double of the triangle DCF; [I. 34]  therefore the parallelogram BG is equal to the parallelogram CE.  Similarly we can prove that AC is also equal to GF, and AE to BF. 
                 
                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 25. 
 
 
If a parallelepipedal solid be cut by a plane which is parallel to the opposite planes, then, as the base is to the base, so will the solid be to the solid. 
 
 
For let the parallelepipedal solid ABCD be cut by the plane FG which is parallel to the opposite planes RA, DH;  I say that, as the base AEFV is to the base EHCF, so is the solid ABFU to the solid EGCD. 
   
   
For let AH be produced in each direction, let any number of straight lines whatever, AK, KL, be made equal to AE, and any number whatever, HM, MN, equal to EH; and let the parallelograms LP, KV, HW, MS and the solids LQ, KR, DM, MT be completed. 
 
 
Then, since the straight lines LK, KA, AE are equal to one another, the parallelograms LP, KV, AF are also equal to one another, KO, KB, AG are equal to one another, and further LX, KQ, AR are equal to one another, for they are opposite. [XI. 24]  For the same reason the parallelograms EC, HW, MS are also equal to one another, HG, HI, IN are equal to one another, and further DH, MY, NT are equal to one another.  Therefore in the solids LQ, KR, AU three planes are equal to three planes.  But the three planes are equal to the three opposite;  therefore the three solids LQ, KR, AU are equal to one another.  For the same reason the three solids ED, DM, MT are also equal to one another.  Therefore, whatever multiple the base LF is of the base AF, the same multiple also is the solid LU of the solid AU.  For the same reason, whatever multiple the base NF is of the base FH, the same multiple also is the solid NU of the solid HU.  And, if the base LF is equal to the base NF, the solid LU is also equal to the solid NU;  if the base LF exceeds the base NF, the solid LU also exceeds the solid NU;  and, if one falls short, the other falls short.  Therefore, there being four magnitudes, the two bases AF, FH, and the two solids AU, UH, equimultiples have been taken of the base AF and the solid AU, namely the base LF and the solid LU, and equimultiples of the base HF and the solid HU, namely the base NF and the solid NU,  and it has been proved that, if the base LF exceeds the base FN, the solid LU also exceeds the solid NU, if the bases are equal, the solids are equal,  and if the base falls short, the solid falls short.  Therefore, as the base AF is to the base FH, so is the solid AU to the solid UH. [V. Def. 5]  Q. E. D. 
                               
                               
PROPOSITION 26. 
 
 
On a given straight line, and at a given point on it, to construct a solid angle equal to a given solid angle. 
 
 
Let AB be the given straight line, A the given point on it, and the angle at D, contained by the angles EDC, EDF, FDC, the given solid angle;  thus it is required to construct on the straight line AB, and at the point A on it, a solid angle equal to the solid angle at D. 
   
   
For let a point F be taken at random on DF, let FG be drawn from F perpendicular to the plane through ED, DC, and let it meet the plane at G, [XI. 11] let DG be joined, let there be constructed on the straight line AB and at the point A on it the angle BAL equal to the angle EDC, and the angle BAK equal to the angle EDG, [I. 23] let AK be made equal to DG, let KH be set up from the point K at right angles to the plane through BA, AL, [XI. 12] let KH be made equal to GF, and let HA be joined;  I say that the solid angle at A, contained by the angles BAL, BAH, HAL is equal to the solid angle at D contained by the angles EDC, EDF, FDC. 
   
   
For let AB, DE be cut off equal to one another, and let HB, KB, FE, GE be joined.  Then, since FG is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference; [XI. Def. 3]  therefore each of the angles FGD, FGE is right.  For the same reason each of the angles HKA, HKB is also right.  And, since the two sides KA, AB are equal to the two sides GD, DE respectively, and they contain equal angles,  therefore the base KB is equal to the base GE. [I. 4]  But KH is also equal to GF, and they contain right angles;  therefore HB is also equal to FE. [I. 4]  Again, since the two sides AK, KH are equal to the two sides DG, GF, and they contain right angles,  therefore the base AH is equal to the base FD. [I. 4]  But AB is also equal to DE;  therefore the two sides HA, AB are equal to the two sides DF, DE.  And the base HB is equal to the base FE;  therefore the angle BAH is equal to the angle EDF. [I. 8]  For the same reason the angle HAL is also equal to the angle FDC.                      And the angle BAL is also equal to the angle EDC. 
                                                   
                                                   
Therefore on the straight line AB, and at the point A on it, a solid angle has been constructed equal to the given solid angle at D.  Q. E. F. 
   
   
PROPOSITION 27. 
 
 
On a given straight line to describe a parallelepipedal solid similar and similarly situated to a given parallelepipedal solid. 
 
 
Let AB be the given straight line and CD the given parallelepipedal solid;  thus it is required to describe on the given straight line AB a parallelepipedal solid similar and similarly situated to the given parallelepipedal solid CD. 
   
   
For on the straight line AB and at the point A on it let the solid angle, contained by the angles BAH, HAK, KAB, be constructed equal to the solid angle at C, so that the angle BAH is equal to the angle ECF, the angle BAK equal to the angle ECG, and the angle KAH to the angle GCF;  and let it be contrived that, as EC is to CG, so is BA to AK, and, as GC is to CF, so is KA to AH. [VI. 12]  Therefore also, ex aequali, as EC is to CF, so is BA to AH. [V. 22]  Let the parallelogram HB and the solid AL be completed. 
       
       
Now since, as EC is to CG, so is BA to AK, and the sides about the equal angles ECG, BAK are thus proportional,  therefore the parallelogram GE is similar to the parallelogram KB.  For the same reason the parallelogram KH is also similar to the parallelogram GF, and further FE to HB;  therefore three parallelograms of the solid CD are similar to three parallelograms of the solid AL.  But the former three are both equal and similar to the three opposite parallelograms, and the latter three are both equal and similar to the three opposite parallelograms;  therefore the whole solid CD is similar to the whole solid AL. [XI. Def. 9] 
           
           
Therefore on the given straight line AB there has been described AL similar and similarly situated to the given parallelepipedal solid CD.  Q. E. F. 
   
   
PROPOSITION 28. 
 
 
If a parallelepipedal solid be cut by a plane through the diagonals of the opposite planes, the solid will be bisected by the plane. 
 
 
For let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF, DE of opposite planes;  I say that the solid AB will be bisected by the plane CDEF. 
   
   
For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH,  therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE;  for they are contained by planes equal both in multitude and in magnitude. [XI. Def. 10]  Hence the whole solid AB is bisected by the plane CDEF.  Q. E. D. 
         
         
PROPOSITION 29. 
 
 
Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are on the same straight lines, are equal to one another. 
 
 
Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AG, AF, LM, LN, CD, CE, BH, BK, be on the same straight lines FN, DK;  I say that the solid CM is equal to the solid CN. 
   
   
For, since each of the figures CH, CK is a parallelogram, CB is equal to each of the straight lines DH, EK, [I. 34]  hence DH is also equal to EK.  Let EH be subtracted from each;  therefore the remainder DE is equal to the remainder HK.  Hence the triangle DCE is also equal to the triangle HBK [I. 8, 4],  and the parallelogram DG to the parallelogram HN. [I. 36]  For the same reason the triangle AFG is also equal to the triangle MLN.  But the parallelogram CF is equal to the parallelogram BM, and CG to BN, for they are opposite;  therefore the prism contained by the two triangles AFG, DCE  and the three parallelograms AD, DG, CG is equal to the prism contained by the two triangles MLN, HBK  and the three parallelograms BM, HN, BN.  Let there be added to each the solid of which the parallelogram AB is the base and GEHM its opposite;  therefore the whole parallelepipedal solid CM is equal to the whole parallelepipedal solid CN. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 30. 
 
 
Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are not on the same straight lines, are equal to one another. 
 
 
Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AF, AG, LM, LN, CD, CE, BH, BK, not be on the same straight lines;  I say that the solid CM is equal to the solid CN. 
   
   
For let NK, DH be produced and meet one another at R, and further let FM, GE be produced to P, Q; let AO, LP, CQ, BR be joined.  Then the solid CM, of which the parallelogram ACBL is the base, and FDHM its opposite, is equal to the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite;  for they are on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AF, AO, LM, LP, CD, CQ, BH, BR, are on the same straight lines FP, DR. [XI. 29]  But the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite, is equal to the solid CN, of which the parallelogram ACBL is the base and GEKN its opposite;  for they are again on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AG, AO, CE, CQ, LN, LP, BK, BR, are on the same straight lines GQ, NR.  Hence the solid CM is also equal to the solid CN. 
           
           
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 31. 
 
 
Parallelepipedal solids which are on equal bases and of the same height are equal to one another. 
 
 
Let the parallelepipedal solids AE, CF, of the same height, be on equal bases AB, CD.  I say that the solid AE is equal to the solid CF. 
   
   
First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, RS, be at right angles to the bases AB, CD; let the straight line RT be produced in a straight line with CR; on the straight line RT, and at the point R on it, let the angle TRU be constructed equal to the angle ALB, [I. 23] let RT be made equal to AL, and RU equal to LB, and let the base RW and the solid XU be completed. 
 
 
Now, since the two sides TR, RU are equal to the two sides AL, LB, and they contain equal angles,  therefore the parallelogram RW is equal and similar to the parallelogram HL.  Since again AL is equal to RT, and LM to RS, and they contain right angles,  therefore the parallelogram RX is equal and similar to the parallelogram AM.  For the same reason LE is also equal and similar to SU;  therefore three parallelograms of the solid AE are equal and similar to three parallelograms of the solid XU.  But the former three are equal and similar to the three opposite, and the latter three to the three opposite; [XI. 24]  therefore the whole parallelepipedal solid AE is equal to the whole parallelepipedal solid XU. [XI. Def. 10]  Let DR, WU be drawn through and meet one another at Y, let aTb be drawn through T parallel to DY, let PD be produced to a, and let the solids YX, RI be completed.  Then the solid XY, of which the parallelogram RX is the base and Yc its opposite, is equal to the solid XU of which the parallelogram RX is the base and UV its opposite,  for they are on the same base RX and of the same height, and the extremities of their sides which stand up, namely RY, RU, Tb, TW, Se, Sd, Xc, XV, are on the same straight lines YW, eV. [XI. 29]  But the solid XU is equal to AE:  therefore the solid XY is also equal to the solid AE.  And, since the parallelogram RUWT is equal to the parallelogram YT for they are on the same base RT and in the same parallels RT, YW, [I. 35]  while RUWT is equal to CD, since it is also equal to AB,  therefore the parallelogram YT is also equal to CD.  But DT is another parallelogram;  therefore, as the base CD is to DT, so is YT to DT. [V. 7]  And, since the parallelepipedal solid CI has been cut by the plane RF which is parallel to opposite planes,  as the base CD is to the base DT, so is the solid CF to the solid RI. [XI. 25]  For the same reason, since the parallelepipedal solid YI has been cut by the plane RX which is parallel to opposite planes,  as the base YT is to the base TD, so is the solid YX to the solid RI. [XI. 25]  But, as the base CD is to DT, so is YT to DT;  therefore also, as the solid CF is to the solid RI, so is the solid YX to RI. [V. 11]  Therefore each of the solids CF, YX has to RI the same ratio;  therefore the solid CF is equal to the solid YX. [V. 9]  But YX was proved equal to AE;  therefore AE is also equal to CF. 
                                                       
                                                       
Next, let the sides standing up, AG, HK, BE, LM, CN, PQ, DF, RS, not be at right angles to the bases AB, CD;  I say again that the solid AE is equal to the solid CF. 
   
   
For from the points K, E, G, M, Q, F, N, S let KO, ET, GU, MV, QW, FX, NY, SI be drawn perpendicular to the plane of reference, and let them meet the plane at the points O, T, U, V, W, X, Y, I, and let OT, OU, UV, TV, WX, WY, YI, IX be joined.  Then the solid KV is equal to the solid QI,  for they are on the equal bases KM, QS and of the same height, and their sides which stand up are at right angles to their bases. [First part of this Prop.]  But the solid KV is equal to the solid AE, and QI to CF;  for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [XI. 30]  Therefore the solid AE is also equal to the solid CF. 
           
           
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 32. 
 
 
Parallelepipedal solids which are of the same height are to one another as their bases. 
 
 
Let AB, CD be parallelepipedal solids of the same height;  I say that the parallelepipedal solids AB, CD are to one another as their bases, that is, that, as the base AE is to the base CF, so is the solid AB to the solid CD. 
   
   
For let FH equal to AE be applied to FG, [I. 45] and on FH as base, and with the same height as that of CD, let the parallelepipedal solid GK be completed.  Then the solid AB is equal to the solid GK;  for they are on equal bases AE, FH and of the same height. [XI. 31]  And, since the parallelepipedal solid CK is cut by the plane DG which is parallel to opposite planes,  therefore, as the base CF is to the base FH, so is the solid CD to the solid DH. [XI. 25]  But the base FH is equal to the base AE, and the solid GK to the solid AB;  therefore also, as the base AE is to the base CF, so is the solid AB to the solid CD. 
             
             
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 33. 
 
 
Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides. 
 
 
Let AB, CD be similar parallelepipedal solids, and let AE be the side corresponding to CF;  I say that the solid AB has to the solid CD the ratio triplicate of that which AE has to CF. 
   
   
For let EK, EL, EM be produced in a straight line with AE, GE, HE, let EK be made equal to CF, EL equal to FN, and further EM equal to FR, and let the parallelogram KL and the solid KP be completed. 
 
 
Now, since the two sides KE, EL are equal to the two sides CF, FN, while the angle KEL is also equal to the angle CFN,  inasmuch as the angle AEG is also equal to the angle CFN because of the similarity of the solids AB, CD,  therefore the parallelogram KL is equal to the parallelogram CN.  For the same reason the parallelogram KM is also equal and similar to CR, and further EP to DF;  therefore three parallelograms of the solid KP are equal and similar to three parallelograms of the solid CD.  But the former three parallelograms are equal and similar to their opposites, and the latter three to their opposites; [XI. 24]  therefore the whole solid KP is equal and similar to the whole solid CD. [XI. Def. 10]  Let the parallelogram GK be completed, and on the parallelograms GK, KL as bases, and with the same height as that of AB, let the solids EO, LQ be completed.  Then since; owing to the similarity of the solids AB, CD, as AE is to CF, so is EG to FN, and EH to FR, while CF is equal to EK, FN to EL, and FR to EM,  therefore, as AE is to EK, so is GE to EL, and HE to EM.  But, as AE is to EK, so is AG to the parallelogram GK, as GE is to EL, so is GK to KL,  and, as HE is to EM, so is QE to KM; [VI. 1]  therefore also, as the parallelogram AG is to GK, so is GK to KL, and QE to KM.  But, as AG is to GK, so is the solid AB to the solid EO, as GK is to KL, so is the solid OE to the solid QL,  and, as QE is to KM, so is the solid QL to the solid KP; [XI. 32]  therefore also, as the solid AB is to EO, so is EO to QL, and QL to KP.  But, if four magnitudes be continuously proportional, the first has to the fourth the ratio triplicate of that which it has to the second; [V. Def. 10]  therefore the solid AB has to KP the ratio triplicate of that which AB has to EO.  But, as AB is to EO, so is the parallelogram AG to GK, and the straight line AE to EK [VI. 1];  hence the solid AB has also to KP the ratio triplicate of that which AE has to EK.  But the solid KP is equal to the solid CD, and the straight line EK to CF;  therefore the solid AB has also to the solid CD the ratio triplicate of that which the corresponding side of it, AE, has to the corresponding side CF. 
                                           
                                           
Therefore etc.  Q. E. D. 
   
   
PORISM.
From this it is manifest that, if four straight lines be proportional, as the first is to the fourth, so will a parallelepipedal solid on the first be to the similar and similarly described parallelepipedal solid on the second, inasmuch as the first has to the fourth the ratio triplicate of that which it has to the second. 
 
 
PROPOSITION 34. 
 
 
In equal parallelepipedal solids the bases are reciprocally proportional to the heights;  and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal. 
   
   
Let AB, CD be equal parallelepipedal solids;  I say that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB. 
   
   
First, let the sides which stand up, namely AG, EF, LB, HK, CM, NO, PD, QR, be at right angles to their bases;  I say that, as the base EH is to the base NQ, so is CM to AG. 
   
   
If now the base EH is equal to the base NQ, while the solid AB is also equal to the solid CD, CM will also be equal to AG.  For parallelepipedal solids of the same height are to one another as the bases; [XI. 32]    and, as the base EH is to NQ, so will CM be to AG, and it is manifest that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights. 
       
       
Next, let the base EH not be equal to the base NQ, but let EH be greater.  Now the solid AB is equal to the solid CD;  therefore CM is also greater than AG.  Let then CT be made equal to AG, and let the parallelepipedal solid VC be completed on NQ as base and with CT as height.  Now, since the solid AB is equal to the solid CD, and CV is outside them, while equals have to the same the same ratio, [V. 7]  therefore, as the solid AB is to the solid CV, so is the solid CD to the solid CV.  But, as the solid AB is to the solid CV, so is the base EH to the base NQ, for the solids AB, CV are of equal height; [XI. 32]  and, as the solid CD is to the solid CV, so is the base MQ to the base TQ [XI. 25] and CM to CT [VI. 1];  therefore also, as the base EH is to the base NQ, so is MC to CT.  But CT is equal to AG;  therefore also, as the base EH is to the base NQ, so is MC to AG.  Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights. 
                       
                       
Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB;  I say that the solid AB is equal to the solid CD. 
   
   
Let the sides which stand up be again at right angles to the bases.  Now, if the base EH is equal to the base NQ,  and, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB,  therefore the height of the solid CD is also equal to the height of the solid AB.  But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31]  therefore the solid AB is equal to the solid CD. 
           
           
Next, let the base EH not be equal to the base NQ, but let EH be greater;  therefore the height of the solid CD is also greater than the height of the solid AB, that is, CM is greater than AG.  Let CT be again made equal to AG, and let the solid CV be similarly completed.  Since, as the base EH is to the base NQ, so is MC to AG, while AG is equal to CT,  therefore, as the base EH is to the base NQ, so is CM to CT.  But, as the base EH is to the base NQ, so is the solid AB to the solid CV, for the solids AB, CV are of equal height; [XI. 32]  and, as CM is to CT, so is the base MQ to the base QT [VI. 1] and the solid CD to the solid CV. [XI. 25]  Therefore also, as the solid AB is to the solid CV, so is the solid CD to the solid CV;  therefore each of the solids AB, CD has to CV the same ratio.  Therefore the solid AB is equal to the solid CD. [V. 9] 
                   
                   
Now let the sides which stand up, FE, BL, GA, HK, ON, DP, MC, RQ, not be at right angles to their bases; let perpendiculars be drawn from the points F, G, B, K, O, M, D, R to the planes through EH, NQ, and let them meet the planes at S, T, U, V, W, X, Y, a, and let the solids FV, Oa be completed;  I say that, in this case too, if the solids AB, CD are equal, the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB. 
   
   
Since the solid AB is equal to the solid CD, while AB is equal to BT,  for they are on the same base FK and of the same height; [XI. 29, 30]  and the solid CD is equal to DX,  for they are again on the same base RO and of the same height; [id.]  therefore the solid BT is also equal to the solid DX.  Therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT. [Part 1.]  But the base FK is equal to the base EH, and the base OR to the base NQ;  therefore, as the base EH is to the base NQ, so is the height of the solid DX to the height of the solid BT.  But the solids DX, BT and the solids DC, BA have the same heights respectively;  therefore, as the base EH is to the base NQ, so is the height of the solid DC to the height of the solid AB.  Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights. 
                     
                     
Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB;  I say that the solid AB is equal to the solid CD. 
   
   
For, with the same construction, since, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB,  while the base EH is equal to the base FK, and NQ to OR,  therefore, as the base FK is to the base OR, so is the height of the solid CD to the height of the solid AB.  But the solids AB, CD and BT, DX have the same heights respectively;  therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT.  Therefore in the parallelepipedal solids BT, DX the bases are reciprocally proportional to the heights;  therefore the solid BT is equal to the solid DX. [Part 1.]  But BT is equal to BA,  for they are on the same base FK and of the same height; [XI. 29, 30]  and the solid DX is equal to the solid DC. [id.]  Therefore the solid AB is also equal to the solid CD.  Q. E. D. 
                       
                       
PROPOSITION 35. 
 
 
If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines be joined to the vertices of the original angles, they will contain, with the elevated straight lines, equal angles. 
 
 
Let the angles BAC, EDF be two equal rectilineal angles, and from the points A, D let the elevated straight lines AG, DM be set up containing, with the original straight lines, equal angles respectively, namely, the angle MDE to the angle GAB and the angle MDF to the angle GAC, let points G, M be taken at random on AG, DM, let GL, MN be drawn from the points G, M perpendicular to the planes through BA, AC and ED, DF, and let them meet the planes at L, N, and let LA, ND be joined;  I say that the angle GAL is equal to the angle MDN. 
   
   
Let AH be made equal to DM, and let HK be drawn through the point H parallel to GL.  But GL is perpendicular to the plane through BA, AC;  therefore HK is also perpendicular to the plane through BA, AC. [XI. 8]  From the points K, N let KC, NF, KB, NE be drawn perpendicular to the straight lines AC, DF, AB, DE, and let HC, CB, MF, FE be joined.  Since the square on HA is equal to the squares on HK, KA, and the squares on KC, CA are equal to the square on KA, [I. 47]  therefore the square on HA is also equal to the squares on HK, KC, CA.  But the square on HC is equal to the squares on HK, KC; [I. 47]  therefore the square on HA is equal to the squares on HC, CA.  Therefore the angle HCA is right. [I. 48]  For the same reason the angle DFM is also right.  Therefore the angle ACH is equal to the angle DFM.  But the angle HAC is also equal to the angle MDF.  Therefore MDF, HAC are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that subtending one of the equal angles, that is, HA equal to MD;  therefore they will also have the remaining sides equal to the remaining sides respectively. [I. 26]  Therefore AC is equal to DF. 
                             
                             
Similarly we can prove that AB is also equal to DE.                      Since then AC is equal to DF, and AB to DE, the two sides CA, AB are equal to the two sides FD, DE.  But the angle CAB is also equal to the angle FDE;  therefore the base BC is equal to the base EF, the triangle to the triangle, and the remaining angles to the remaining angles; [I. 4]  therefore the angle ACB is equal to the angle DFE.  But the right angle ACK is also equal to the right angle DFN;  therefore the remaining angle BCK is also equal to the remaining angle EFN.  For the same reason the angle CBK is also equal to the angle FEN.  Therefore BCK, EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, BC equal to EF;  therefore they will also have the remaining sides equal to the remaining sides. [I. 26]  Therefore CK is equal to FN.  But AC is also equal to DF;  therefore the two sides AC, CK are equal to the two sides DF, FN;  and they contain right angles.  Therefore the base AK is equal to the base DN. [I. 4]  And, since AH is equal to DM, the square on AH is also equal to the square on DM.  But the squares on AK, KH are equal to the square on AH, for the angle AKH is right; [I. 47]  and the squares on DN, NM are equal to the square on DM, for the angle DNM is right; [I. 47]  therefore the squares on AK, KH are equal to the squares on DN, NM; and of these the square on AK is equal to the square on DN;  therefore the remaining square on KH is equal to the square on NM;  therefore HK is equal to MN.  And, since the two sides HA, AK are equal to the two sides MD, DN respectively, and the base HK was proved equal to the base MN, therefore the angle HAK is equal to the angle MDN. [I. 8] 
                                                               
                                                               
Therefore etc.   
   
   
PORISM.
From this it is manifest that, if there be two equal plane angles, and if there be set up on them elevated straight lines which are equal and contain equal angles with the original straight lines respectively, the perpendiculars drawn from their extremities to the planes in which are the original angles are equal to one another. 
Q. E. D. 
   
   
PROPOSITION 36. 
 
 
If three straight lines be proportional, the parallelepipedal solid formed out of the three is equal to the parallelepipedal solid on the mean which is equilateral, but equiangular with the aforesaid solid. 
 
 
Let A, B, C be three straight lines in proportion, so that, as A is to B, so is B to C;  I say that the solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid. 
   
   
Let there be set out the solid angle at E contained by the angles DEG, GEF, FED, let each of the straight lines DE, GE, EF be made equal to B, and let the parallelepipedal solid EK be completed, let LM be made equal to A, and on the straight line LM, and at the point L on it, let there be constructed a solid angle equal to the solid angle at E, namely that contained by NLO, OLM, MLN; let LO be made equal to B, and LN equal to C.  Now, since, as A is to B, so is B to C, while A is equal to LM, B to each of the straight lines LO, ED, and C to LN, therefore, as LM is to EF, so is DE to LN.  Thus the sides about the equal angles NLM, DEF are reciprocally proportional;  therefore the parallelogram MN is equal to the parallelogram DF. [VI. 14]  And, since the angles DEF, NLM are two plane rectilineal angles, and on them the elevated straight lines LO, EG are set up which are equal to one another and contain equal angles with the original straight lines respectively,  therefore the perpendiculars drawn from the points G, O to the planes through NL, LM and DE, EF are equal to one another; [XI. 35, Por.]  hence the solids LH, EK are of the same height.  But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31]  therefore the solid HL is equal to the solid EK.  And LH is the solid formed out of A, B, C, and EK the solid on B;  therefore the parallelepipedal solid formed out of A, B, C is equal to the solid on B which is equilateral, but equiangular with the aforesaid solid.  Q. E. D. 
                       
                       
PROPOSITION 37. 
 
 
If four straight lines be proportional, the parallelepipedal solids on them which are similar and similarly described will also be proportional; and, if the parallelepipedal solids on them which are similar and similarly described be proportional, the straight lines will themselves also be proportional. 
 
 
Let AB, CD, EF, GH be four straight lines in proportion, so that, as AB is to CD, so is EF to GH; and let there be described on AB, CD, EF, GH the similar and similarly situated parallelepipedal solids KA, LC, ME, NG;  I say that, as KA is to LC, so is ME to NG. 
   
   
For, since the parallelepipedal solid KA is similar to LC,  therefore KA has to LC the ratio triplicate of that which AB has to CD. [XI. 33]  For the same reason ME also has to NG the ratio triplicate of that which EF has to GH. [id.]  And, as AB is to CD, so is EF to GH.  Therefore also, as AK is to LC, so is ME to NG. 
         
         
Next, as the solid AK is to the solid LC, so let the solid ME be to the solid NG;  I say that, as the straight line AB is to CD, so is EF to GH. 
   
   
For since, again, KA has to LC the ratio triplicate of that which AB has to CD, [XI. 33] and ME also has to NG the ratio triplicate of that which EF has to GH, [id.] and, as KA is to LC, so is ME to NG, therefore also, as AB is to CD, so is EF to GH. 
 
 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 38. 
 
 
If the sides of the opposite planes of a cube be bisected, and planes be carried through the points of section, the common section of the planes and the diameter of the cube bisect one another. 
 
 
For let the sides of the opposite planes CF, AH of the cube AF be bisected at the points K, L, M, N, O, Q, P, R, and through the points of section let the planes KN, OR be carried; let US be the common section of the planes, and DG the diameter of the cube AF.  I say that UT is equal to TS, and DT to TG. 
   
   
For let DU, UE, BS, SG be joined.  Then, since DO is parallel to PE, the alternate angles DOU, UPE are equal to one another. [I. 29]  And, since DO is equal to PE, and OU to UP, and they contain equal angles,  therefore the base DU is equal to the base UE, the triangle DOU is equal to the triangle PUE, and the remaining angles are equal to the remaining angles; [I. 4]  therefore the angle OUD is equal to the angle PUE.  For this reason DUE is a straight line. [I. 14]  For the same reason, BSG is also a straight line, and BS is equal to SG.  Now, since CA is equal and parallel to DB, while CA is also equal and parallel to EG,  therefore DB is also equal and parallel to EG. [XI. 9]  And the straight lines DE, BG join their extremities;  therefore DE is parallel to BG. [I. 33]  Therefore the angle EDT is equal to the angle BGT, for they are alternate; [I. 29]  and the angle DTU is equal to the angle GTS. [I. 15]  Therefore DTU, GTS are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, DU equal to GS,  for they are the halves of DE, BG;  therefore they will also have the remaining sides equal to the remaining sides. [I. 26]  Therefore DT is equal to TG, and UT to TS. 
                                 
                                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 39. 
 
 
If there be two prisms of equal height, and one have a parallelogram as base and the other a triangle, and if the parallelogram be double of the triangle, the prisms will be equal. 
 
 
Let ABCDEF, GHKLMN be two prisms of equal height, let one have the parallelogram AF as base, and the other the triangle GHK, and let the parallelogram AF be double of the triangle GHK;  I say that the prism ABCDEF is equal to the prism GHKLMN. 
   
   
For let the solids AO, GP be completed.  Since the parallelogram AF is double of the triangle GHK, while the parallelogram HK is also double of the triangle GHK, [I. 34] therefore the parallelogram AF is equal to the parallelogram HK.  But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [XI. 31]  therefore the solid AO is equal to the solid GP.  And the prism ABCDEF is half of the solid AO, and the prism GHKLMN is half of the solid GP; [XI. 28]  therefore the prism ABCDEF is equal to the prism GHKLMN. 
           
           
Therefore etc.  Q. E. D. 
   
   
BOOK XII. 
 
 
PROPOSITION 1. 
 
 
Similar polygons inscribed in circles are to one another as the squares on the diameters. 
 
 
Let ABC, FGH be circles, let ABCDE, FGHKL be similar polygons inscribed in them, and let BM, GN be diameters of the circles;  I say that, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL. 
   
   
For let BE, AM, GL, FN be joined.  Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL,  and, as BA is to AE, so is GF to FL. [VI. Def. I]  Thus BAE, GFL are two triangles which have one angle equal to one angle, namely the angle BAE to the angle GFL, and the sides about the equal angles proportional;  therefore the triangle ABE is equiangular with the triangle FGL. [VI. 6]  Therefore the angle AEB is equal to the angle FLG.  But the angle AEB is equal to the angle AMB,  for they stand on the same circumference; [III. 27]  and the angle FLG to the angle FNG;  therefore the angle AMB is also equal to the angle FNG.  But the right angle BAM is also equal to the right angle GFN; [III. 31]  therefore the remaining angle is equal to the remaining angle. [I. 32]  Therefore the triangle ABM is equiangular with the triangle FGN.  Therefore, proportionally, as BM is to GN, so is BA to GF. [VI. 4]  But the ratio of the square on BM to the square on GN is duplicate of the ratio of BM to GN,  and the ratio of the polygon ABCDE to the polygon FGHKL is duplicate of the ratio of BA to GF; [VI. 20]  therefore also, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL. 
                                 
                                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 2. 
 
 
Circles are to one another as the squares on the diameters. 
 
 
Let ABCD, EFGH be circles, and BD, FH their diameters;  I say that, as the circle ABCD is to the circle EFGH, so is the square on BD to the square on FH. 
   
   
For, if the square on BD is not to the square on FH as the circle ABCD is to the circle EFGH, then, as the square on BD is to the square on FH, so will the circle ABCD be either to some less area than the circle EFGH, or to a greater.  First, let it be in that ratio to a less area S.  Let the square EFGH be inscribed in the circle EFGH;  then the inscribed square is greater than the half of the circle EFGH,  inasmuch as, if through the points E, F, G, H we draw tangents to the circle,  the square EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square;  hence the inscribed square EFGH is greater than the half of the circle EFGH.  Let the circumferences EF, FG, GH, HE be bisected at the points K, L, M, N, and let EK, KF, FL, LG, GM, MH, HN, NE be joined;  therefore each of the triangles EKF, FLG, GMH, HNE is also greater than the half of the segment of the circle about it,  inasmuch as, if through the points K, L, M, N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, HE,  each of the triangles EKF, FLG, GMH, HNE will be half of the parallelogram about it,  while the segment about it is less than the parallelogram;  hence each of the triangles EKF, FLG, GMH, HNE is greater than the half of the segment of the circle about it.  Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S.  For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.  Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, NE be less than the excess by which the circle EFGH exceeds the area S.  Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.  Let there be inscribed, also, in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHN;  therefore, as the square on BD is to the square on FH, so is the polygon AOBPCQDR to the polygon EKFLGMHN. [XII. 1]  But, as the square on BD is to the square on FH, so also is the circle ABCD to the area S;  therefore also, as the circle ABCD is to the area S, so is the polygon AOBPCQDR to the polygon EKFLGMHN; [V. 11]  therefore, alternately, as the circle ABCD is to the polygon inscribed in it, so is the area S to the polygon EKFLGMHN. [V. 16]  But the circle ABCD is greater than the polygon inscribed in it;  therefore the area S is also greater than the polygon EKFLGMHN.  But it is also less: which is impossible.  Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area less than the circle EFGH.  Similarly we can prove that neither is the circle EFGH to any area less than the circle ABCD as the square on FH is to the square on BD. 
                                                     
                                                     
I say next that neither is the circle ABCD to any area greater than the circle EFGH as the square on BD is to the square on FH. 
 
 
For, if possible, let it be in that ratio to a greater area S.  Therefore, inversely, as the square on FH is to the square on DB, so is the area S to the circle ABCD.  But, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD;  therefore also, as the square on FH is to the square on BD, so is the circle EFGH to some area less than the circle ABCD: [V. 11]  which was proved impossible.  Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area greater than the circle EFGH.  And it was proved that neither is it in that ratio to any area less than the circle EFGH;  therefore, as the square on BD is to the square on FH, so is the circle ABCD to the circle EFGH. 
               
               
Therefore etc.  Q. E. D. 
   
   
LEMMA.
I say that, the area S being greater than the circle EFGH, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD. 
For let it be contrived that, as the area S is to the circle ABCD, so is the circle EFGH to the area T.  I say that the area T is less than the circle ABCD.  For since, as the area S is to the circle ABCD, so is the circle EFGH to the area T,  therefore, alternately, as the area S is to the circle EFGH, so is the circle ABCD to the area T. [V. 16]  But the area S is greater than the circle EFGH;  therefore the circle ABCD is also greater than the area T.  Hence, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD.  Q. E. D. 
                 
                 
PROPOSITION 3. 
 
 
Any pyramid which has a triangular base is divided into two pyramids equal and similar to one another, similar to the whole and having triangular bases, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid. 
 
 
Let there be a pyramid of which the triangle ABC is the base and the point D the vertex;  I say that the pyramid ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid. 
   
   
For let AB, BC, CA, AD, DB, DC be bisected at the points E, F, G, H, K, L, and let HE, EG, GH, HK, KL, LH, KF, FG be joined.  Since AE is equal to EB, and AH to DH, therefore EH is parallel to DB. [VI. 2]  For the same reason HK is also parallel to AB.  Therefore HEBK is a parallelogram;  therefore HK is equal to EB. [I. 34]  But EB is equal to EA;  therefore AE is also equal to HK.  But AH is also equal to HD;  therefore the two sides EA, AH are equal to the two sides KH, HD respectively,  and the angle EAH is equal to the angle KHD;  therefore the base EH is equal to the base KD. [I. 4]  Therefore the triangle AEH is equal and similar to the triangle HKD.  For the same reason the triangle AHG is also equal and similar to the triangle HLD.  Now, since two straight lines EH, HG meeting one another are parallel to two straight lines KD, DL meeting one another, and are not in the same plane, they will contain equal angles. [XI. 10]  Therefore the angle EHG is equal to the angle KDL.  And, since the two straight lines EH, HG are equal to the two KD, DL respectively, and the angle EHG is equal to the angle KDL,  therefore the base EG is equal to the base KL; [I. 4]  therefore the triangle EHG is equal and similar to the triangle KDL.  For the same reason the triangle AEG is also equal and similar to the triangle HKL.  Therefore the pyramid of which the triangle AEG is the base and the point H the vertex is equal and similar to the pyramid of which the triangle HKL is the base and the point D the vertex. [XI. Def. 10]  And, since HK has been drawn parallel to AB, one of the sides of the triangle ADB, the triangle ADB is equiangular to the triangle DHK, [I. 29] and they have their sides proportional;  therefore the triangle ADB is similar to the triangle DHK. [VI. Def. 1]  For the same reason the triangle DBC is also similar to the triangle DKL, and the triangle ADC to the triangle DLH.  Now, since the two straight lines BA, AC meeting one another are parallel to the two straight lines KH, HL meeting one another, not in the same plane, they will contain equal angles. [XI. 10]  Therefore the angle BAC is equal to the angle KHL.  And, as BA is to AC, so is KH to HL;  therefore the triangle ABC is similar to the triangle HKL.  Therefore also the pyramid of which the triangle ABC is the base and the point D the vertex is similar to the pyramid of which the triangle HKL is the base and the point D the vertex.  But the pyramid of which the triangle HKL is the base and the point D the vertex was proved similar to the pyramid of which the triangle AEG is the base and the point H the vertex.  Therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD. 
                                                           
                                                           
Next, since BF is equal to FC, the parallelogram EBFG is double of the triangle GFC.  And since, if there be two prisms of equal height, and one have a parallelogram as base, and the other a triangle, and if the parallelogram be double of the triangle, the prisms are equal, [XI. 39]  therefore the prism contained by the two triangles BKF, EHG, and the three parallelograms EBFG, EBKH, HKFG is equal to the prism contained by the two triangles GFC, HKL and the three parallelograms KFCL, LCGH, HKFG.  And it is manifest that each of the prisms, namely that in which the parallelogram EBFG is the base and the straight line HK is its opposite, and that in which the triangle GFC is the base and the triangle HKL its opposite, is greater than each of the pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices,  inasmuch as, if we join the straight lines EF, EK, the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle EBF is the base and the point K the vertex.  But the pyramid of which the triangle EBF is the base and the point K the vertex is equal to the pyramid of which the triangle AEG is the base and the point H the vertex;  for they are contained by equal and similar planes.  Hence also the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is greater than the pyramid of which the triangle AEG is the base and the point H the vertex.  But the prism in which the parallelogram EBFG is the base and the straight line HK its opposite is equal to the prism in which the triangle GFC is the base and the triangle HKL its opposite,  and the pyramid of which the triangle AEG is the base and the point H the vertex is equal to the pyramid of which the triangle HKL is the base and the point D the vertex.  Therefore the said two prisms are greater than the said two pyramids of which the triangles AEG, HKL are the bases and the points H, D the vertices. 
                     
                     
Therefore the whole pyramid, of which the triangle ABC is the base and the point D the vertex, has been divided into two pyramids equal to one another and into two equal prisms, and the two prisms are greater than the half of the whole pyramid.  Q. E. D. 
   
   
PROPOSITION 4. 
 
 
If there be two pyramids of the same height which have triangular bases, and each of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of the one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid. 
 
 
Let there be two pyramids of the same height which have the triangular bases ABC, DEF, and vertices the points G, H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [XII. 3]  I say that, as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH, 
   
   
For, since BO is equal to OC, and AL to LC,  therefore LO is parallel to AB, and the triangle ABC is similar to the triangle LOC.  For the same reason the triangle DEF is also similar to the triangle RVF.  And, since BC is double of CO, and EF of FV,  therefore, as BC is to CO, so is EF to FV.  And on BC, CO are described the similar and similarly situated rectilineal figures ABC, LOC, and on EF, FV the similar and similarly situated figures DEF, RVF;  therefore, as the triangle ABC is to the triangle LOC, so is the triangle DEF to the triangle RVF; [VI. 22]  therefore, alternately, as the triangle ABC is to the triangle DEF, so is the triangle LOC to the triangle RVF. [V. 16]  But, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite;  [Lemma following] therefore also, as the triangle ABC is to the triangle DEF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite.  But, as the said prisms are to one another, so is the prism in which the parallelogram KBOL is the base and the straight line PM its opposite to the prism in which the parallelogram QEVR is the base and the straight line ST its opposite. [XI. 39; cf. XII. 3]  Therefore also the two prisms, that in which the parallelogram KBOL is the base and PM its opposite, and that in which the triangle LOC is the base and PMN its opposite, are to the prisms in which QEVR is the base and the straight line ST its opposite and in which the triangle RVF is the base and STU its opposite in the same ratio [V. 12]  Therefore also, as the base ABC is to the base DEF, so are the said two prisms to the said two prisms. 
                         
                         
And similarly, if the pyramids PMNG, STUH be divided into two prisms and two pyramids, as the base PMN is to the base STU, so will the two prisms in the pyramid PMNG be to the two prisms in the pyramid STUH.  But, as the base PMN is to the base STU, so is the base ABC to the base DEF;  for the triangles PMN, STU are equal to the triangles LOC, RVF respectively.  Therefore also, as the base ABC is to the base DEF, so are the four prisms to the four prisms.  And similarly also, if we divide the remaining pyramids into two pyramids and into two prisms, then, as the base ABC is to the base DEF, so will all the prisms in the pyramid ABCG be to all the prisms, being equal in multitude, in the pyramid DEFH.  Q. E. D. 
           
           
LEMMA.
But that, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite, we must prove as follows. 
 
 
For in the same figure let perpendiculars be conceived drawn from G, H to the planes ABC, DEF;  these are of course equal because, by hypothesis, the pyramids are of equal height.  Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC, PMN,  they will be cut in the same ratios. [XI. 17]  And GC is bisected by the plane PMN at N;  therefore the perpendicular from G to the plane ABC will also be bisected by the plane PMN.  For the same reason the perpendicular from H to the plane DEF will also be bisected by the plane STU.  And the perpendiculars from G, H to the planes ABC, DEF are equal;  therefore the perpendiculars from the triangles PMN, STU to the planes ABC, DEF are also equal.  Therefore the prisms in which the triangles LOC, RVF are bases, and PMN, STU their opposites, are of equal height.  Hence also the parallelepipedal solids described from the said prisms are of equal height and are to one another as their bases; [XI. 32]  therefore their halves, namely the said prisms, are to one another as the base LOC is to the base RVF.  Q. E. D. 
                         
                         
PROPOSITION 5. 
 
 
Pyramids which are of the same height and have triangular bases are to one another as the bases. 
 
 
Let there be pyramids of the same height, of which the triangles ABC, DEF are the bases and the points G, H the vertices;  I say that, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH. 
   
   
For, if the pyramid ABCG is not to the pyramid DEFH as the base ABC is to the base DEF,  then, as the base ABC is to the base DEF, so will the pyramid ABCG be either to some solid less than the pyramid DEFH or to a greater.  Let it, first, be in that ratio to a less solid W, and let the pyramid DEFH be divided into two pyramids equal to one another and similar to the whole and into two equal prisms;  then the two prisms are greater than the half of the whole pyramid. [XII. 3]  Again, let the pyramids arising from the division be similarly divided,  and let this be done continually until there are left over from the pyramid DEFH some pyramids which are less than the excess by which the pyramid DEFH exceeds the solid W. [X. I]  Let such be left, and let them be, for the sake of argument, DQRS, STUH;  therefore the remainders, the prisms in the pyramid DEFH, are greater than the solid W.  Let the pyramid ABCG also be divided similarly, and a similar number of times, with the pyramid DEFH;  therefore, as the base ABC is to the base DEF, so are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH. [XII. 4]  But, as the base ABC is to the base DEF, so also is the pyramid ABCG to the solid W;  therefore also, as the pyramid ABCG is to the solid W, so are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH; [V. II]  therefore, alternately, as the pyramid ABCG is to the prisms in it, so is the solid W to the prisms in the pyramid DEFH. [V. 16]  But the pyramid ABCG is greater than the prisms in it;  therefore the solid W is also greater than the prisms in the pyramid DEFH.  But it is also less: which is impossible.  Therefore the prism9 ABCG is not to any solid less than the pyramid DEFH as the base ABC is to the base DEF.  Similarly it can be proved that neither is the pyramid DEFH to any solid less than the pyramid ABCG as the base DEF is to the base ABC. 
                                   
                                   
I say next that neither is the pyramid ABCG to any solid greater than the pyramid DEFH as the base ABC is to the base DEF. 
 
 
For, if possible, let it be in that ratio to a greater solid W;  therefore, inversely, as the base DEF is to the base ABC, so is the solid W to the pyramid ABCG.  But, as the solid W is to the solid ABCG, so is the pyramid DEFH to some solid less than the pyramid ABCG, as was before proved; [XII. 2, Lemma]  therefore also, as the base DEF is to the base ABC, so is the pyramid DEFH to some solid less than the pyramid ABCG: [V. II] which was proved absurd.  Therefore the pyramid ABCG is not to any solid greater than the pyramid DEFH as the base ABC is to the base DEF.  But it was proved that neither is it in that ratio to a less solid.  Therefore, as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH.  Q. E. D. 
               
               
PROPOSITION 6. 
 
 
Pyramids which are of the same height and have polygonal bases are to one another as the bases. 
 
 
Let there be pyramids of the same height of which the polygons ABCDE, FGHKL are the bases and the points M, N the vertices;  I say that, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN. 
   
   
For let AC, AD, FH, FK be joined.  Since then ABCM, ACDM are two pyramids which have triangular bases and equal height, they are to one another as the bases; [XII. 5]  therefore, as the base ABC is to the base ACD, so is the pyramid ABCM to the pyramid ACDM.  And, componendo, as the base ABCD is to the base ACD, so is the pyramid ABCDM to the pyramid ACDM. [V. 18]  But also, as the base ACD is to the base ADE, so is the pyramid ACDM to the pyramid ADEM. [XII. 5]  Therefore, ex aequali, as the base ABCD is to the base ADE, so is the pyramid ABCDM to the pyramid ADEM. [V. 22]  And again componendo, as the base ABCDE is to the base ADE, so is the pyramid ABCDEM to the pyramid ADEM. [V. 18]  Similarly also it can be proved that, as the base FGHKL is to the base FGH, so is the pyramid FGHKLN to the pyramid FGHN.  And, since ADEM, FGHN are two pyramids which have triangular bases and equal height,  therefore, as the base ADE is to the base FGH, so is the pyramid ADEM to the pyramid FGHN. [XII. 5]  But, as the base ADE is to the base ABCDE, so was the pyramid ADEM to the pyramid ABCDEM.  Therefore also, ex aequali, as the base ABCDE is to the base FGH, so is the pyramid ABCDEM to the pyramid FGHN. [V. 22]  But further, as the base FGH is to the base FGHKL, so also was the pyramid FGHN to the pyramid FGHKLN.  Therefore also, ex aequali, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN. [V. 22]  Q. E. D. 
                             
                             
PROPOSITION 7. 
 
 
Any prism which has a triangular base is divided into three pyramids equal to one another which have triangular bases. 
 
 
Let there be a prism in which the triangle ABC is the base and DEF its opposite;  I say that the prism ABCDEF is divided into three pyramids equal to one another, which have triangular bases. 
   
   
For let BD, EC, CD be joined.  Since ABED is a parallelogram, and BD is its diameter, therefore the triangle ABD is equal to the triangle EBD; [I. 34]  therefore also the pyramid of which the triangle ABD is the base and the point C the vertex is equal to the pyramid of which the triangle DEB is the base and the point C the vertex. [XII. 5]  But the pyramid of which the triangle DEB is the base and the point C the vertex is the same with the pyramid of which the triangle EBC is the base and the point D the vertex;  for they are contained by the same planes.  Therefore the pyramid of which the triangle ABD is the base and the point C the vertex is also equal to the pyramid of which the triangle EBC is the base and the point D the vertex.  Again, since FCBE is a parallelogram, and CE is its diameter, the triangle CEF is equal to the triangle CBE. [I. 34]  Therefore also the pyramid of which the triangle BCE is the base and the point D the vertex is equal to the pyramid of which the triangle ECF is the base and the point D the vertex. [XII. 5]  But the pyramid of which the triangle BCE is the base and the point D the vertex was proved equal to the pyramid of which the triangle ABD is the base and the point C the vertex;  therefore also the pyramid of which the triangle CEF is the base and the point D the vertex is equal to the pyramid of which the triangle ABD is the base and the point C the vertex;  therefore the prism ABC DEF has been divided into three pyramids equal to one another which have triangular bases. 
                     
                     
And, since the pyramid of which the triangle ABD is the base and the point C the vertex is the same with the pyramid of which the triangle CAB is the base and the point D the vertex, for they are contained by the same planes,  while the pyramid of which the triangle ABD is the base and the point C the vertex was proved to be a third of the prism in which the triangle ABC is the base and DEF its opposite,  therefore also the pyramid of which the triangle ABC is the base and the point D the vertex is a third of the prism which has the same base, the triangle ABC, and DEF as its opposite. 
     
     
PORISM.
From this it is manifest that any pyramid is a third part of the prism which has the same base with it and equal height. 
Q. E. D. 
   
   
PROPOSITION 8. 
 
 
Similar pyramids which have triangular bases are in the triplicate ratio of their corresponding sides. 
 
 
Let there be similar and similarly situated pyramids of which the triangles ABC, DEF, are the bases and the points G, H the vertices;  I say that the pyramid ABCG has to the pyramid DEFH the ratio triplicate of that which BC has to EF. 
   
   
For let the parallelepipedal solids BGML, EHQP be completed.  Now, since the pyramid ABCG is similar to the pyramid DEFH,  therefore the angle ABC is equal to the angle DEF, the angle GBC to the angle HEF, and the angle ABG to the angle DEH;  and, as AB is to DE, so is BC to EF, and BG to EH.  And since, as AB is to DE, so is BC to EF, and the sides are proportional about equal angles,  therefore the parallelogram BM is similar to the parallelogram EQ.  For the same reason BN is also similar to ER, and BK to EO;  therefore the three parallelograms MB, BK, BN are similar to the three EQ, EO, ER.  But the three parallelograms MB, BK, BN are equal and similar to their three opposites,  and the three EQ, EO, ER are equal and similar to their three opposites. [XI. 24]  Therefore the solids BGML, EHQP are contained by similar planes equal in multitude.  Therefore the solid BGML is similar to the solid EHQP.  But similar parallelepipedal solids are in the triplicate ratio of their corresponding sides. [XI. 33]  Therefore the solid BGML has to the solid EHQP the ratio triplicate of that which the corresponding side BC has to the corresponding side EF.  But, as the solid BGML is to the solid EHQP, so is the pyramid ABCG to the pyramid DEFH,  inasmuch as the pyramid is a sixth part of the solid, because the prism which is half of the parallelepipedal solid [XI. 28] is also triple of the pyramid. [XII. 7]  Therefore the pyramid ABCG also has to the pyramid DEFH the ratio triplicate of that which BC has to EF.  Q. E. D. 
                                   
                                   
PORISM.
From this it is manifest that similar pyramids which have polygonal bases are also to one another in the triplicate ratio of their corresponding sides. 
For, if they are divided into the pyramids contained in them which have triangular bases, by virtue of the fact that the similar polygons forming their bases are also divided into similar triangles equal in multitude and corresponding to the wholes, [VI. 20]  then, as the one pyramid which has a triangular base in the one complete pyramid is to the one pyramid which has a triangular base in the other complete pyramid,  so also will all the pyramids which have triangular bases contained in the one pyramid be to all the pyramids which have triangular bases contained in the other pyramid, [V. 12]  that is, the pyramid itself which has a polygonal base to the pyramid which has a polygonal base.  But the pyramid which has a triangular base is to the pyramid which has a triangular base in the triplicate ratio of the corresponding sides;  therefore also the pyramid which has a polygonal base has to the pyramid which has a similar base the ratio triplicate of that which the side has to the side. 
             
             
PROPOSITION 9. 
 
 
In equal pyramids which have triangular bases the bases are reciprocally proportional to the heights;  and those pyramids in which the bases are reciprocally proportional to the heights are equal. 
   
   
For let there be equal pyramids which have the triangular bases ABC, DEF and vertices the points G, H;  I say that in the pyramids ABCG, DEFH the bases are reciprocally proportional to the heights, that is, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG. 
   
   
For let the parallelepipedal solids BGML, EHQP be completed.  Now, since the pyramid ABCG is equal to the pyramid DEFH, and the solid BGML is six times the pyramid ABCG, and the solid EHQP six times the pyramid DEFH,  therefore the solid BGML is equal to the solid EHQP.  But in equal parallelepipedal solids the bases are reciprocally proportional to the heights; [XI. 34]  therefore, as the base BM is to the base EQ, so is the height of the solid EHQP to the height of the solid BGML.  But, as the base BM is to EQ, so is the triangle ABC to the triangle DEF. [I. 34]  Therefore also, as the triangle ABC is to the triangle DEF, so is the height of the solid EHQP to the height of the solid BGML. [V. 11]  But the height of the solid EHQP is the same with the height of the pyramid DEFH, and the height of the solid BGML is the same with the height of the pyramid ABCG,  therefore, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG.  Therefore in the pyramids ABCG, DEFH the bases are reciprocally proportional to the heights. 
                   
                   
Next, in the pyramids ABCG, DEFH let the bases be reciprocally proportional to the heights;  that is, as the base ABC is to the base DEF, so let the height of the pyramid DEFH be to the height of the pyramid ABCG;  I say that the pyramid ABCG is equal to the pyramid DEFH. 
     
     
For, with the same construction, since, as the base ABC is to the base DEF, so is the height of the pyramid DEFH to the height of the pyramid ABCG,  while, as the base ABC is to the base DEF, so is the parallelogram BM to the parallelogram EQ,  therefore also, as the parallelogram BM is to the parallelogram EQ, so is the height of the pyramid DEFH to the height of the pyramid ABCG. [V. 11]  But the height of the pyramid DEFH is the same with the height of the parallelepiped EHQP, and the height of the pyramid ABCG is the same with the height of the parallelepiped BGML;  therefore, as the base BM is to the base EQ, so is the height of the parallelepiped EHQP to the height of the parallelepiped BGML.  But those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal; [XI. 34]  therefore the parallelepipedal solid BGML is equal to the parallelepipedal solid EHQP.  And the pyramid ABCG is a sixth part of BGML, and the pyramid DEFH a sixth part of the parallelepiped EHQP;  therefore the pyramid ABCG is equal to the pyramid DEFH. 
                 
                 
Therefore, etc.  Q. E. D. 
   
   
PROPOSITION 10. 
 
 
Any cone is a third part of the cylinder which has the same base with it and equal height. 
 
 
For let a cone have the same base, namely the circle ABCD, with a cylinder and equal height;  I say that the cone is a third part of the cylinder, that is, that the cylinder is triple of the cone. 
   
   
For if the cylinder is not triple of the cone, the cylinder will be either greater than triple or less than triple of the cone.  First let it be greater than triple, and let the square ABCD be inscribed in the circle ABCD; [IV. 6]  then the square ABCD is greater than the half of the circle ABCD.  From the square ABCD let there be set up a prism of equal height with the cylinder.  Then the prism so set up is greater than the half of the cylinder,  inasmuch as, if we also circumscribe a square about the circle ABCD, [IV. 7]  the square inscribed in the circle ABCD is half of that circumscribed about it,  and the solids set up from them are parallelepipedal prisms of equal height,  while parallelepipedal solids which are of the same height are to one another as their bases; [XI. 32]  therefore also the prism set up on the square ABCD is half of the prism set up from the square circumscribed about the circle ABCD; [cf. XI. 28, or XII. 6 and 7, Por.]  and the cylinder is less than the prism set up from the square circumscribed about the circle ABCD;  therefore the prism set up from the square ABCD and of equal height with the cylinder is greater than the half of the cylinder.  Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined;  then each of the triangles AEB, BFC, CGD, DHA is greater than the half of that segment of the circle ABCD which is about it, as we proved before. [XII. 2]  On each of the triangles AEB, BFC, CGD, DHA let prisms be set up of equal height with the cylinder;  then each of the prisms so set up is greater than the half part of that segment of the cylinder which is about it,  inasmuch as, if we draw through the points E, F, G, H parallels to AB, BC, CD, DA,  complete the parallelograms on AB, BC, CD, DA,  and set up from them parallelepipedal solids of equal height with the cylinder, the prisms on the triangles AEB, BFC, CGD, DHA are halves of the several solids set up;  and the segments of the cylinder are less than the parallelepipedal solids set up;  hence also the prisms on the triangles AEB, BFC, CGD, DHA are greater than the half of the segments of the cylinder about them.  Thus, bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles prisms of equal height with the cylinder, and doing this continually, we shall leave some segments of the cylinder which will be less than the excess by which the cylinder exceeds the triple of the cone. [X. 1]  Let such segments be left, and let them be AE, EB, BF, FC, CG, GD, DH, HA;  therefore the remainder, the prism of which the polygon AEBFCGDH is the base and the height is the same as that of the cylinder, is greater than triple of the cone.  But the prism of which the polygon AEBFCGDH is the base and the height the same as that of the cylinder is triple of the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone; [XII. 7, Por.]  therefore also the pyramid of which the polygon AEBFCGDH is the base and the vertex is the same as that of the cone is greater than the cone which has the circle ABCD as base.  But it is also less, for it is enclosed by it: which is impossible.  Therefore the cylinder is not greater than triple of the cone. 
                                                       
                                                       
I say next that neither is the cylinder less than triple of the cone, 
 
 
For, if possible, let the cylinder be less than triple of the cone,  therefore, inversely, the cone is greater than a third part of the cylinder.  Let the square ABCD be inscribed in the circle ABCD;  therefore the square ABCD is greater than the half of the circle ABCD.  Now let there be set up from the square ABCD a pyramid having the same vertex with the cone;  therefore the pyramid so set up is greater than the half part of the cone,  seeing that, as we proved before, if we circumscribe a square about the circle, the square ABCD will be half of the square circumscribed about the circle,  and if we set up from the squares parallelepipedal solids of equal height with the cone, which are also called prisms,  the solid set up from the square ABCD will be half of that set up from the square circumscribed about the circle;  for they are to one another as their bases. [XI. 32]  Hence also the thirds of them are in that ratio;  therefore also the pyramid of which the square ABCD is the base is half of the pyramid set up from the square circumscribed about the circle.  And the pyramid set up from the square about the circle is greater than the cone, for it encloses it.  Therefore the pyramid of which the square ABCD is the base and the vertex is the same with that of the cone is greater than the half of the cone.  Let the circumferences AB, BC, CD, DA be bisected at the points E, F, G, H, and let AE, EB, BF, FC, CG, GD, DH, HA be joined;  therefore also each of the triangles AEB, BFC, CGD, DHA is greater than the half part of that segment of the circle ABCD which is about it.  Now, on each of the triangles AEB, BFC, CGD, DHA let pyramids be set up which have the same vertex as the cone;  therefore also each of the pyramids so set up is, in the same manner, greater than the half part of that segment of the cone which is about it.  Thus, by bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles a pyramid which has the same vertex as the cone,  and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone exceeds the third part of the cylinder. [X. 1]  Let such be left, and let them be the segments on AE, EB, BF, FC, CG, GD, DH, HA;  therefore the remainder, the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone, is greater than a third part of the cylinder.  But the pyramid of which the polygon AEBFCGDH is the base and the vertex the same with that of the cone is a third part of the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder;  therefore the prism of which the polygon AEBFCGDH is the base and the height is the same with that of the cylinder is greater than the cylinder of which the circle ABCD is the base.  But it is also less, for it is enclosed by it: which is impossible.  Therefore the cylinder is not less than triple of the cone.  But it was proved that neither is it greater than triple;  therefore the cylinder is triple of the cone;  hence the cone is a third part of the cylinder. 
                                                         
                                                         
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 11. 
 
 
Cones and cylinders which are of the same height are to one another as their bases. 
 
 
Let there be cones and cylinders of the same height, let the circles ABCD, EFGH be their bases, KL, MN their axes and AC, EG the diameters of their bases;  I say that, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN. 
   
   
For, if not, then, as the circle ABCD is to the circle EFGH, so will the cone AL be either to some solid less than the cone EN or to a greater.  First, let it be in that ratio to a less solid O, and let the solid X be equal to that by which the solid O is less than the cone EN;  therefore the cone EN is equal to the solids O, X.  Let the square EFGH be inscribed in the circle EFGH;  therefore the square is greater than the half of the circle.  Let there be set up from the square EFGH a pyramid of equal height with the cone;  therefore the pyramid so set up is greater than the half of the cone,  inasmuch as, if we circumscribe a square about the circle,  and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid,  for they are to one another as their bases, [XII. 6]  while the cone is less than the circumscribed pyramid.  Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let HP, PE, EQ, QF, FR, RG, GS, SH be joined.  Therefore each of the triangles HPE, EQF, FRG, GSH is greater than the half of that segment of the circle which is about it.  On each of the triangles HPE, EQF, FRG, GSH let there be set up a pyramid of equal height with the cone;  therefore, also, each of the pyramids so set up is greater than the half of that segment of the cone which is about it.  Thus, bisecting the circumferences which are left, joining straight lines, setting up on each of the triangles pyramids of equal height with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the solid X. [X. 1]  Let such be left, and let them be the segments on HP, PE, EQ, QF, FR, RG, GS, SH;  therefore the remainder, the pyramid of which the polygon HPEQFRGS is the base and the height the same with that of the cone, is greater than the solid O.  Let there also be inscribed in the circle ABCD the polygon DTAUBVCW similar and similarly situated to the polygon HPEQFRGS, and on it let a pyramid be set up of equal height with the cone AL.  Since then, as the square on AC is to the square on EG, so is the polygon DTAUBVCW to the polygon HPEQFRGS, [XII. 1]  while, as the square on AC is to the square on EG, so is the circle ABCD to the circle EFGH, [XII. 2]  therefore also, as the circle ABCD is to the circle EFGH, so is the polygon DTAUBVCW to the polygon HPEQFRGS.  But, as the circle ABCD is to the circle EFGH, so is the cone AL to the solid O,  and, as the polygon DTAUBVCW is to the polygon HPEQFRGS, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex. [XII. 6]  Therefore also, as the cone AL is to the solid O, so is the pyramid of which the polygon DTAUBVCW is the base and the point L the vertex to the pyramid of which the polygon HPEQFRGS is the base and the point N the vertex; [V. 11]  therefore, alternately, as the cone AL is to the pyramid in it, so is the solid O to the pyramid in the cone EN. [V. 16]  But the cone AL is greater than the pyramid in it;  therefore the solid O is also greater than the pyramid in the cone EN.  But it is also less: which is absurd.  Therefore the cone AL is not to any solid less than the cone EN as the circle ABCD is to the circle EFGH.  Similarly we can prove that neither is the cone EN to any solid less than the cone AL as the circle EFGH is to the circle ABCD. 
                                                             
                                                             
I say next that neither is the cone AL to any solid greater than the cone EN as the circle ABCD is to the circle EFGH. 
 
 
For, if possible, let it be in that ratio to a greater solid O;  therefore, inversely, as the circle EFGH is to the circle ABCD, so is the solid O to the cone AL.  But, as the solid O is to the cone AL, so is the cone EN to some solid less than the cone AL;  therefore also, as the circle EFGH is to the circle ABCD, so is the cone EN to some solid less than the cone AL: which was proved impossible.  Therefore the cone AL is not to any solid greater than the cone EN as the circle ABCD is to the circle EFGH.  But it was proved that neither is it in this ratio to a less solid;  therefore, as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN. 
             
             
But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [XII. 10]  Therefore also, as the circle ABCD is to the circle EFGH, so are the cylinders on them which are of equal height. 
   
   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 12. 
 
 
Similar cones and cylinders are to one another in the triplicate ratio of the diameters in their bases. 
 
 
Let there be similar cones and cylinders, let the circles ABCD, EFGH be their bases, BD, FH the diameters of the bases, and KL, MN the axes of the cones and cylinders;  I say that the cone of which the circle ABCD is the base and the point L the vertex has to the cone of which the circle EFGH is the base and the point N the vertex the ratio triplicate of that which BD has to FH. 
   
   
For, if the cone ABCDL has not to the cone EFGHN the ratio triplicate of that which BD has to FH, the cone ABCDL will have that triplicate ratio either to some solid less than the cone EFGHN or to a greater.  First, let it have that triplicate ratio to a less solid O.  Let the square EFGH be inscribed in the circle EFGH; [IV. 6]  therefore the square EFGH is greater than the half of the circle EFGH.  Now let there be set up on the square EFGH a pyramid having the same vertex with the cone;  therefore the pyramid so set up is greater than the half part of the cone.  Let the circumferences EF, FG, GH, HE be bisected at the points P, Q, R, S, and let EP, PF, FQ, QG, GR, RH, HS, SE be joined.  Therefore each of the triangles EPF, FQG, GRH, HSE is also greater than the half part of that segment of the circle EFGH which is about it.  Now on each of the triangles EPF, FQG, GRH, HSE let a pyramid be set up having the same vertex with the cone;  therefore each of the pyramids so set up is also greater than the half part of that segment of the cone which is about it.  Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids having the same vertex with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone EFGHN exceeds the solid O. [X. 1]  Let such be left, and let them be the segments on EP, PF, FQ, QG, GR, RH, HS, SE;  therefore the remainder, the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex, is greater than the solid O.  Let there be also inscribed in the circle ABCD the polygon ATBUCVDW similar and similarly situated to the polygon EPFQGRHS,  and let there be set up on the polygon ATBUCVDW a pyramid having the same vertex with the cone;  of the triangles containing the pyramid of which the polygon ATBUCVDW is the base and the point L the vertex let LBT be one,  and of the triangles containing the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex let NFP be one; and let KT, MP be joined.  Now, since the cone ABCDL is similar to the cone EFGHN, therefore, as BD is to FH, so is the axis KL to the axis MN. [XI. Def. 24]  But, as BD is to FH, so is BK to FM;  therefore also, as BK is to FM, so is KL to MN.  And, alternately, as BK is to KL, so is FM to MN. [V. 16]  And the sides are proportional about equal angles, namely the angles BKL, FMN;  therefore the triangle BKL is similar to the triangle FMN. [VI. 6]  Again, since, as BK is to KT, so is FM to MP, and they are about equal angles, namely the angles BKT, FMP,  inasmuch as, whatever part the angle BKT is of the four right angles at the centre K, the same part also is the angle FMP of the four right angles at the centre M;  since then the sides are proportional about equal angles, therefore the triangle BKT is similar to the triangle FMP. [VI. 6]  Again, since it was proved that, as BK is to KL, so is FM to MN, while BK is equal to KT, and FM to PM, therefore, as TK is to KL, so is PM to MN;  and the sides are proportional about equal angles, namely the angles TKL, PMN, for they are right;  therefore the triangle LKT is similar to the triangle NMP. [VI. 6]  And since, owing to the similarity of the triangles LKB, NMF, as LB is to BK, so is NF to FM,  and, owing to the similarity of the triangles BKT, FMP, as KB is to BT, so is MF to FP,  therefore, ex aequali, as LB is to BT, so is NF to FP. [V. 22]  Again since, owing to the similarity of the triangles LTK, NPM, as LT is to TK, so is NP to PM,  and, owing to the similarity of the triangles TKB, PMF, as KT is to TB, so is MP to PF;  therefore, ex aequali, as LT is to TB, so is NP to PF. [V. 22]  But it was also proved that, as TB is to BL, so is PF to FN.  Therefore, ex aequali, as TL is to LB, so is PN to NF. [V. 22]  Therefore in the triangles LTB, NPF the sides are proportional;  therefore the triangles LTB, NPF are equiangular; [VI. 5]  hence they are also similar. [VI. Def. I]  Therefore the pyramid of which the triangle BKT is the base and the point L the vertex is also similar to the pyramid of which the triangle FMP is the base and the point N the vertex, for they are contained by similar planes equal in multitude. [XI. Def. 9]  But similar pyramids which have triangular bases are to one another in the triplicate ratio of their corresponding sides. [XII. 8]  Therefore the pyramid BKTL has to the pyramid FMPN the ratio triplicate of that which BK has to FM.  Similarly, by joining straight lines from A, W, D, V, C, U to K, and from E, S, H, R, G, Q to M, and setting up on each of the triangles pyramids which have the same vertex with the cones, we can prove that each of the similarly arranged pyramids will also have to each similarly arranged pyramid the ratio triplicate of that which the corresponding side BK has to the corresponding side FM, that is, which BD has to FH.  And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12]  therefore also, as the pyramid BKTL is to the pyramid FMPN, so is the whole pyramid of which the polygon ATBUCVDW is the base and the point L the vertex to the whole pyramid of which the polygon EPFQGRHS is the base and the point N the vertex;  hence also the pyramid of which ATBUCVDW is the base and the point L the vertex has to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex the ratio triplicate of that which BD has to FH.  But, by hypothesis, the cone of which the circle ABCD is the base and the point L the vertex has also to the solid O the ratio triplicate of that which BD has to FH;  therefore, as the cone of which the circle ABCD is the base and the point L the vertex is to the solid O, so is the pyramid of which the polygon ATBUCVDW is the base and L the vertex to the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex;  therefore, alternately, as the cone of which the circle ABCD is the base and L the vertex is to the pyramid contained in it of which the polygon ATBUCVDW is the base and L the vertex, so is the solid O to the pyramid of which the polygon EPFQGRHS is the base and N the vertex. [V. 16]  But the said cone is greater than the pyramid in it; for it encloses it.  Therefore the solid O is also greater than the pyramid of which the polygon EPFQGRHS is the base and N the vertex.  But it is also less: which is impossible.  Therefore the cone of which the circle ABCD is the base and L the vertex has not to any solid less than the cone of which the circle EFGH is the base and the point N the vertex the ratio triplicate of that which BD has to FH:  Similarly we can prove that neither has the cone EFGHN to any solid less than the cone ABCDL the ratio triplicate of that which FH has to BD. 
                                                                                                             
                                                                                                             
I say next that neither has the cone ABCDL to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH. 
 
 
For, if possible, let it have that ratio to a greater solid O.  Therefore, inversely, the solid O has to the cone ABCDL the ratio triplicate of that which FH has to BD.  But, as the solid O is to the cone ABCDL, so is the cone EFGHN to some solid less than the cone ABCDL.  Therefore the cone EFGHN also has to some solid less than the cone ABCDL the ratio triplicate of that which FH has to BD: which was proved impossible.  Therefore the cone ABCDL has not to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH.  But it was proved that neither has it this ratio to a less solid than the cone EFGHN.  Therefore the cone ABCDL has to the cone EFGHN the ratio triplicate of that which BD has to FH. 
             
             
But, as the cone is to the cone, so is the cylinder to the cylinder,  for the cylinder which is on the same base as the cone and of equal height with it is triple of the cone; [XII. 10]  therefore the cylinder also has to the cylinder the ratio triplicate of that which BD has to FH. 
     
     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 13. 
 
 
If a cylinder be cut by a plane which is parallel to its opposite planes, then, as the cylinder is to the cylinder, so will the axis be to the axis. 
 
 
For let the cylinder AD be cut by the plane GH which is parallel to the opposite planes AB, CD, and let the plane GH meet the axis at the point K;  I say that, as the cylinder BG is to the cylinder GD, so is the axis EK to the axis KF. 
   
   
For let the axis EF be produced in both directions to the points L, M, and let there be set out any number whatever of axes EN, NL equal to the axis EK, and any number whatever FO, OM equal to FK;  and let the cylinder PW on the axis LM be conceived of which the circles PQ, VW are the bases.  Let planes be carried through the points N, O parallel to AB, CD and to the bases of the cylinder PW, and let them produce the circles RS, TU about the centres N, O.  Then, since the axes LN, NE, EK are equal to one another, therefore the cylinders QR, RB, BG are to one another as their bases. [XII. 11]  But the bases are equal; therefore the cylinders QR, RB, BG are also equal to one another.  Since then the axes LN, NE, EK are equal to one another, and the cylinders QR, RB, BG are also equal to one another,  and the multitude of the former is equal to the multitude of the latter,  therefore, whatever multiple the axis KL is of the axis EK, the same multiple also will the cylinder QG be of the cylinder GB.  For the same reason, whatever multiple the axis MK is of the axis KF, the same multiple also is the cylinder WG of the cylinder GD.  And, if the axis KL is equal to the axis KM, the cylinder QG will also be equal to the cylinder GW, if the axis is greater than the axis, the cylinder will also be greater than the cylinder, and if less, less.  Thus, there being four magnitudes, the axes EK, KF and the cylinders BG, GD, there have been taken equimultiples of the axis EK and of the cylinder BG,  namely the axis LK and the cylinder QG, and equimultiples of the axis KF and of the cylinder GD, namely the axis KM and the cylinder GW;  and it has been proved that, if the axis KL is in excess of the axis KM, the cylinder QG is also in excess of the cylinder GW, if equal, equal, and if less, less.  Therefore, as the axis EK is to the axis KF, so is the cylinder BG to the cylinder GD. [V. Def. 5]  Q. E. D. 
                             
                             
PROPOSITION 14. 
 
 
Cones and cylinders which are on equal bases are to one another as their heights. 
 
 
For let EB, FD be cylinders on equal bases, the circles AB, CD;  I say that, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL. 
   
   
For let the axis KL be produced to the point N, let LN be made equal to the axis GH, and let the cylinder CM be conceived about LN as axis.  Since then the cylinders EB, CM are of the same height, they are to one another as their bases. [XII. 11]  But the bases are equal to one another;  therefore the cylinders EB, CM are also equal.  And, since the cylinder FM has been cut by the plane CD which is parallel to its opposite planes,  therefore, as the cylinder CM is to the cylinder FD, so is the axis LN to the axis KL. [XII. 13]  But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH;  therefore, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL.  But, as the cylinder EB is to the cylinder FD, so is the cone ABG to the cone CDK. [XII. 10]  Therefore also, as the axis GH is to the axis KL, so is the cone ABG to the cone CDK and the cylinder EB to the cylinder FD.  Q. E. D. 
                     
                     
PROPOSITION 15. 
 
 
In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal. 
 
 
Let there be equal cones and cylinders of which the circles ABCD, EFGH are the bases; let AC, EG be the diameters of the bases, and KL, MN the axes, which are also the heights of the cones or cylinders; let the cylinders AO, EP be completed.  I say that in the cylinders AO, EP the bases are reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so is the height MN to the height KL. 
   
   
For the height LK is either equal to the height MN or not equal.  First, let it be equal.  Now the cylinder AO is also equal to the cylinder EP.  But cones and cylinders which are of the same height are to one another as their bases; [XII. 11]  therefore the base ABCD is also equal to the base EFGH.  Hence also, reciprocally, as the base ABCD is to the base EFGH, so is the height MN to the height KL.  Next, let the height LK not be equal to MN, but let MN be greater; from the height MN let QN be cut off equal to KL,  through the point Q let the cylinder EP be cut by the plane TUS parallel to the planes of the circles EFGH, RP,  and let the cylinder ES be conceived erected from the circle EFGH as base and with height NQ.  Now, since the cylinder AO is equal to the cylinder EP,  therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 7]  But, as the cylinder AO is to the cylinder ES, so is the base ABCD to the base EFGH,  for the cylinders AO, ES are of the same height; [XII. 11]  and, as the cylinder EP is to the cylinder ES, so is the height MN to the height QN,  for the cylinder EP has been cut by a plane which is parallel to its opposite planes. [XII. 13]  Therefore also, as the base ABCD is to the base EFGH, so is the height MN to the height QN. [V. 11]  But the height QN is equal to the height KL;  therefore, as the base ABCD is to the base EFGH, so is the height MN to the height KL.  Therefore in the cylinders AO, EP the bases are reciprocally proportional to the heights. 
                                     
                                     
Next, in the cylinders AO, EP let the bases be reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so let the height MN be to the height KL;  I say that the cylinder AO is equal to the cylinder EP. 
   
   
For, with the same construction, since, as the base ABCD is to the base EFGH, so is the height MN to the height KL,  while the height KL is equal to the height QN, therefore, as the base ABCD is to the base EFGH, so is the height MN to the height QN  But, as the base ABCD is to the base EFGH, so is the cylinder AO to the cylinder ES, for they are of the same height; [XII. 11]  and, as the height MN is to QN, so is the cylinder EP to the cylinder ES; [XII. 13]  therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [V. 11]  Therefore the cylinder AO is equal to the cylinder EP. [V. 9]  And the same is true for the cones also.  Q. E. D. 
               
               
PROPOSITION 16. 
 
 
Given two circles about the same centre, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle. 
 
 
Let ABCD, EFGH be the two given circles about the same centre K;  thus it is required to inscribe in the greater circle ABCD an equilateral polygon with an even number of sides which does not touch the circle EFGH. 
   
   
For let the straight line BKD be drawn through the centre K, and from the point G let GA be drawn at right angles to the straight line BD and carried through to C;  therefore AC touches the circle EFGH. [III. 16, Por.]  Then, bisecting the circumference BAD, bisecting the half of it, and doing this continually, we shall leave a circumference less than AD. [X. 1]  Let such be left, and let it be LD; from L let LM be drawn perpendicular to BD and carried through to N, and let LD, DN be joined;  therefore LD is equal to DN. [III. 3, I. 4]  Now, since LN is parallel to AC, and AC touches the circle EFGH,  therefore LN does not touch the circle EFGH;  therefore LD, DN are far from touching the circle EFGH.  If then we fit into the circle ABCD straight lines equal to the straight line LD and placed continuously, there will be inscribed in the circle ABCD an equilateral polygon with an even number of sides which does not touch the lesser circle EFGH.  Q. E. F. 
                   
                   
PROPOSITION 17. 
 
 
Given two spheres about the same centre, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. 
 
 
Let two spheres be conceived about the same centre A;  thus it is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. 
   
   
Let the spheres be cut by any plane through the centre;  then the sections will be circles, inasmuch as the sphere was produced by the diameter remaining fixed and the semicircle being carried round it; [XI. Def. 14]  hence, in whatever position we conceive the semicircle to be, the plane carried through it will produce a circle on the circumference of the sphere.  And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere.  Let then BCDE be the circle in the greater sphere, and FGH the circle in the lesser sphere;  let two diameters in them, BD, CE, be drawn at right angles to one another;  then, given the two circles BCDE, FGH about the same centre, let there be inscribed in the greater circle BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle FGH,  let BK, KL, LM ME be its sides in the quadrant BE,  let KA be joined and carried through to N,  let AO be set up from the point A at right angles to the plane of the circle BCDE, and let it meet the surface of the sphere at O,  and through AO and each of the straight lines BD, KN let planes be carried;  they will then make greatest circles on the surface of the sphere, for the reason stated.  Let them make such, and in them let BOD, KON be the semicircles on BD, KN.  Now, since OA is at right angles to the plane of the circle BCDE,  therefore all the planes through OA are also at right angles to the plane of the circle BCDE; [XI. 18]  hence the semicircles BOD, KON are also at right angles to the plane of the circle BCDE.  And, since the semicircles BED, BOD, KON are equal,  for they are on the equal diameters BD, KN,  therefore the quadrants BE, BO, KO are also equal to one another.  Therefore there are as many straight lines in the quadrants BO, KO equal to the straight lines BK, KL, LM, ME as there are sides of the polygon in the quadrant BE.  Let them be inscribed, and let them be BP, PQ, QR, RO and KS, ST, TU, UO, let SP, TQ, UR be joined, and from P, S let perpendiculars be drawn to the plane of the circle BCDE; [XI. 11]  these will fall on BD, KN, the common sections of the planes, inasmuch as the planes of BOD, KON are also at right angles to the plane of the circle BCDE. [cf. XI. Def. 4]  Let them so fall, and let them be PV, SW, and let WV be joined.  Now since, in the equal semicircles BOD, KON, equal straight lines BP, KS have been cut off, and the perpendiculars PV, SW have been drawn,  therefore PV is equal to SW, and BV to KW. [III. 27, I. 26]  But the whole BA is also equal to the whole KA;  therefore the remainder VA is also equal to the remainder WA;  therefore, as BV is to VA, so is KW to WA;  therefore WV is parallel to KB. [VI. 2]  And, since each of the straight lines PV, SW is at right angles to the plane of the circle BCDE,  therefore PV is parallel to SW. [XI. 6]  But it was also proved equal to it;  therefore WV, SP are also equal and parallel. [I. 33]  And, since WV is parallel to SP, while WV is parallel to KB,  therefore SP is also parallel to KB. [XI. 9]  And BP, KS join their extremities;  therefore the quadrilateral KBPS is in one plane,  inasmuch as, if two straight lines be parallel, and points be taken at random on each of them,  the straight line joining the points is in the same plane with the parallels. [XI. 7]  For the same reason each of the quadrilaterals SPQT, TQRU is also in one plane.  But the triangle URO is also in one plane. [XI. 2]  If then we conceive straight lines joined from the points P, S, Q, T, R, U to A,  there will be constructed a certain polyhedral solid figure between the circumferences BO, KO,  consisting of pyramids of which the quadrilaterals KBPS, SPQT, TQRU and the triangle URO are the bases and the point A the vertex.  And, if we make the same construction in the case of each of the sides KL, LM, ME as in the case of BK, and further in the case of the remaining three quadrants,  there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids,  of which the said quadrilaterals and the triangle URO, and the others corresponding to them, are the bases and the point A the vertex. 
                                                                                             
                                                                                             
I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle FGH is. 
 
 
Let AX be drawn from the point A perpendicular to the plane of the quadrilateral KBPS, and let it meet the plane at the point X; [XI. 11]  let XB, XK be joined.  Then, since AX is at right angles to the plane of the quadrilateral KBPS,  therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. [XI. Def. 3]  Therefore AX is at right angles to each of the straight lines BX, XK.  And, since AB is equal to AK, the square on AB is also equal to the square on AK.  And the squares on AX, XB are equal to the square on AB, for the angle at X is right; [I. 47]  and the squares on AX, XK are equal to the square on AK. [id.]  Therefore the squares on AX, XB are equal to the squares on AX, XK.  Let the square on AX be subtracted from each;  therefore the remainder, the square on BX, is equal to the remainder, the square on XK;  therefore BX is equal to XK.  Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK.  Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also,  and KBPS will be a quadrilateral in a circle. 
                             
                             
Now, since KB is greater than WV, while WV is equal to SP,  therefore KB is greater than SP.  But KB is equal to each of the straight lines KS, BP;  therefore each of the straight lines KS, BP is greater than SP.  And, since KBPS is a quadrilateral in a circle, and KB, BP, KS are equal, and PS less,  and BX is the radius of the circle,  therefore the square on KB is greater than double of the square on BX.  Let KZ be drawn from K perpendicular to BV.  Then, since BD is less than double of DZ, and, as BD is to DZ, so is the rectangle DB, BZ to the rectangle DZ, ZB,  if a square be described upon BZ and the parallelogram on ZD be completed, then the rectangle DB, BZ is also less than double of the rectangle DZ, ZB.  And, if KD be joined, the rectangle DB, BZ is equal to the square on BK, and the rectangle DZ, ZB equal to the square on KZ; [III. 31, VI. 8 and Por.]  therefore the square on KB is less than double of the square on KZ.  But the square on KB is greater than double of the square on BX;  therefore the square on KZ is greater than the square on BX.  And, since BA is equal to KA, the square on BA is equal to the square on AK.  And the squares on BX, XA are equal to the square on BA, and the squares on KZ, ZA equal to the square on KA; [I. 47]  therefore the squares on BX, XA are equal to the squares on KZ, ZA, and of these the square on KZ is greater than the square on BX;  therefore the remainder, the square on ZA, is less than the square on XA.  Therefore AX is greater than AZ;  therefore AX is much greater than AG.  And AX is the perpendicular on one base of the polyhedron, and AG on the surface of the lesser sphere;  hence the polyhedron will not touch the lesser sphere on its surface. 
                                           
                                           
Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface.  Q. E. F. 
   
   
PORISM.
But if in another sphere also a polyhedral solid be inscribed similar to the solid in the sphere BCDE, the polyhedral solid in the sphere BCDE has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere BCDE has to the diameter of the other sphere. 
For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar.  But similar pyramids are to one another in the triplicate ratio of their corresponding sides; [XII. 8, Por.]  therefore the pyramid of which the quadrilateral KBPS is the base, and the point A the vertex, has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius AB of the sphere about A as centre has to the radius of the other sphere.  Similarly also each pyramid of those in the sphere about A as centre has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere.  And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12]  hence the whole polyhedral solid in the sphere about A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere, that is, of that which the diameter BD has to the diameter of the other sphere.  Q. E. D. 
               
               
PROPOSITION 18. 
 
 
Spheres are to one another in the triplicate ratio of their respective diameters. 
 
 
Let the spheres ABC, DEF be conceived, and let BC, EF be their diameters;  I say that the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. 
   
   
For, if the sphere ABC has not to the sphere DEF the ratio triplicate of that which BC has to EF,  then the sphere ABC will have either to some less sphere than the sphere DEF, or to a greater, the ratio triplicate of that which BC has to EF.  First, let it have that ratio to a less sphere GHK, let DEF be conceived about the same centre with GHK, let there be inscribed in the greater sphere DEF a polyhedral solid which does not touch the lesser sphere GHK at its surface, [XII. 17] and let there also be inscribed in the sphere ABC a polyhedral solid similar to the polyhedral solid in the sphere DEF;  therefore the polyhedral solid in ABC has to the polyhedral solid in DEF the ratio triplicate of that which BC has to EF. [XII. 17, Por.]  But the sphere ABC also has to the sphere GHK the ratio triplicate of that which BC has to EF;  therefore, as the sphere ABC is to the sphere GHK, so is the polyhedral solid in the sphere ABC to the polyhedral solid in the sphere DEF;  and, alternately, as the sphere ABC is to the polyhedron in it, so is the sphere GHK to the polyhedral solid in the sphere DEF. [V. 16]  But the sphere ABC is greater than the polyhedron in it;  therefore the sphere GHK is also greater than the polyhedron in the sphere DEF.  But it is also less, for it is enclosed by it.  Therefore the sphere ABC has not to a less sphere than the sphere DEF the ratio triplicate of that which the diameter BC has to EF.  Similarly we can prove that neither has the sphere DEF to a less sphere than the sphere ABC the ratio triplicate of that which EF has to BC. 
                       
                       
I say next that neither has the sphere ABC to any greater sphere than the sphere DEF the ratio triplicate of that which BC has to EF. 
 
 
For, if possible, let it have that ratio to a greater, LMN;  therefore, inversely, the sphere LMN has to the sphere ABC the ratio triplicate of that which the diameter EF has to the diameter BC.  But, inasmuch as LMN is greater than DEF, therefore, as the sphere LMN is to the sphere ABC, so is the sphere DEF to some less sphere than the sphere ABC, as was before proved. [XII. 2, Lemma]  Therefore the sphere DEF also has to some less sphere than the sphere ABC the ratio triplicate of that which EF has to BC: which was proved impossible.  Therefore the sphere ABC has not to any sphere greater than the sphere DEF the ratio triplicate of that which BC has to EF.  But it was proved that neither has it that ratio to a less sphere.  Therefore the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF.  Q. E. D. 
               
               
Book XIII 
 
 
PROPOSITION 1. 
 
 
If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half. 
 
 
For let the straight line AB be cut in extreme and mean ratio at the point C, and let AC be the greater segment; let the straight line AD be produced in a straight line with CA, and let AD be made half of AB;  I say that the square on CD is five times the square on AD. 
   
   
For let the squares AE, DF be described on AB, DC, and let the figure in DF be drawn; let FC be carried through to G.  Now, since AB has been cut in extreme and mean ratio at C,  therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]  And CE is the rectangle AB, BC, and FH the square on AC;  therefore CE is equal to FH.  And, since BA is double of AD, while BA is equal to KA, and AD to AH,  therefore KA is also double of AH.  But, as KA is to AH, so is CK to CH; [VI. 1]  therefore CK is double of CH.  But LH, HC are also double of CH.  Therefore KC is equal to LH, HC.  But CE was also proved equal to HF;  therefore the whole square AE is equal to the gnomon MNO.  And, since BA is double of AD, the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH.  But AE is equal to the gnomon MNO;  therefore the gnomon MNO is also quadruple of AP;  therefore the whole DF is five times AP.  And DF is the square on DC, and AP the square on DA;  therefore the square on CD is five times the square on DA. 
                                     
                                     
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 2. 
 
 
If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line. 
 
 
For let the square on the straight line AB be five times the square on the segment AC of it, and let CD be double of AC;  I say that, when CD is cut in extreme and mean ratio, the greater segment is CB. 
   
   
Let the squares AF, CG be described on AB, CD respectively, let the figure in AF be drawn, and let BE be drawn through.  Now, since the square on BA is five times the square on AC, AF is five times AH.  Therefore the gnomon MNO is quadruple of AH.  And, since DC is double of CA,  therefore the square on DC is quadruple of the square on CA, that is, CG is quadruple of AH.  But the gnomon MNO was also proved quadruple of AH;  therefore the gnomon MNO is equal to CG.  And, since DC is double of CA, while DC is equal to CK, and AC to CH,  therefore KB is also double of BH. [VI. 1]  But LH, HB are also double of HB;  therefore KB is equal to LH, HB.  But the whole gnomon MNO was also proved equal to the whole CG;  therefore the remainder HF is equal to BG.  And BG is the rectangle CD, DB, for CD is equal to DG;  and HF is the square on CB;  therefore the rectangle CD, DB is equal to the square on CB.  Therefore, as DC is to CB, so is CB to BD.  But DC is greater than CB;  therefore CB is also greater than BD.  Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment. 
                                       
                                       
Therefore etc.  Q. E. D. 
   
   
LEMMA.
That the double of AC is greater than BC is to be proved thus. 
 
 
If not, let BC be, if possible, double of CA.  Therefore the square on BC is quadruple of the square on CA;  therefore the squares on BC, CA are five times the square on CA.  But, by hypothesis, the square on BA is also five times the square on CA;  therefore the square on BA is equal to the squares on BC, CA: which is impossible. [II. 4]  Therefore CB is not double of AC.  Similarly we can prove that neither is a straight line less than CB double of CA;  for the absurdity is much greater. 
               
               
Therefore the double of AC is greater than CB.  Q. E. D. 
   
   
PROPOSITION 3. 
 
 
If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment. 
 
 
For let any straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AC be bisected at D;  I say that the square on BD is five times the square on DC. 
   
   
For let the square AE be described on AB, and let the figure be drawn double.  Since AC is double of DC,  therefore the square on AC is quadruple of the square on DC, that is, RS is quadruple of FG.  And, since the rectangle AB, BC is equal to the square on AC, and CE is the rectangle AB, BC,  therefore CE is equal to RS.  But RS is quadruple of FG;  therefore CE is also quadruple of FG.  Again, since AD is equal to DC, HK is also equal to KF.  Hence the square GF is also equal to the square HL.  Therefore GK is equal to KL, that is, MN to NE;  hence MF is also equal to FE.  But MF is equal to CG;  therefore CG is also equal to FE.  Let CN be added to each;  therefore the gnomon OPQ is equal to CE.  But CE was proved quadruple of GF;  therefore the gnomon OPQ is also quadruple of the square FG.  Therefore the gnomon OPQ and the square FG are five times FG.  But the gnomon OPQ and the square FG are the square DN.  And DN is the square on DB, and GF the square on DC.  Therefore the square on DB is five times the square on DC.  Q. E. D. 
                                           
                                           
PROPOSITION 4. 
 
 
If a straight line be cut in extreme and mean ratio, the square on the whole and the square on the lesser segment together are triple of the square on the greater segment. 
 
 
Let AB be a straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment;  I say that the squares on AB, BC are triple of the square on CA. 
   
   
For let the square ADEB be described on AB, and let the figure be drawn.  Since then AB has been cut in extreme and mean ratio at C, and AC is the greater segment,  therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]  And AK is the rectangle AB, BC, and HG the square on AC;  therefore AK is equal to HG.  And, since AF is equal to FE, let CK be added to each;  therefore the whole AK is equal to the whole CE;  therefore AK, CE are double of AK.  But AK, CE are the gnomon LMN and the square CK;  therefore the gnomon LMN and the square CK are double of AK.  But, further, AK was also proved equal to HG;  therefore the gnomon LMN  and the squares CK, HG are triple of the square HG.  And the gnomon LMN and the squares CK, HG are the whole square AE and CK, which are the squares on AB, BC, while HG is the square on AC.  Therefore the squares on AB, BC are triple of the square on AC.  Q. E. D. 
                               
                               
PROPOSITION 5. 
 
 
If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment. 
 
 
For let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC.  I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment. 
   
   
For let the square AE be described on AB, and let the figure be drawn.  Since AB has been cut in extreme and mean ratio at C,  therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]  And CE is the rectangle AB, BC, and CH the square on AC;  therefore CE is equal to HC.  But HE is equal to CE, and DH is equal to HC;  therefore DH is also equal to HE.  Therefore the whole DK is equal to the whole AE.  And DK is the rectangle BD, DA, for AD is equal to DL;  and AE is the square on AB;  therefore the rectangle BD, DA is equal to the square on AB.  Therefore, as DB is to BA, so is BA to AD. [VI. 17]  And DB is greater than BA;  therefore BA is also greater than AD. [V. 14] 
                           
                           
Therefore DB has been cut in extreme and mean ratio at A, and AB is the greater segment.  Q. E. D. 
   
   
PROPOSITION 6. 
 
 
If a rational straight line be cut in extreme and mean ratio, each of the segments is the irrational straight line called apotome. 
 
 
Let AB be a rational straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment;  I say that each of the straight lines AC, CB is the irrational straight line called apotome. 
   
   
For let BA be produced, and let AD be made half of BA.  Since then the straight line AB has been cut in extreme and mean ratio, and to the greater segment AC is added AD which is half of AB,  therefore the square on CD is five times the square on DA. [XIII. 1]  Therefore the square on CD has to the square on DA the ratio which a number has to a number;  therefore the square on CD is commensurable with the square on DA. [X. 6]  But the square on DA is rational, for DA is rational, being half of AB which is rational;  therefore the square on CD is also rational; [X. Def. 4]  therefore CD is also rational.  And, since the square on CD has not to the square on DA the ratio which a square number has to a square number,  therefore CD is incommensurable in length with DA; [X. 9]  therefore CD, DA are rational straight lines commensurable in square only;  therefore AC is an apotome. [X. 73]  Again, since AB has been cut in extreme and mean ratio, and AC is the greater segment,  therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]  Therefore the square on the apotome AC, if applied to the rational straight line AB, produces BC as breadth.  But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome; [X. 97]  therefore CB is a first apotome.  And CA was also proved to be an apotome. 
                                   
                                   
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 7. 
 
 
If three angles of an equilateral pentagon, taken either in order or not in order, be equal, the pentagon will be equiangular. 
 
 
For in the equilateral pentagon ABCDE let, first, three angles taken in order, those at A, B, C, be equal to one another;  I say that the pentagon ABCDE is equiangular. 
   
   
For let AC, BE, FD be joined.  Now, since the two sides CB, BA are equal to the two sides BA, AE respectively,  and the angle CBA is equal to the angle BAE,  therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE,  and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4]  that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB;  hence the side AF is also equal to the side BF. [I. 6]  But the whole AC was also proved equal to the whole BE;  therefore the remainder FC is also equal to the remainder FE.  But CD is also equal to DE.  Therefore the two sides FC, CD are equal to the two sides FE, ED;  and the base FD is common to them;  therefore the angle FCD is equal to the angle FED. [I. 8]  But the angle BCA was also proved equal to the angle AEB;  therefore the whole angle BCD is also equal to the whole angle AED.  But, by hypothesis, the angle BCD is equal to the angles at A, B;  therefore the angle AED is also equal to the angles at A, B.  Similarly we can prove that the angle CDE is also equal to the angles at A, B, C;  therefore the pentagon ABCDE is equiangular. 
                                     
                                     
Next, let the given equal angles not be angles taken in order, but let the angles at the points A, C, D be equal;  I say that in this case too the pentagon ABCDE is equiangular. 
   
   
For let BD be joined.  Then, since the two sides BA, AE are equal to the two sides BC, CD, and they contain equal angles,  therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD,  and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]  therefore the angle AEB is equal to the angle CDB.  But the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD. [I. 5]  Therefore the whole angle AED is equal to the whole angle CDE.  But the angle CDE is, by hypothesis, equal to the angles at A, C;  therefore the angle AED is also equal to the angles at A, C.  For the same reason the angle ABC is also equal to the angles at A, C, D.  Therefore the pentagon ABCDE is equiangular.  Q. E. D. 
                       
                       
PROPOSITION 8. 
 
 
If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon. 
 
 
For in the equilateral and equiangular pentagon ABCDE let the straight lines AC, BE, cutting one another at the point H, subtend two angles taken in order, the angles at A, B;  I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments are equal to the side of the pentagon. 
   
   
For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14]  Then, since the two straight lines EA, AB are equal to the two AB, BC, and they contain equal angles,  therefore the base BE is equal to the base AC, the triangle ABE is equal to the triangle ABC,  and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [I. 4]  Therefore the angle BAC is equal to the angle ABE;  therefore the angle AHE is double of the angle BAH. [I. 32]  But the angle EAC is also double of the angle BAC,  inasmuch as the circumference EDC is also double of the circumference CB; [III. 28, VI. 33]  therefore the angle HAE is equal to the angle AHE;  hence the straight line HE is also equal to EA, that is, to AB. [I. 6]  And, since the straight line BA is equal to AE, the angle ABE is also equal to the angle AEB. [I. 5]  But the angle ABE was proved equal to the angle BAH;  therefore the angle BEA is also equal to the angle BAH.  And the angle ABE is common to the two triangles ABE and ABH;  therefore the remaining angle BAE is equal to the remaining angle AHB; [I. 32]  therefore the triangle ABE is equiangular with the triangle ABH;  therefore, proportionally, as EB is to BA, so is AB to BH. [VI. 4]  But BA is equal to EH;  therefore, as BE is to EH, so is EH to HB.  And BE is greater than EH;  therefore EH is also greater than HB. [V. 14]  Therefore BE has been cut in extreme and mean ratio at H, and the greater segment HE is equal to the side of the pentagon.  Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH is equal to the side of the pentagon.  Q. E. D. 
                                               
                                               
PROPOSITION 9. 
 
 
If the side of the hexagon and that of the decagon inscribed in the same circle be added together, the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon. 
 
 
Let ABC be a circle; of the figures inscribed in the circle ABC let BC be the side of a decagon, CD that of a hexagon, and let them be in a straight line;  I say that the whole straight line BD has been cut in extreme and mean ratio, and CD is its greater segment. 
   
   
For let the centre of the circle, the point E, be taken, let EB, EC, ED be joined, and let BE be carried through to A.  Since BC is the side of an equilateral decagon,  therefore the circumference ACB is five times the circumference BC;  therefore the circumference AC is quadruple of CB.  But, as the circumference AC is to CB, so is the angle AEC to the angle CEB; [VI. 33]  therefore the angle AEC is quadruple of the angle CEB.  And, since the angle EBC is equal to the angle ECB, [I. 5]  therefore the angle AEC is double of the angle ECB. [I. 32]  And, since the straight line EC is equal to CD, for each of them is equal to the side of the hexagon inscribed in the circle ABC, [IV. 15, Por.]  the angle CED is also equal to the angle CDE; [I. 5]  therefore the angle ECB is double of the angle EDC. [I. 32]  But the angle AEC was proved double of the angle ECB;  therefore the angle AEC is quadruple of the angle EDC.  But the angle AEC was also proved quadruple of the angle BEC;  therefore the angle EDC is equal to the angle BEC.  But the angle EBD is common to the two triangles BEC and BED;  therefore the remaining angle BED is also equal to the remaining angle ECB; [I. 32]  therefore the triangle EBD is equiangular with the triangle EBC.  Therefore, proportionally, as DB is to BE, so is EB to BC. [VI. 4]  But EB is equal to CD.  Therefore, as BD is to DC, so is DC to CB.  And BD is greater than DC;  therefore DC is also greater than CB.  Therefore the straight line BD has been cut in extreme and mean ratio, and DC is its greater segment.  Q. E. D. 
                                                 
                                                 
PROPOSITION 10. 
 
 
If an equilateral pentagon be inscribed in a circle, the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle. 
 
 
Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE.  I say that the square on the side of the pentagon ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle ABCDE. 
   
   
For let the centre of the circle, the point F, be taken, let AF be joined and carried through to the point G, let FB be joined,  let FH be drawn from F perpendicular to AB and be carried through to K, let AK, KB be joined,  let FL be again drawn from F perpendicular to AK, and be carried through to M, and let KN be joined. 
     
     
Since the circumference ABCG is equal to the circumference AEDG, and in them ABC is equal to AED,  therefore the remainder, the circumference CG, is equal to the remainder GD.  But CD belongs to a pentagon; therefore CG belongs to a decagon.  And, since FA is equal to FB, and FH is perpendicular,  therefore the angle AFK is also equal to the angle KFB. [I. 5, I. 26]  Hence the circumference AK is also equal to KB; [III. 26]  therefore the circumference AB is double of the circumference BK;  therefore the straight line AK is a side of a decagon.  For the same reason AK is also double of KM.  Now, since the circumference AB is double of the circumference BK, while the circumference CD is equal to the circumference AB,  therefore the circumference CD is also double of the circumference BK.  But the circumference CD is also double of CG;  therefore the circumference CG is equal to the circumference BK.  But BK is double of KM, since KA is so also;  therefore CG is also double of KM.  But, further, the circumference CB is also double of the circumference BK,  for the circumference CB is equal to BA.  Therefore the whole circumference GB is also double of BM;  hence the angle GFB is also double of the angle BFM. [VI. 33]  But the angle GFB is also double of the angle FAB,  for the angle FAB is equal to the angle ABF.  Therefore the angle BFN is also equal to the angle FAB.  But the angle ABF is common to the two triangles ABF and BFN;  therefore the remaining angle AFB is equal to the remaining angle BNF; [I. 32]  therefore the triangle ABF is equiangular with the triangle BFN.  Therefore, proportionally, as the straight line AB is to BF, so is FB to BN; [VI. 4]  therefore the rectangle AB, BN is equal to the square on BF. [VI. 17]  Again, since AL is equal to LK, while LN is common and at right angles,  therefore the base KN is equal to the base AN; [I. 4]  therefore the angle LKN is also equal to the angle LAN.  But the angle LAN is equal to the angle KBN;  therefore the angle LKN is also equal to the angle KBN.  And the angle at A is common to the two triangles AKB and AKN.  Therefore the remaining angle AKB is equal to the remaining angle KNA; [I. 32]  therefore the triangle KBA is equiangular with the triangle KNA.  Therefore, proportionally, as the straight line BA is to AK, so is KA to AN; [VI. 4]  therefore the rectangle BA, AN is equal to the square on AK. [VI. 17]  But the rectangle AB, BN was also proved equal to the square on BF;  therefore the rectangle AB, BN together with the rectangle BA, AN, that is, the square on BA, [II. 2]  is equal to the square on BF together with the square on AK.  And BA is a side of the pentagon, BF of the hexagon [IV. 15, Por.], and AK of the decagon. 
                                                                                 
                                                                                 
Therefore etc.  Q. E. D. 
   
   
PROPOSITION 11. 
 
 
If in a circle which has its diameter rational an equilateral pentagon be inscribed, the side of the pentagon is the irrational straight line called minor. 
 
 
For in the circle ABCDE which has its diameter rational let the equilateral pentagon ABCDE be inscribed;  I say that the side of the pentagon is the irrational straight line called minor. 
   
   
For let the centre of the circle, the point F, be taken, let AF, FB be joined and carried through to the points, G, H, let AC be joined, and let FK be made a fourth part of AF.  Now AF is rational; therefore FK is also rational.  But BF is also rational; therefore the whole BK is rational.  And, since the circumference ACG is equal to the circumference ADG, and in them ABC is equal to AED,  therefore the remainder CG is equal to the remainder GD.  And, if we join AD, we conclude that the angles at L are right, and CD is double of CL.  For the same reason the angles at M are also right, and AC is double of CM.  Since then the angle ALC is equal to the angle AMF, and the angle LAC is common to the two triangles ACL and AMF,  therefore the remaining angle ACL is equal to the remaining angle MFA; [I. 32]  therefore the triangle ACL is equiangular with the triangle AMF;  therefore, proportionally, as LC is to CA, so is MF to FA.  And the doubles of the antecedents may be taken;  therefore, as the double of LC is to CA, so is the double of MF to FA.  But, as the double of MF is to FA, so is MF to the half of FA;  therefore also, as the double of LC is to CA, so is MF to the half of FA.  And the halves of the consequents may be taken;  therefore, as the double of LC is to the half of CA, so is MF to the fourth of FA.  And DC is double of LC, CM is half of CA, and FK a fourth part of FA;  therefore, as DC is to CM, so is MF to FK.  Componendo also, as the sum of DC, CM is to CM, so is MK to KF; [V. 18]  therefore also, as the square on the sum of DC, CM is to the square on CM, so is the square on MK to the square on KF.  And since, when the straight line subtending two sides of the pentagon, as AC, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to DC, [XIII. 8]  while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, [XIII. 1]  and CM is half of the whole AC,  therefore the square on DC, CM taken as one straight line is five times the square on CM.  But it was proved that, as the square on DC, CM taken as one straight line is to the square on CM, so is the square on MK to the square on KF;  therefore the square on MK is five times the square on KF.  But the square on KF is rational, for the diameter is rational;  therefore the square on MK is also rational; therefore MK is rational  And, since BF is quadruple of FK, therefore BK is five times KF;  therefore the square on BK is twenty-five times the square on KF.  But the square on MK is five times the square on KF;  therefore the square on BK is five times the square on KM;  therefore the square on BK has not to the square on KM the ratio which a square number has to a square number;  therefore BK is incommensurable in length with KM. [X. 9]  And each of them is rational.  Therefore BK, KM are rational straight lines commensurable in square only.  But, if from a rational straight line there be subtracted a rational straight line which is commensurable with the whole in square only, the remainder is irrational, namely an apotome;  therefore MB is an apotome and MK the annex to it. [X. 73]  I say next that MB is also a fourth apotome. 
                                                                               
                                                                               
Let the square on N be equal to that by which the square on BK is greater than the square on KM;  therefore the square on BK is greater than the square on KM by the square on N.  And, since KF is commensurable with FB, componendo also, KB is commensurable with FB. [X. 15]  But BF is commensurable with BH;  therefore BK is also commensurable with BH. [X. 12]  And, since the square on BK is five times the square on KM,  therefore the square on BK has to the square on KM the ratio which 5 has to 1.  Therefore, convertendo, the square on BK has to the square on N the ratio which 5 has to 4 [V. 19, Por.], and this is not the ratio which a square number has to a square number;  therefore BK is incommensurable with N; [X. 9]  therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK.  Since then the square on the whole BK is greater than the square on the annex KM by the square on a straight line incommensurable with BK,  and the whole BK is commensurable with the rational straight line, BH, set out, therefore MB is a fourth apotome. [X. Deff. III. 4]  But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. [X. 94]  But the square on AB is equal to the rectangle HB, BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and, as HB is to BA, so is AB to BM. 
                           
                           
Therefore the side AB of the pentagon is the irrational straight line called minor.  Q. E. D. 
   
   
PROPOSITION 12. 
 
 
If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle. 
 
 
Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it;  I say that the square on one side of the triangle ABC is triple of the square on the radius of the circle. 
   
   
For let the centre D of the circle ABC be taken, let AD be joined and carried through to E, and let BE be joined. 
 
 
Then, since the triangle ABC is equilateral,  therefore the circumference BEC is a third part of the circumference of the circle ABC.  Therefore the circumference BE is a sixth part of the circumference of the circle;  therefore the straight line BE belongs to a hexagon;  therefore it is equal to the radius DE. [IV. 15, Por.]  And, since AE is double of DE, the square on AE is quadruple of the square on ED, that is, of the square on BE.  But the square on AE is equal to the squares on AB, BE; [III. 31, I. 47]  therefore the squares on AB, BE are quadruple of the square on BE.  Therefore, separando, the square on AB is triple of the square on BE.  But BE is equal to DE;  therefore the square on AB is triple of the square on DE. 
                     
                     
Therefore the square on the side of the triangle is triple of the square on the radius.  Q. E. D. 
   
   
PROPOSITION 13. 
 
 
To construct a pyramid, to comprehend it in a given sphere, and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. 
 
 
Let the diameter AB of the given sphere be set out, and let it be cut at the point C so that AC is double of CB;  let the semicircle ADB be described on AB, let CD be drawn from the point C at right angles to AB, and let DA be joined;  let the circle EFG which has its radius equal to DC be set out, let the equilateral triangle EFG be inscribed in the circle EFG, [IV. 2]  let the centre of the circle, the point H, be taken, [III. 1]  let EH, HF, HG be joined;  from the point H let HK be set up at right angles to the plane of the circle EFG, [XI. 12]  let HK equal to the straight line AC be cut off from HK,  and let KE, KF, KG be joined.  Now, since KH is at right angles to the plane of the circle EFG,  therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle EFG. [XI. Def. 3]  But each of the straight lines HE, HF, HG meets it:  therefore HK is at right angles to each of the straight lines HE, HF, HG.  And, since AC is equal to HK, and CD to HE, and they contain right angles,  therefore the base DA is equal to the base KE. [I. 4]  For the same reason each of the straight lines KF, KG is also equal to DA;  therefore the three straight lines KE, KF, KG are equal to one another.  And, since AC is double of CB, therefore AB is triple of BC.  But, as AB is to BC, so is the square on AD to the square on DC, as will be proved afterwards.  Therefore the square on AD is triple of the square on DC.  But the square on FE is also triple of the square on EH, [XIII. 12]  and DC is equal to EH;  therefore DA is also equal to EF.  But DA was proved equal to each of the straight lines KE, KF, KG;  therefore each of the straight lines EF, FG, GE is also equal to each of the straight lines KE, KF, KG;  therefore the four triangles EFG, KEF, KFG, KEG are equilateral.  Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex. 
                                                   
                                                   
It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. 
 
 
For let the straight line HL be produced in a straight line with KH, and let HL be made equal to CB.  Now, since, as AC is to CD, so is CD to CB, [VI. 8, Por.]  while AC is equal to KH, CD to HE, and CB to HL,  therefore, as KH is to HE, so is EH to HL;  therefore the rectangle KH, HL is equal to the square on EH. [VI. 17]  And each of the angles KHE, EHL is right;  therefore the semicircle described on KL will pass through E also. [cf. VI. 8, III. 31.]  If then, KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, since, if FL, LG be joined, the angles at F, G similarly become right angles;  and the pyramid will be comprehended in the given sphere.  For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB. 
                   
                   
I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid 
 
 
For, since AC is double of CB, therefore AB is triple of BC;  and, convertendo, BA is one and a half times AC.  But, as BA is to AC, so is the square on BA to the square on AD.  Therefore the square on BA is also one and a half times the square on AD.  And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid. 
         
         
Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.  Q. E. D. 
   
   
LEMMA.
It is to be proved that, as AB is to BC, so is the square on AD to the square on DC. 
 
 
For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed.  Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4] therefore the rectangle BA, AC is equal to the square on AD. [VI. 17]  And since, as AB is to BC, so is EB to BF, [VI. 1]  and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB,  therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB.  And the rectangle BA, AC is equal to the square on AD, and the rectangle AC, CB to the square on DC,  for the perpendicular DC is a mean proportional between the segments AC, CB of the base, because the angle ADB is right. [VI. 8, Por.] 
             
             
Therefore, as AB is to BC, so is the square on AD to the square on DC.  Q. E. D. 
   
   
PROPOSITION 14. 
 
 
To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron. 
 
 
Let the diameter AB of the given sphere be set out, and let it be bisected at C;  let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, let DB be joined;  let the square EFGH, having each of its sides equal to DB, be set out, let HF, EG be joined,  from the point K let the straight line KL be set up at right angles to the plane of the square EFGH, [XI. 12]  and let it be carried through to the other side of the plane, as KM;  from the straight lines KL, KM let KL, KM be respectively cut off equal to one of the straight lines EK, FK, GK, HK, and let LE, LF, LG, LH, ME, MF, MG, MH be joined. 
           
           
Then, since KE is equal to KH, and the angle EKH is right,  therefore the square on HE is double of the square on EK. [I. 47]  Again, since LK is equal to KE, and the angle LKE is right,  therefore the square on EL is double of the square on EK. [id.]  But the square on HE was also proved double of the square on EK;  therefore the square on LE is equal to the square on EH;  therefore LE is equal to EH.  For the same reason LH is also equal to HE;  therefore the triangle LEH is equilateral.  Similarly we can prove that each of the remaining triangles  of which the sides of the square EFGH are the bases, and the points L, M the vertices, is equilateral;  therefore an octahedron has been constructed which is contained by eight equilateral triangles. 
                       
                       
It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron. 
 
 
For, since the three straight lines LK, KM, KE are equal to one another,  therefore the semicircle described on LM will also pass through E.  And for the same reason, if, LM remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved,  it will also pass through the points F, G, H, and the octahedron will have been comprehended in a sphere.  I say next that it is also comprehended in the given sphere. 
         
         
For, since LK is equal to KM, while KE is common, and they contain right angles,  therefore the base LE is equal to the base EM. [I. 4]  And, since the angle LEM is right, for it is in a semicircle, [III. 31]  therefore the square on LM is double of the square on LE. [I. 47]  Again, since AC is equal to CB, AB is double of BC.  But, as AB is to BC, so is the square on AB to the square on BD;  therefore the square on AB is double of the square on BD.  But the square on LM was also proved double of the square on LE.  And the square on DB is equal to the square on LE, for EH was made equal to DB.  Therefore the square on AB is also equal to the square on LM;  therefore AB is equal to LM.  And AB is the diameter of the given sphere;  therefore LM is equal to the diameter of the given sphere. 
                         
                         
Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double of the square on the side of the octahedron.  Q. E. D. 
   
   
PROPOSITION 15. 
 
 
To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. 
 
 
Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is double of CB;  let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, and let DB be joined;  let the square EFGH having its side equal to DB be set out, from E, F, G, H let EK, FL, GM, HN be drawn at right angles to the plane of the square EFGH,  from EK, FL, GM, HN let EK, FL, GM, HN respectively be cut off equal to one of the straight lines EF, FG, GH, HE, and let KL, LM, MN, NK be joined;  therefore the cube FN has been constructed which is contained by six equal squares.  It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. 
           
           
For let KG, EG be joined.  Then, since the angle KEG is right, because KE is also at right angles to the plane EG and of course to the straight line EG also, [XI. Def. 3]  therefore the semicircle described on KG will also pass through the point E.  Again, since GF is at right angles to each of the straight lines FL, FE, GF is also at right angles to the plane FK;  hence also, if we join FK, GF will be at right angles to FK;  and for this reason again the semicircle described on GK will also pass through F.  Similarly it will also pass through the remaining angular points of the cube.  If then, KG remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved,  the cube will be comprehended in a sphere.  I say next that it is also comprehended in the given sphere. 
                   
                   
For, since GF is equal to FE, and the angle at F is right,  therefore the square on EG is double of the square on EF.  But EF is equal to EK;  therefore the square on EG is double of the square on EK;  hence the squares on GE, EK, that is the square on GK, [I. 47]  is triple of the square on EK.  And, since AB is triple of BC, while, as AB is to BC, so is the square on AB to the square on BD,  therefore the square on AB is triple of the square on BD.  But the square on GK was also proved triple of the square on KE.  And KE was made equal to DB;  therefore KG is also equal to AB.  And AB is the diameter of the given sphere;  therefore KG is also equal to the diameter of the given sphere. 
                         
                         
Therefore the cube has been comprehended in the given sphere; and it has been demonstrated at the same time that the square on the diameter of the sphere is triple of the square on the side of the cube.  Q. E. D. 
   
   
PROPOSITION 16. 
 
 
To construct an icosahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor. 
 
 
Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is quadruple of CB,  let the semicircle ADB be described on AB,  let the straight line CD be drawn from C at right angles to AB, and let DB be joined;  let the circle EFGHK be set out and let its radius be equal to DB,  let the equilateral and equiangular pentagon EFGHK be inscribed in the circle EFGHK,  let the circumferences EF, FG, GH, HK, KE be bisected at the points L, M, N, O, P, and let LM, MN, NO, OP, PL, EP be joined.  Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon.  Now from the points E, F, G, H, K let the straight lines EQ, FR, GS, HT, KU be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle EFGHK, let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, PQ be joined. 
               
               
Now, since each of the straight lines EQ, KU is at right angles to the same plane, therefore EQ is parallel to KU. [XI. 6]  But it is also equal to it;  and the straight lines joining those extremities of equal and parallel straight lines which are in the same direction are equal and parallel. [I. 33]  Therefore QU is equal and parallel to EK.  But EK belongs to an equilateral pentagon;  therefore QU also belongs to the equilateral pentagon inscribed in the circle EFGHK.  For the same reason each of the straight lines QR, RS, ST, TU also belongs to the equilateral pentagon inscribed in the circle EFGHK;  therefore the pentagon QRSTU is equilateral.  And, since QE belongs to a hexagon, and EP to a decagon, and the angle QEP is right,  therefore QP belongs to a pentagon;  for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [XIII. 10]  For the same reason PU is also a side of a pentagon.  But QU also belongs to a pentagon;  therefore the triangle QPU is equilateral.  For the same reason each of the triangles QLR, RMS, SNT, TOU is also equilateral.  And, since each of the straight lines QL, QP was proved to belong to a pentagon, and LP also belongs to a pentagon,  therefore the triangle QLP is equilateral.  For the same reason each of the triangles LRM, MSN, NTO, OUP is also equilateral. 
                                   
                                   
Let the centre of the circle EFGHK the point V, be taken;  from V let VZ be set up at right angles to the plane of the circle, let it be produced in the other direction, as VX, let there be cut off VW, the side of a hexagon, and each of the straight lines VX, WZ, being sides of a decagon, and let QZ, QW, UZ, EV, LV, LX, XM be joined. 
   
   
Now, since each of the straight lines VW, QE is at right angles to the plane of the circle, therefore VW is parallel to QE. [XI. 6]  But they are also equal; therefore EV, QW are also equal and parallel. [I. 33]  But EV belongs to a hexagon; therefore QW also belongs to a hexagon.  And, since QW belongs to a hexagon, and WZ to a decagon, and the angle QWZ is right, therefore QZ belongs to a pentagon. [XIII. 10]  For the same reason UZ also belongs to a pentagon,  inasmuch as, if we join VK, WU, they will be equal and opposite, and VK, being a radius, belongs to a hexagon; [IV. 15, Por.]  therefore WU also belongs to a hexagon.  But WZ belongs to a decagon, and the angle UWZ is right;  therefore UZ belongs to a pentagon. [XIII. 10]  But QU also belongs to a pentagon;  therefore the triangle QUZ is equilateral.  For the same reason each of the remaining triangles  of which the straight lines QR, RS, ST, TU are the bases, and the point Z the vertex, is also equilateral.  Again, since VL belongs to a hexagon, and VX to a decagon, and the angle LVX is right,  therefore LX belongs to a pentagon. [XIII. 10]  For the same reason, if we join MV, which belongs to a hexagon, MX is also inferred to belong to a pentagon.  But LM also belongs to a pentagon;  therefore the triangle LMX is equilateral.  Similarly it can be proved that each of the remaining triangles of which MN, NO, OP, PL are the bases, and the point X the vertex, is also equilateral.  Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles. 
                                       
                                       
It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor. 
 
 
For, since VW belongs to a hexagon, and WZ to a decagon,  therefore VZ has been cut in extreme and mean ratio at W, and VW is its greater segment; [XIII. 9]  therefore, as ZV is to VW, so is VW to WZ.  But VW is equal to VE, and WZ to VX;  therefore, as ZV is to VE, so is EV to VX.  And the angles ZVE, EVX are right;  therefore, if we join the straight line EZ, the angle XEZ will be right because of the similarity of the triangles XEZ, VEZ.  For the same reason, since, as ZV is to VW, so is VW to WZ, and ZV is equal to XW, and VW to WQ,  therefore, as XW is to WQ, so is QW to WZ.  And for this reason again, if we join QX, the angle at Q will be right; [VI. 8]  therefore the semicircle described on XZ will also pass through Q. [III. 31]  And if, XZ remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved,  it will also pass through Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere.  I say next that it is also comprehended in the given sphere. 
                           
                           
For let VW be bisected at A'.  Then, since the straight line VZ has been cut in extreme and mean ratio at W, and ZW is its lesser segment,  therefore the square on ZW added to the half of the greater segment, that is WA', is five times the square on the half of the greater segment; [XIII. 3]  therefore the square on ZA' is five times the square on .  And ZX is double of ZA', and VW double of ;  therefore the square on ZX is five times the square on WV.  And, since AC is quadruple of CB, therefore AB is five times BC.  But, as AB is to BC, so is the square on AB to the square on BD; [VI. 8, V. Def. 9]  therefore the square on AB is five times the square on BD.  But the square on ZX was also proved to be five times the square on VW.  And DB is equal to VW,  for each of them is equal to the radius of the circle EFGHK;  therefore AB is also equal to XZ.  And AB is the diameter of the given sphere;  therefore XZ is also equal to the diameter of the given sphere.  Therefore the icosahedron has been comprehended in the given sphere. 
                               
                               
I say next that the side of the icosahedron is the irrational straight line called minor.  For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle EFGHK,  therefore the radius of the circle EFGHK is also rational;  hence its diameter is also rational.  But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [XIII. 11]  And the side of the pentagon EFGHK is the side of the icosahedron.  Therefore the side of the icosahedron is the irrational straight line called minor. 
             
             
PORISM.
From this it is manifest that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. 
Q. E. D. 
   
   
PROPOSITION 17. 
 
 
To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome. 
 
 
Let ABCD, CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides AB, BC, CD, DA, EF, EB, FC be bisected at G, H, K, L, M, N, O respectively, let GK, HL, MH, NO be joined, let the straight lines NP, PO, HQ be cut in extreme and mean ratio at the points R, S, T respectively, and let RP, PS, TQ be their greater segments; from the points R, S, T let RU, SV, TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to RP, PS, TQ, and let UB, BW, WC, CV, VU be joined.  I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular. 
   
   
For let RB, SB, VB be joined.  Then, since the straight line NP has been cut in extreme and mean ratio at R, and RP is the greater segment,  therefore the squares on PN, NR are triple of the square on RP. [XIII. 4]  But PN is equal to NB, and PR to RU;  therefore the squares on BN, NR are triple of the square on RU.  But the square on BR is equal to the squares on BN, NR; [I. 47]  therefore the square on BR is triple of the square on RU;  hence the squares on BR, RU are quadruple of the square on RU.  But the square on BU is equal to the squares on BR, RU;  therefore the square on BU is quadruple of the square on RU;  therefore BU is double of RU.  But VU is also double of UR, inasmuch as SR is also double of PR, that is, of RU;  therefore BU is equal to UV.  Similarly it can be proved that each of the straight lines BW, WC, CV is also equal to each of the straight lines BU, UV.  Therefore the pentagon BUVCW is equilateral. 
                             
                             
I say next that it is also in one plane. 
 
 
For let PX be drawn from P parallel to each of the straight lines RU, SV and towards the outside of the cube, and let XH, HW be joined;  I say that XHW is a straight line. 
   
   
For, since HQ has been cut in extreme and mean ratio at T, and QT is its greater segment,  therefore, as HQ is to QT, so is QT to TH.  But HQ is equal to HP, and QT to each of the straight lines TW, PX;  therefore, as HP is to PX, so is WT to TH.  And HP is parallel to TW,  for each of them is at right angles to the plane BD; [XI. 6]  and TH is parallel to PX,  for each of them is at right angles to the plane BF. [id.]  But if two triangles, as XPH, HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel,  the remaining straight lines will be in a straight line; [VI. 32]  therefore XH is in a straight line with HW.  But every straight line is in one plane; [XI. 1]  therefore the pentagon UBWCV is in one plane. 
                         
                         
I say next that it is also equiangular. 
 
 
For, since the straight line NP has been cut in extreme and mean ratio at R, and PR is the greater segment, while PR is equal to PS,  therefore NS has also been cut in extreme and mean ratio at P, and NP is the greater segment; [XIII. 5]  therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]  But NP is equal to NB, and PS to SV;  therefore the squares on NS, SV are triple of the square on NB;  hence the squares on VS, SN, NB are quadruple of the square on NB.  But the square on SB is equal to the squares on SN, NB;  therefore the squares on BS, SV, that is, the square on BV — for the angle VSB is right — is quadruple of the square on NB;  therefore VB is double of BN.  But BC is also double of BN;  therefore BV is equal to BC.  And, since the two sides BU, UV are equal to the two sides BW, WC, and the base BV is equal to the base BC,  therefore the angle BUV is equal to the angle BWC. [I. 8]  Similarly we can prove that the angle UVC is also equal to the angle BWC;  therefore the three angles BWC, BUV, UVC are equal to one another.  But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [XIII. 7]  therefore the pentagon BUVCW is equiangular.  And it was also proved equilateral;  therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube.  Therefore, if we make the same construction in the case of each of the twelve sides of the cube,  a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron. 
                                         
                                         
It is then required to comprehend it in the given sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome. 
 
 
For let XP be produced, and let the produced straight line be XZ;  therefore PZ meets the diameter of the cube, and they bisect one another,  for this has been proved in the last theorem but one of the eleventh book. [XI. 38]  Let them cut at Z;  therefore Z is the centre of the sphere which comprehends the cube, and ZP is half of the side of the cube.  Let UZ be joined.  Now, since the straight line NS has been cut in extreme and mean ratio at P, and NP is its greater segment,  therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]  But NS is equal to XZ, inasmuch as NP is also equal to PZ, and XP to PS.  But further PS is also equal to XU, since it is also equal to RP;  therefore the squares on ZX, XU are triple of the square on NP.  But the square on UZ is equal to the squares on ZX, XU;  therefore the square on UZ is triple of the square on NP.  But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube,  for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. [XIII. 15]  But, if whole is so related to whole, so is half to half also;  and NP is half of the side of the cube;  therefore UZ is equal to the radius of the sphere which comprehends the cube.  And Z is the centre of the sphere which comprehends the cube;  therefore the point U is on the surface of the sphere.  Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere;  therefore the dodecahedron has been comprehended in the given sphere. 
                                           
                                           
I say next that the side of the dodecahedron is the irrational straight line called apotome. 
 
 
For since, when NP has been cut in extreme and mean ratio, RP is the greater segment,  and, when PO has been cut in extreme and mean ratio, PS is the greater segment,  therefore, when the whole NO is cut in extreme and mean ratio, RS is the greater segment.  [Thus, since, as NP is to PR, so is PR to RN, the same is true of the doubles also,  for parts have the same ratio as their equimultiples; [V. 15]  therefore as NO is to RS, so is RS to the sum of NR, SO.  But NO is greater than RS;  therefore RS is also greater than the sum of NR, SO;  therefore NO has been cut in extreme and mean ratio, and RS is its greater segment.]  But RS is equal to UV;  therefore, when NO is cut in extreme and mean ratio, UV is the greater segment.  And, since the diameter of the sphere is rational, and the square on it is triple of the square on the side of the cube,  therefore NO, being a side of the cube, is rational.  [But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.] 
                           
                           
Therefore UV, being a side of the dodecahedron, is an irrational apotome. [XIII. 6] 
 
 
PORISM.
From this it is manifest that, when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. 
Q. E. D. 
   
   
PROPOSITION 18. 
 
 
To set out the sides of the five figures and to compare them with one another. 
 
 
Let AB, the diameter of the given sphere, be set out,  and let it be cut at C so that AC is equal to CB,  and at D so that AD is double of DB;  let the semicircle AEB be described on AB,  from C, D let CE, DF be drawn at right angles to AB,  and let AF, FB, EB be joined.  Then, since AD is double of DB,  therefore AB is triple of BD.  Convertendo, therefore, BA is one and a half times AD.  But, as BA is to AD, so is the square on BA to the square on AF, [V. Def. 9, VI. 8]  for the triangle AFB is equiangular with the triangle AFD;  therefore the square on BA is one and a half times the square on AF.  But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [XIII. 13]  And AB is the diameter of the sphere;  therefore AF is equal to the side of the pyramid. 
                             
                             
Again, since AD is double of DB, therefore AB is triple of BD.  But, as AB is to BD, so is the square on AB to the square on BF; [VI. 8, V. Def. 9]  therefore the square on AB is triple of the square on BF.  But the square on the diameter of the sphere is also triple of the square on the side of the cube. [XIII. 15]  And AB is the diameter of the sphere;  therefore BF is the side of the cube. 
           
           
And, since AC is equal to CB, therefore AB is double of BC.  But, as AB is to BC, so is the square on AB to the square on BE;  therefore the square on AB is double of the square on BE.  But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [XIII. 14]  And AB is the diameter of the given sphere;  therefore BE is the side of the octahedron. 
           
           
Next, let AG be drawn from the point A at right angles to the straight line AB, let AG be made equal to AB, let GC be joined, and from H let HK be drawn perpendicular to AB.  Then, since GA is double of AC, for GA is equal to AB, and, as GA is to AC, so is HK to KC,  therefore HK is also double of KC.  Therefore the square on HK is quadruple of the square on KC;  therefore the squares on HK, KC, that is, the square on HC, is five times the square on KC.  But HC is equal to CB;  therefore the square on BC is five times the square on CK.  And, since AB is double of CB, and, in them, AD is double of DB,  therefore the remainder BD is double of the remainder DC.  Therefore BC is triple of CD;  therefore the square on BC is nine times the square on CD.  But the square on BC is five times the square on CK;  therefore the square on CK is greater than the square on CD;  therefore CK is greater than CD.  Let CL be made equal to CK, from L let LM be drawn at right angles to AB, and let MB be joined.  Now, since the square on BC is five times the square on CK,  and AB is double of BC, and KL double of CK,  therefore the square on AB is five times the square on KL.  But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [XIII. 16, Por.]  And AB is the diameter of the sphere;  therefore KL is the radius of the circle from which the icosahedron has been described;  therefore KL is a side of the hexagon in the said circle. [IV. 15, Por.]  And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [XIII. 16, Por.]  and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK is equal to LB,  therefore each of the straight lines AK, LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.  And, since LB belongs to a decagon, and ML to a hexagon,  for ML is equal to KL, since it is also equal to HK,  being the same distance from the centre,  and each of the straight lines HK, KL is double of KC,  therefore MB belongs to a pentagon. [XIII. 10]  But the side of the pentagon is the side of the icosahedron; [XIII. 16]  therefore MB belongs to the icosahedron. 
                                                               
                                                               
Now, since FB is a side of the cube, let it be cut in extreme and mean ratio at N, and let NB be the greater segment;  therefore NB is a side of the dodecahedron. [XIII. 17, Por.] 
   
   
And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side AF of the pyramid,  double of the square on the side BE of the octahedron and triple of the side FB of the cube,  therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.  Therefore the square on the side of the pyramid is fourthirds of the square on the side of the octahedron, and double of the square on the side of the cube;  and the square on the side of the octahedron is one and a half times the square on the side of the cube.  The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.  But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides;  for they are irrational, the one being minor [XIII. 16] and the other an apotome [XIII. 17]. 
               
               
That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus. 
 
 
For, since the triangle FDB is equiangular with the triangle FAB, [VI. 8]  proportionally, as DB is to BF, so is BF to BA. [VI. 4]  And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9, VI. 20, Por.]  therefore, as DB is to BA, so is the square on DB to the square on BF;  therefore, inversely, as AB is to BD, so is the square on FB to the square on BD.  But AB is triple of BD;  therefore the square on FB is triple of the square on BD.  But the square on AD is also quadruple of the square on DB,  for AD is double of DB;  therefore the square on AD is greater than the square on FB;  therefore AD is greater than FB;  therefore AL is by far greater than FB.  And, when AL is cut in extreme and mean ratio, KL is the greater segment,  inasmuch as LK belongs to a hexagon, and KA to a decagon; [XIII. 9]  and, when FB is cut in extreme and mean ratio, NB is the greater segment;  therefore KL is greater than NB.  But KL is equal to LM;  therefore LM is greater than NB.  Therefore MB, which is a side of the icosahedron, is by far greater than NB which is a side of the dodecahedron.  Q. E. D. 
                                       
                                       
I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another. 
 
 
For a solid angle cannot be constructed with two triangles, or indeed planes.  With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron;  but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point,  for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible,  for any solid angle is contained by angles less than four right angles. [XI. 21]  For the same reason, neither can a solid angle be constructed by more than six plane angles.  By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained,  for they will again be four right angles.  By three equilateral and equiangular pentagons the angle of the dodecahedron is contained;  but by four such it is impossible for any solid angle to be contained,  for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.  Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity. 
                       
                       
Therefore etc.  Q. E. D. 
   
   
LEMMA.
But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus. 
 
 
Let ABCDE be an equilateral and equiangular pentagon, let the circle ABCDE be circumscribed about it,  let its centre F be taken, and let FA, FB, FC, FD, FE be joined.  Therefore they bisect the angles of the pentagon at A, B, C, D, E.  And, since the angles at F are equal to four right angles and are equal,  therefore one of them, as the angle AFB, is one right angle less a fifth;  therefore the remaining angles FAB, ABF consist of one right angle and a fifth.  But the angle FAB is equal to the angle FBC;  therefore the whole angle ABC of the pentagon consists of one right angle and a fifth.  Q. E. D.