Bibliotheca Polyglotta

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,¹ 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,² 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.

1. 19. primi 2. 19. primi

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,³ 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;

3. 8. primi.

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。

1. Probably read 己戊甲 so that chi corresponds to the argument

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.⁴

4. For ‘Quod etat faciendum.’ see two records above.

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,⁵ 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.

5. 1. tertii.

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 ducuntur⁶ ) erunt anguli EAD, ECD, aequales: Est autem angulus EDA, angulo ECD, maior, externus interno opposito,⁷ 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,⁸ 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.

6. 5. primi. 7. 16. primi. 8. 19. primi.

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.⁹ Est autem angulus CDA, angulo CBA, maior, externus interno;¹⁰ Igitur idem angulus CDA, angulo CAD, maior erit. Quare cum CA, sit ducta a centro ad circumferentiam,¹¹ 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.

9. 5. primi. 10. 16. primi. 11. 19. primi.

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,¹² 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.

12. 8. primi.

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.

s

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.¹³ 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.¹⁴

13. 5. primi. 14. 26. primi.

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.¹⁵ 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.

15. 47. primi.

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, ¹⁶ 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.

16. 20. primi.

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,¹⁷ erunt anguli DAB, DBA, aequales: Est autem externus angulus BDE,¹⁸ 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.

17. 5. primi. 18. 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,¹⁹ aequalis est duobus internis DAC, DCA. Hiautem duo²⁰ inter se sunt aequales, quod latera DA, DC sint aequalia; erit angulus BDC, duplus alterius corum, nempe anguli BAC. Quod est propositum.

19. 32. primi. 20. 5. primi.

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.²¹ Quando enim totum totius est duplum, et ablatum ablati; est et reliquum reliqui duplum. In circulo igitur angulus ad centrum duplex est, etc.

21. 20. primi.

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.

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