Bibliotheca Polyglotta

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.

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