Bibliotheca Polyglotta

BOOK VΙ.

幾何原本第六卷之首

DEFINITIONS.

界說六則

1. Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional.

第一界
凡形相當之各角等。而各等角旁兩線之比例。俱等。為相似之形。
甲乙丙、丁戊己、兩角形之甲角、與丁角等。乙與戊、丙與己、各等。其甲角旁之甲乙、與甲丙、兩線之比例。若丁角旁之丁戊與丁己兩線。而甲乙與乙丙。若丁戊與戊己。甲丙與丙乙。若丁己與己戊。則此兩角形、為相似之形。依顯凡平邊形、皆相似之形。如庚辛壬、癸子丑、俱平邊角形。其各角俱等。而各邊之比例亦等者、是也。四邊、五邊、以上諸形。俱倣此。

2. [When two sides of one figure together with two sides of another figure form antecedents and consequents in a proportion, the figures are reciprocally related.

第二界
兩形之各兩邊線。互為前後率。相與為比例而等。為互相視之形。
甲乙丙丁、戊己庚辛、兩方形。其甲乙、乙丙、邊。與戊己、己庚、邊。相與為比例等。而彼此互為前、後。如甲乙與。戊己。若己庚與乙丙也。則此兩形為互相視之形。依顯壬癸子、丑寅卯、兩角形之壬子與丑寅。若丑卯與壬癸。或壬癸與丑寅。若丑卯與壬子。亦互相視之形也。

3. A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.

第三界
理分中末線者。一線兩分之。其全與大分之比例。若大分與小分之比例。(p. 二七九)
甲乙線。兩分之於丙。而甲乙與大分甲丙之比例。若大分甲丙與小分丙乙。此為理分中末線。其分法。見本卷三十題。而與二卷十一題理同名異。此線為用甚廣、至量體。尤所必須。十三卷諸題多賴之。古人目為神分線也。

4. The height of any figure is the perpendicular drawn from the vertex to the base.

第四界
度各形之高。皆以垂線之亘為度。
甲乙丙角形。從甲頂、向乙丙底。作甲庚垂線。卽甲庚為甲乙丙之高。又丁戊己角形。作丁辛垂線。卽丁辛為丁戊己之高。若兩形相視。兩垂線等。卽兩形之高、必等。如上兩形在兩平行線之內者是也。若以丙、己、為頂。以甲乙、丁戊、為底。則不等。自餘諸形之度高、俱倣此。(p. 二八○)
凡度物高。以頂底為界。以垂線為度。蓋物之定度。止有一。不得有二。自頂至底。垂線一而己。偏線無數也。

A ratio is said to be compounded of ratios when the sizes of the ratios multiplied together make som (?ratio, or size). A parallelogram not “filling” a line is a figure lesser than the line. If it has a surplus and the line is not enough, it is a figure greater than the line.¹

1. Translation from the Chinese by Engelfriet, p. 204.

第五界
比例以比例相結者。以多比例之命數、相乘、除、而結為一比例之命數。
此各比例、不同理、而相聚為一比例者。則用相結之法。合各比例之命數。求首尾一比例之命數也。曷為比例之命數。謂大幾何、所倍於小幾何若干。或小幾何、在大幾何內若干也。如大幾何、四倍於小。或(p. 二八一)小幾何、為大四分之一。卽各以四為命比例之數也。五卷界　\\　說三今言以彼多比例之命數、相乘、除、而結為此一比例之命數者。如十二倍之此比例。則以彼二倍、六倍、兩比例相結也。二六相乘為十二、故也。或以彼三倍、四倍、兩比例相結也。三四相乘亦十二、故也。又如三十倍之此比例。則以彼二倍、三倍、五倍、三比例相結也。二乘三為六、六乘五為三十、故也。其曰相結者。相結之理。蓋在中率。凡中率為前比例之後。後比例之前。故以二比例合為一比例。則中率為輳合之因。如兩爿合。此為之膠。如兩襟合。此為之紐矣。第五卷第十界、言數幾何為同理之比例。則第一與第三、為再加之比例。再加者。以前中二率之命數。再加為前後二率之命數。亦以中率為紐也。但彼所言者。多比例同理。故止以第一比例之命數累加之。此題所言。則不同理之多比例。不得以第一比例之命數累加之。故用此乘除相結之理。于不同理之中。求其同理。別為累加之法。其紐結之義。頗相類焉。下文仍發明借象(p. 二八二)之術、以需後用也。
五卷言多比例同理者。第一、與第三為再加。與第四為三加。與第五為四加。以至無窮。今此相結之理。亦以三率為始。三率。則兩比例相乘除、而中率為紐也。若四率。則先以前三率之兩比例、相乘除、而結為一比例。復以此初結之比例、與第三比例乘除、相結為一比例也。若五率。則先以前三率之兩比例、乘除相結。復以此再結之比例、與第三比例、乘除相結。又以三結之比例、與第四比例、乘除相結、為一比例也。或以第一第二第三率之兩比例、乘除相結。以第三第四第五之兩比例、乘除相結。又以此二所結比例、乘除相結、而為一比例也。自六以上。倣此以至無窮。
設三幾何、為二比例、不同理、而合為一比例。則以第一與二、第二與三、兩比例相結也。如上圖。三幾何、(p. 二八三)二比例。皆以大不等者。其甲乙與丙丁為二倍大丙丁與戊己為三倍大。則甲乙與戊己、為六倍大。二乘三為六也。若以小不等。戊己為第一。甲乙為第三。三乘二亦六。則戊己與甲乙、為反六倍大也。
甲乙與丙丁。旣二倍大。試以甲乙二平分之。為甲庚、庚乙。必各與丙丁等。丙丁與戊己。旣三倍大。而甲庚、庚乙、各與丙丁等。卽甲庚亦三倍大于戊己。庚乙亦三倍大於戊己。而甲乙必六倍大於戊己。

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

PROPOSITION 1.

幾何原本第六卷　本篇論線面之比例　計三十三題
第一題

Triangles and parallelograms which are under the same height are to one another as their bases.

等高之三角形、方形。自相與為比例。與其底之比例等。

Let ABC, ACD be triangles and EC, CF parallelograms under the same height; I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.

For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH be made equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined.

Then, since CB, BG, GH are equal to one another, the triangles ABC, AGB, AHG are also equal to one another. [I. 38] Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC. For the same reason, whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ACL, and, if less, less. Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and the triangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC; and it has been proved that, if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ALC; if equal, equal; and, if less, less. Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. Def. 5]

Next, since the parallelogram EC is double of the triangle ABC, [I. 41] and the parallelogram FC is double of the triangle ACD, while parts have the same ratio as the same multiples of them, [V. 15] therefore, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram FC. Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF, therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]

Therefore etc. Q. E. D.

PROPOSITION 2.

第二題　二支

If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle.

三角形。任依一邊作平行線。卽此線分兩餘邊以為比例。必等。三角形內。有一線分兩邊以為比例、而等。卽此線與餘邊為平行。

For let DE be drawn parallel to BC, one of the sides of the triangle ABC; I say that, as BD is to DA, so is CE to EA.

For let BE, CD be joined.

Therefore the triangle BDE is equal to the triangle CDE; for they are on the same base DE and in the same parallels DE, BC. [I. 38] And the triangle ADE is another area. But equals have the same ratio to the same; [V. 7] therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. But, as the triangle BDE is to ADE, so is BD to DA; for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1] For the same reason also, as the triangle CDE is to ADE, so is CE to EA. Therefore also, as BD is to DA, so is CE to EA. [V. 11]

Again, let the sides AB, AC of the triangle ABC be cut proportionally, so that, as BD is to DA, so is CE to EA; and let DE be joined. I say that DE is parallel to BC.

For, with the same construction, since, as BD is to DA, so is CE to EA, but, as BD is to DA, so is the triangle BDE to the triangle ADE, and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1] therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11] Therefore each of the triangles BDE, CDE has the same ratio to ADE. Therefore the triangle BDE is equal to the triangle CDE; [V. 9] and they are on the same base DE. But equal triangles which are on the same base are also in the same parallels. [I. 39] Therefore DE is parallel to BC.

Therefore etc. Q. E. D.

PROPOSITION 3.

第三題　二支

If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.

三角形。任以直線、分一角為兩平分。而分對角邊為兩分。則兩分之比例。若餘兩邊之比例。三角形分角(p. 二九一)之線。所分對角邊之比例。若餘兩邊。則所分角為兩平分。

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC.

For let CE be drawn through C parallel to DA, and let BA be carried through and meet it at E.

Then, since the straight line AC falls upon the parallels AD, EC, the angle ACE is equal to the angle CAD. [I. 29] But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE. Again, since the straight line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. [I. 29] But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6] And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE. But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC.

Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D.

For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE: for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11] Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5] But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.] therefore the angle BAD is also equal to the angle CAD. Therefore the angle BAC has been bisected by the straight line AD.

Therefore etc. Q. E. D.

PROPOSITION 4.

第四題

In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

幾何原本　卷六
凡等角三角形。其在等角旁之各兩腰線。相與為比例、必等。而對等角之邊、為相似之邊。

Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

For let BC be placed in a straight line with CE. Then, since the angles ABC, ACB are less than two right angles, [I. 17] and the angle ACB is equal to the angle DEC, therefore the angles ABC, DEC are less than two right angles; therefore BA, ED, when produced, will meet. [I. Post. 5] Let them be produced and meet at F.

Now, since the angle DCE is equal to the angle ABC, BF is parallel to CD. [I. 28] Again, since the angle ACB is equal to the angle DEC, AC is parallel to FE. [I. 28] Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34] And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2] But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE. [V. 16] Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2] But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16] Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]

Therefore etc. Q. E. D.

PROPOSITION 5.

第五題

If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

兩三角形。其各兩邊之比例等。卽兩形為等角形。而對各相似邊之角、各等。

Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF.

For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]

Therefore the triangle ABC is equiangular with the triangle GEF. Therefore in the triangles ABC, GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [VI. 4] therefore, as AB is to BC, so is GE to EF. But, as AB is to BC, so by hypothesis is DE to EF; therefore, as DE is to EF, so is GE to EF. [V. 11] Therefore each of the straight lines DE, GE has the same ratio to EF; therefore DE is equal to GE. [V. 9] For the same reason DF is also equal to GF. Since then DE is equal to EG, and EF is common, the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle DEF is equal to the angle GEF, [I. 8] and the triangle DEF is equal to the triangle GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle DFE is also equal to the angle GFE, and the angle EDF to the angle EGF. And, since the angle FED is equal to the angle GEF, while the angle GEF is equal to the angle ABC, therefore the angle ABC is also equal to the angle DEF. For the same reason the angle ACB is also equal to the angle DFE, and further, the angle at A to the angle at D; therefore the triangle ABC is equiangular with the triangle DEF.

Therefore etc. Q. E. D.

PROPOSITION 6.

第六題

If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

兩三角形之一角等。而等角旁之各兩邊、比例等。卽兩形為等角形。而對各相似邊之角、各等。

Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE.

For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23] therefore the remaining angle at B is equal to the remaining angle at G. [I. 32]

Therefore the triangle ABC is equiangular with the triangle DGF. Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4] But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11] Therefore ED is equal to DG; [V. 9] and DF is common; therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle DFG is equal to the angle DFE, and the angle DGF to the angle DEF. But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE. And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32] therefore the triangle ABC is equiangular with the triangle DEF.

Therefore etc. Q. E. D.

PROPOSITION 7.

第七題

If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional.

兩三角形之第一角等。而第二相當角、各兩旁之邊、比例等。其第三相當角。或俱小于直角。或俱不小于直角。卽兩形為等角形。而對各相似邊之角、各等。

Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle; I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F.

For, if the angle ABC is unequal to the angle DEF, one of them is greater. Let the angle ABC be greater; and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23]

Then, since the angle A is equal to D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32] Therefore the triangle ABG is equiangular with the triangle DEF. Therefore, as AB is to BG, so is DE to EF [VI. 4] But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9] so that the angle at C is also equal to the angle BGC. [I. 5] But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13] And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle. But it is by hypothesis less than a right angle: which is absurd. Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it. But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF.

But, again, let each of the angles at C, F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF.

For, with the same construction, we can prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC. [I. 5] But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle. Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17] Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it. But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF.

Therefore etc. Q. E. D.

PROPOSITION 8.

第八題

If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another.

直角三邊形。從直角向對邊。作一垂線。分本形為兩直角三邊形。卽兩形皆與全形相似。亦自相似。

Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another.

For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the two triangles ABC and ABD, therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32] therefore the triangle ABC is equiangular with the triangle ABD. Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4] Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional. Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1] Similarly we can prove that the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC.

I say next that the triangles ABD, ADC are also similar to one another.

For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC. Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]

Therefore etc.

PORISM.
From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Q. E. D.

PROPOSITION 9.

PROBL. 1 PROPOS. 9

第九題

From a given straight line to cut off a prescribed part.

ADATA recta linea imperatam partem auferre.

一直線。求截所取之分。

Let AB be the given straight line; thus it is required to cut off from AB a prescribed part.

IMPERETVR, ut ex linea AB, auferamus partem tertiam.

法曰。甲乙直線。求截取三分之一。

Let the third part be that prescribed. Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3] Let BC be joined, and through D let DF be drawn parallel to it. [I. 31]

Ex A, ducatur recta AC, utcunque faciens angulum CAB; et ex AC, abscindantur tot partes aequales cuiuslibet magnitudinis, quota pars detrahenda est ex AB, ut in proposito exemplo tres AD, DE, EF. Deinde ex F, ad B, recta ducatur FB, cui per D, parallela agatur DG.

先從甲、任作一甲丙線、為丙甲乙角。次從甲向丙。任作所命分之平度。如甲丁、丁戊、戊己、為三分也。次作己乙直線。末作丁庚線。與己乙平行。卽甲庚為甲乙三分之一。

Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2] But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF.

Dico AG, esse partem tertiam imperatam rectae AB. Nam cum in triangulo ABF, lateri FB, parallela sit recta DG;¹ erit ut FD, ad DA, ita BG, ad GA. ² Componendo igitur , ut FA, ad DA, ita BA, erit ad GA: sed FA, ipsius AD, est tripla, ex constructione. Igitur et BA, ipsis AG, erit tripla, ideoque AG, tertia pars erit ipsius AB, quae imperabatur.

1. 2. sexti. 2. 18. quinti.

論曰。甲乙己角形內之丁庚線。旣與乙己邊平行。卽己丁與丁甲之比例。若乙庚與庚甲也。本篇二合之。己甲與甲丁。若乙甲與庚甲也。五卷十八而甲丁旣為己甲三分之一。卽庚甲亦為乙甲三分之一也。

Therefore from the given straight line AB the prescribed third part AF has been cut off. Q. E. F.

A data ergo recta linea imperatam partem abstulimus. Quod faciendum erat.

SCHOLIUM. QVOD si ex AB, auferenda sit pars non aliquota, sed quae plures aliquotas non efficientes unam complectatur, nimirum quae contineat quatuor undecimas ipsius AB, sumendae erunt ex AC, undecim partes aquales usque ad D, punctum, ex quo ad B recta ducatur DB; et huic parallela EF, ex E, termino quatuor partium. Nam AF, erit pars imperata. Erit enim rursus ut DA, ad AE, ita BA, ad AF. Quare, et convertendo ut AE, ad AD, ita AF, ad AB: Est autem AE, pars continens quatuor undecimas ipsius AD, ex constructione. Igitur et AF, eadem pars erit rectae AB. Quod est propositum. Non aliter detrahetur ex AB, pars complectens quotcunque partes ipsius aliquotas non facientes unam.

注曰。甲乙線。欲截取十一分之四。先作甲丙線、為丙甲乙角。從甲向丙。任平分十一分、至丁。次作丁乙線。末從甲取四分、得戊。作戊己線。與丁乙平行。卽甲己為十一分甲乙之四。何者。依上論、丁甲與戊甲之比例。若乙甲與己甲也。反之。甲戊與甲丁。若甲己與甲乙也。五卷四甲戊為十一分甲丁之四。則甲己亦十一分甲乙之四矣。依此可推不盡分之數。蓋四不為十一之盡分故。

PROPOSITION 10.

第十題

To cut a given uncut straight line similarly to a given cut straight line.

一直線。求截各分。如所設之截分

Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31]

Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34] Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2] But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF. Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2] But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA.

Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC. Q. E. F.

PROPOSITION 11.

PROBL. 3. PROPOS. 11.

第十一題

To two given straight lines to find a third proportional.

DUABUS datis rectis lineis tertiam proportionalem adinuenire.

兩直線。求別作一線。相與為連比例。

Let BA, AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to BA, AC. For let them be produced to the points D, E, and let BD be made equal to AC; [I. 3] let BC be joined, and through D let DE be drawn parallel to it. [I. 31]

SINT duae rectae AB, AC, ita dispositae, ut efficiant angulum A, quemcunque, fitque inuenienda illis tertia proportionalis, sicut quidem AB, ad AC, ita AC, ad tertiam. Producatur AB, quam volumus esse antecedentem, et capiatur BD, aequalis ipsi AC, quae consequens esse debet, sive media. Deinde ducta recta BC, agatur illi ex D, parallela DE, occurrens ipsi AC, productae in E.

法曰。甲乙、甲丙、兩線。求別作一線。相與為連比例者。合兩線。任作甲角。而甲乙與甲戊之比例。若甲丙與他線也。先于甲乙引長之、為乙丁。與甲丙等。次作丙乙線相聯。次從丁作丁戊線。與丙乙平行。末于甲丙引長之、遇于戊。卽丙戊為所求線。如以甲丙為前率。倣此。

Since, then, BC has been drawn parallel to DE, one of the sides of the triangle ADE, proportionally, as AB is to BD, so is AC to CE. [VI. 2] But BD is equal to AC; therefore, as AB is to AC, so is AC to CE.

Dico CE, esse tertiam proportionalem: hoc est, esse ut AB, ad AC, ita AC, ad CE. Cum enim in triangulo ADE, lateri DE, parallela sit recta BC;³ erit ut AB, ad BD, ita AC, ad CE: ⁴ Sed ut AB, ad BD, ita eadem AB, ad AC, aequalem ipsi BD. Ut igitur AB, ad AC, ita AC, ad CE. quod est propositum.

3. 4. sexti 4. 7. quinti.

論曰。甲丁戊角形內之丙乙線。旣與戊丁邊平行。卽甲乙與乙丁之比例。若甲丙與丙戊也。本篇二而乙丁、甲丙、元等。卽甲乙與甲丙。若甲丙與丙戊也。五卷七

Therefore to two given straight lines AB, AC a third proportional to them, CE, has been found. Q. E. F.

Duabus ergo datis rectis lineis, tertiam proportionalem adinvenimus. Quod erat faciendum.

SCHOLIUM. ALITER idem demonstrabimus, hoc modo. Duae rectae datae AB, BC, constituantur ad angulum rectum ABC, et coniungatur recta AC. Producta autem AB, antecedente, ducatur ex C, ad AC, perpendicularis CD, occurrens ipsi AB, productae in D. Dico BD, esse tertiam proportionalem.

注曰。別有一法。以甲乙、乙丙、兩線。列作甲乙丙直角。次以甲丙線聯之。而甲乙引長之。末從丙作丙丁。為甲丙之垂線。遇引長線于丁。卽乙丁為所求線。

Cum enim in triangulo ACD, angulus ACD, sit rectus, et ab eo ad basin AD, deducta perpendicularis CB; erit por corollarium propositio 8 huius liber BC, media proportionalis inter AB, et BD, hoc est, ut AB, ad BC, ita erit BC, ad BD. Quod est propositum.

論曰。甲丙丁角形之甲丙丁。旣為直角。而從直角至甲。丁底。有丙乙垂線。卽丙乙為甲乙、乙丁、比例之中率。本篇八之系則甲乙與乙丙。若乙丙與乙丁也。

INVENTA autem tertia linea continue proportionali, si primam omiseris, et alijs duabus tertiam inueneris, habebis quatuor lineas continue proportionales. Ut si lineis A, et B, adinueniatur tertia proportionalis C, et duabus B, et C, tertia proportionalis D, erunt quatuor lineae A, B, C, D, continue proportionales. Eadem artereperietur quinta proportionalis, sexta, septima, octaua; et sic in infinitum.INVENTA autem tertia linea continue proportionali, si primam omiseris, et alijs duabus tertiam inueneris, habebis quatuor lineas continue proportionales. Ut si lineis A, et B, adinueniatur tertia proportionalis C, et duabus B, et C, tertia proportionalis D, erunt quatuor lineae A, B, C, D, continue proportionales. Eadem artereperietur quinta proportionalis, sexta, septima, octaua; et sic in infinitum.

旣從一二得三。卽從二、三、求四、以上、至于無窮。俱倣此。

PROPOSITION 12.

第十二題

To three given straight lines to find a fourth proportional.

三直線。求別作一線。相與為斷比例。

Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C.

Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [I. 31]

Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [VI. 2] But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF.

Therefore to the three given straight lines A, B, C a fourth proportional HF has been found. Q. E. F.

PROPOSITION 13.

第十三題

To two given straight lines to find a mean proportional.

兩直線。求別作一線。為連比例之中率。

Let AB, BC be the two given straight lines; thus it is required to find a mean proportional to AB, BC.

Let them be placed in a straight line, and let the semicircle ADC be described on AC; let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined.

Since the angle ADC is an angle in a semicircle, it is right. [III. 31] And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8, Por.]

Therefore to the two given straight lines AB, BC a mean proportional DB has been found. Q. E. F.

PROPOSITION 14.

第十四題二支

In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.

兩平行方形等。一角又等。卽等角旁之兩邊。為互相視之邊。兩平行方形之一角等。而等角旁兩邊、為互相視之邊。卽兩形等。

Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14] I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF.

For let the parallelogram FE be completed. Since, then, the parallelogram AB is equal to the parallelogram BC, and FE is another area, therefore, as AB is to FE, so is BC to FE. [V. 7] But, as AB is to FE, so is DB to BE, [VI. 1] and, as BC is to FE, so is GB to BF. [id.] therefore also, as DB is to BE, so is GB to BF. [V. 11] Therefore in the parallelograms AB, BC the sides about the equal angles are reciprocally proportional.

Next, let GB be to BF as DB to BE; I say that the parallelogram AB is equal to the parallelogram BC.

For since, as DB is to BE, so is GB to BF, while, as DB is to BE, so is the parallelogram AB to the parallelogram FE, [VI. 1] and, as GB is to BF, so is the parallelogram BC to the parallelogram FE, [VI. 1] therefore also, as AB is to FE, so is BC to FE; [V. 11] therefore the parallelogram AB is equal to the parallelogram BC. [V. 9]

Therefore etc. Q. E. D.

PROPOSITION 15.

第十五題二支

In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal.

相等兩三角形之一角等。卽等角旁之各兩邊、互相視。兩三角形之一角等。而等角旁之各兩邊、互相視。卽兩三角形等。

Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE; I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB.

For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [I. 14] Let BD be joined.

Since then the triangle ABC is equal to the triangle ADE, and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 7] But, as CAB is to BAD, so is CA to AD, [VI. 1] and, as EAD is to BAD, so is EA to AB. [id.] Therefore also, as CA is to AD, so is EA to AB. [V. 11] Therefore in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional.

Next, let the sides of the triangles ABC, ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD; I say that the triangle ABC is equal to the triangle ADE.

For, if BD be again joined, since, as CA is to AD, so is EA to AB, while, as CA is to AD, so is the triangle ABC to the triangle BAD, and, as EA is to AB, so is the triangle EAD to the triangle BAD, [VI. 1] therefore, as the triangle ABC is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 11] Therefore each of the triangles ABC, EAD has the same ratio to BAD. Therefore the triangle ABC is equal to the triangle EAD. [V. 9]

Therefore etc. Q. E. D.

PROPOSITION 16.

第十六題二支

If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional.

四直線為斷比例。卽首尾兩線、矩內直角形。與中兩線、矩內直角形、等。首尾兩線、與中兩線、兩矩內直角形等。卽四線為斷比例。

Let the four straight lines AB, CD, E, F be proportional, so that, as AB is to CD, so is E to F; I say that the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E. Let the parallelograms BG, DH be completed.

Then since, as AB is to CD, so is E to F, while E is equal to CH, and F to AG, therefore, as AB is to CD, so is CH to AG. Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional. But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [VI. 14] therefore the parallelogram BG is equal to the parallelogram DH. And BG is the rectangle AB, F, for AG is equal to F; and DH is the rectangle CD, E, for E is equal to CH; therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E; I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F.

For, with the same construction, since the rectangle AB, F is equal to the rectangle CD, E, and the rectangle AB, F is BG, for AG is equal to F, and the rectangle CD, E is DH, for CH is equal to E, therefore BG is equal to DH. And they are equiangular. But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [VI. 14] Therefore, as AB is to CD, so is CH to AG. But CH is equal to E, and AG to F; therefore, as AB is to CD, so is E to F.

Therefore etc. Q. E. D.

PROPOSITION 17

第十七題二支

If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional.

三直線為連比例。卽首尾兩線、矩內直角形。與中線上直角方形、等。首尾線矩內直角形、與中線上直角方形、等。卽三線為連比例。

Let the three straight lines A, B, C be proportional, so that, as A is to B, so is B to C; I say that the rectangle contained by A, C is equal to the square on B.

Let D be made equal to B.

Then, since, as A is to B, so is B to C, and B is equal to D, therefore, as A is to B, so is D to C. But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [VI. 16] Therefore the rectangle A, C is equal to the rectangle B, D. But the rectangle B, D is the square on B, for B is equal to D; therefore the rectangle contained by A, C is equal to the square on B.

Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C.

For, with the same construction, since the rectangle A, C is equal to the square on B, while the square on B is the rectangle B, D, for B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D. But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [VI. 16] Therefore, as A is to B, so is D to C. But B is equal to D; therefore, as A is to B, so is B to C.

Therefore etc. Q. E. D.

PROPOSITION 18.

第十八題

On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.

直線上。求作直線形。與所設直線形、相似而體勢等。

Let AB be the given straight line and CE the given rectilineal figure; thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE.

Let DF be joined, and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C, and the angle ABG equal to the angle CDF. [I. 23] Therefore the remaining angle CFD is equal to the angle AGB; [I. 32] therefore the triangle FCD is equiangular with the triangle GAB. Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB. Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE, and the angle GBH equal to the angle FDE. [I. 23] Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4] But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB; therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB. And, since the angle CFD is equal to the angle AGB, and the angle DFE to the angle BGH, therefore the whole angle CFE is equal to the whole angle AGH. For the same reason the angle CDE is also equal to the angle ABH. And the angle at C is also equal to the angle at A, and the angle at E to the angle at H. Therefore AH is equiangular with CE; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1]

Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE. Q. E. F.

PROPOSITION 19.

第十九

Similar triangles are to one another in the duplicate ratio of the corresponding sides.

相似三角形之比例。為其相似。邊再加之比例。

Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF.

For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11] and let AG be joined.

Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16] But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11] Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional. But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15] therefore the triangle ABG is equal to the triangle DEF. Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9] therefore BC has to BG a ratio duplicate of that which CB has to EF. But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF. But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF.

Therefore etc.

PORISM.
From this it is manifest that, if three straight lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second.

Q. E. D.

PROPOSITION 20.

Theor. 20 Propos. 20

第二十題　三支

Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side.

Si sint tres magnitudines, et aliae ipsis aequales numero, quae binae, et in eadem ratione sumantur; ex aequo autem prima quam tertia, maior fuerit; Erit et quarta quam sexta, maior. Quod si prima quam tertia fuerit aequalis, erit et quarta aequalis sextae; sin illa minor, haec quoque minor erit.

以三角形、分相似之多邊直線形。則分數必等。而相當之各三角形、各相似。其各相當兩三角形之比例。若兩元形之比例。其元形之比例。為兩相似邊再加之比例。

Let ABCDE, FGHKL be similar polygons, and let AB correspond to FG; I say that the polygons ABCDE, FGHKL are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG.

SINT tres magnitudines A, B, C, et totidem D, E, F, sitque A, ad B, ut D, ad E; et B, ad C, ut E, ad F, sit autem primum A, prima maior quam C, tertia. Dico et D, quartam esse maiorem F, sexta. Cum enim A, maior sit quam C, erit maior proportio A, ad B, 8. quinti. quam C, ad B. Estautem ut A, ad B, ita D, ad E. Maior igitur 13. quinti. proportio quoque erit D, ad E, quam C, ad B. At ut C, ad B, ita est F, ad E. (Cum enim sit B, ad C, ut E, ad F, erit conuertendo ut C, ad B, ita F, ad E.) Maiorigitur quoque proportio erit D, ad E, quam 10. quinti. F, ad E. Quare D, maior erit, quam F. Quod est propositum.

先解曰。此甲乙丙丁戊、彼己庚辛壬癸、兩多邊直線形。其乙甲戊、庚己癸、兩角等。餘相當之各角俱等。而各等角旁各兩邊之比例各等。題先言各以角形分之。其角形之分數必等。而相當之各角形各相似。

Let BE, EC, GL, LH be joined.

論曰。試從乙甲戊、庚己癸、兩角。向各對角、俱作直線。為甲丙、甲丁、己辛、己壬。

Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. [VI. Def. 1] Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6] so that it is also similar; [VI. 4 and Def. 1] therefore the angle ABE is equal to the angle FGL. But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH. And, since, because of the similarity of the triangles ABE, FGL, as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH, therefore, ex aequali, as EB is to BC, so is LG to GH; [V. 22] that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH, [VI. 6] so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1] For the same reason the triangle ECD is also similar to the triangle LHK. Therefore the similar polygons ABCDE, FGHKL have been divided into similar triangles, and into triangles equal in multitude.

SIT deinde A, aequalis ipsi C. Dico et D, aequalem esse ipsi F. Cum enim A, sit ipsi C, aequalis, erit A, ad B, ut C, ad B. Est autem ut A, ad B, ita 7. quinti. D, ad EIgitur erit et D, ad E, ut C, ad B: At ut C, ad B, ita est F, ad E, per inuersam rationem, 11. quinti. uti prius. Quare erit quoque D, ad E, ut F, ad E; Ideoque aequales erunt D, et F. Quod est pro9. quinti. positum. SIT tertio A, minor quam C. Dico et D, minorem esse, quam F. Cum enim A, minor sit quam C, erit minor proportio A, ad 8. quinti. B, quam C, ad B. sed ut A, ad B, ita est D, ad E. Minor ergo quoque proportio est D, ad 13. quinti. E, quam C, ad B. Est autem conuertendo, ut prius, ut C, ad B, ita F, ad E. Igitur minor est quoque proportio D, ad E, quam F, ad E, proptereaque D, minor erit quam F. Quod 10. quinti. est propositum. si sint itaque tres magnitudines, et aliae ipsis aequales numero, etc.

其元形旣相似。卽角數等。而所分角形之數亦等。又乙角旣與庚角等。而角旁各兩邊之比例亦等。卽甲乙丙、與己庚辛、兩角形必相似。本篇六乙甲丙、與庚己辛、兩角。甲丙乙、與己辛庚、兩角。各等。而各等角旁、各兩邊之比例、各等。本篇四依顯甲戊丁、己癸壬、兩角形亦相似。又甲丙與丙乙之比例。旣若己辛與辛庚。而丙乙與丙丁。若辛庚與辛壬。兩元形相似故平之。卽甲丙與丙丁。若己辛與辛壬也。五卷廿二又乙丙丁角。旣與庚辛壬角等。而各減一相等之甲丙乙角、己辛庚角。卽所存甲丙丁角、與己辛壬角、必等。則甲丙丁與己辛壬兩角形。亦等角形。亦相似矣。本篇六

I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents, and that the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG.

次解曰。題又言各相當角形之比例。若兩元形之比例。

For let AC, FH be joined. Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH, and, as AB is to BC, so is FG to GH, the triangle ABC is equiangular with the triangle FGH; [VI. 6] therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF. And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangle FGN. Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH. Therefore, proportionally, as AM is to MB, so is FN to NG, and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH. But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC; for they are to one another as their bases. [VI. 1] Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle AMB is to BMC, so is ABE to CBE. But, as AMB is to BMC, so is AM to MC; therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC. For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH. And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH. Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK. And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL. But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG; for similar triangles are in the duplicate ratio of the corresponding sides. [VI. 19] Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG.

論曰。甲乙丙、己庚辛、兩角形旣相似。卽兩形之比例。為甲丙、己辛、兩相似邊再加之比例。本篇十九依顯甲丙丁、己辛壬、之比例。亦為甲丙、己辛、再加之比例。則甲乙丙與己庚辛兩角形之比例。若甲丙丁與己辛壬兩角形之比例。依顯甲丁戊與己壬癸之比例。亦若甲丙丁與己辛壬之比例。則此形中諸角形之比例。若彼形中諸角形之比例。此諸形為前率。彼諸形為後率。而一前與一後之比例。又若幷前與幷後之比例。五卷十二卽此一角形、與相當彼一角形之比例。若此元形、與彼元形之比例矣。

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

先解曰。甲線、倍乙線。題言甲上方形、與乙上方形。為四倍大之比例。甲乙

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

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

Therefore etc.

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

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

此系與本篇第十九題之系同論。

PORRO propositione 22. ostendet Euclides, A, et D, magnitudines non solum esse maiores, vel aequales, vel minores duabus magnitudinibus C, et E, ut hic demonstrauit, sed etiam illas ad has eandem habere proportionem ex aequalitate: quod quidem demonstrare non poterat, nisi prius theorema hoc ostendisset, ut ex eadem propositione 22. erit perspicuum.

PROPOSITION 21.

第二十一題

Figures which are similar to the same rectilineal figure are also similar to one another.

兩直線形。各與他直線形相似。則自相似。

For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B.

For, since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. [VI. Def. 1] Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional. Therefore each of the figures A, B is equiangular with C and with C has the sides about the equal angles proportional; therefore A is similar to B. Q. E. D.

PROPOSITION 22.

第二十二題二支

If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional.

四直線為斷比例。則兩比例線上、各任作自相似之直線形。亦為斷比例。兩比例線上、各任作自相似之直線形、為斷比例。則四直線亦為斷比例。

Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH.

For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11] Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P. [V. 22] But, as AB is to O, so is KAB to LCD, [VI. 19, Por.] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [V. 11]

Next, let MF be to NH as KAB is to LCD; I say also that, as AB is to CD, so is EF to GH. For, if EF is not to GH as AB to CD, let EF be to QR as AB to CD, [VI. 12] and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH. [VI. 18]

Since then, as AB is to CD, so is EF to QR, and there have been described on AB, CD the similar and similarly situated figures KAB, LCD, and on EF, QR the similar and similarly situated figures MF, SR, therefore, as KAB is to LCD, so is MF to SR. But also, by hypothesis, as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [V. 11] Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR. [V. 9] But it is also similar and similarly situated to it; therefore GH is equal to QR. And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH.

Therefore etc. Q. E. D.

PROPOSITION 23.

第二十三題

Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides.

等角兩平行方形之比例。以兩形之各兩邊兩比例相結。

Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG; I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides.

For let them be placed so that BC is in a straight line with CG; therefore DC is also in a straight line with CE.

Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG, so is K to L, and, as DC is to CE, so is L to M. [VI. 12]

Then the ratios of K to L and of L to M are the same as the ratios of the sides, namely of BC to CG and of DC to CE. But the ratio of K to M is compounded of the ratio of K to L and of that of L to M; so that K has also to M the ratio compounded of the ratios of the sides. Now since, as BC is to CG, so is the parallelogram AC to the parallelogram CH, [VI. 1] while, as BC is to CG, so is K to L, therefore also, as K is to L, so is AC to CH. [V. 11] Again, since, as DC is to CE, so is the parallelogram CH to CF, [VI. 1] while, as DC is to CE, so is L to M, therefore also, as L is to M, so is the parallelogram CH to the parallelogram CF. [V. 11] Since then it was proved that, as K is to L, so is the parallelogram AC to the parallelogram CH, and, as L is to M, so is the parallelogram CH to the parallelogram CF, therefore, ex aequali, as K is to M, so is AC to the parallelogram CF. But K has to M the ratio compounded of the ratios of the sides; therefore AC also has to CF the ratio compounded of the ratios of the sides.

Therefore etc. Q. E. D.

PROPOSITION 24.

第二十四題

In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.

平行線方形之兩角線方形。自相似。亦與全形相似。

Let ABCD be a parallelogram, and AC its diameter, and let EG, HK be parallelograms about AC; I say that each of the parallelograms EG, HK is similar both to the whole ABCD and to the other.

For, since EF has been drawn parallel to BC, one of the sides of the triangle ABC, proportionally, as BE is to EA, so is CF to FA. [VI. 2] Again, since FG has been drawn parallel to CD, one of the sides of the triangle ACD, proportionally, as CF is to FA, so is DG to GA. [VI. 2] But it was proved that, as CF is to FA, so also is BE to EA; therefore also, as BE is to EA, so is DG to GA, and therefore, componendo, as BA is to AE, so is DA to AG, [V. 18] and, alternately, as BA is to AD, so is EA to AG. [V. 16] Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional. And, since GF is parallel to DC, the angle AFG is equal to the angle DCA; and the angle DAC is common to the two triangles ADC, AGF; therefore the triangle ADC is equiangular with the triangle AGF. For the same reason the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram ABCD is equiangular with the parallelogram EG. Therefore, proportionally, as AD is to DC, so is AG to GF, as DC is to CA, so is GF to FA, as AC is to CB, so is AF to FE, and further, as CB is to BA, so is FE to EA. And, since it was proved that, as DC is to CA, so is GF to FA, and, as AC is to CB, so is AF to FE, therefore, ex aequali, as DC is to CB, so is GF to FE. [V. 22] Therefore in the parallelograms ABCD, EG the sides about the equal angles are proportional; therefore the parallelogram ABCD is similar to the parallelogram EG. [VI. Def. 1] For the same reason the parallelogram ABCD is also similar to the parallelogram KH; therefore each of the parallelograms EG, HK is similar to ABCD. But figures similar to the same rectilineal figure are also similar to one another; [VI. 21] therefore the parallelogram EG is also similar to the parallelogram HK.

Therefore etc. Q. E. D.

PROPOSITION 25.

第二十五題

To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.

兩直線形、求作他直線形。與一形相似。與一形相等。

Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal; thus it is required to construct one and the same figure similar to ABC and equal to D.

Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45] Therefore BC is in a straight line with CF, and LE with EM. Now let GH be taken a mean proportional to BC, CF [VI. 13], and on GH let KGH be described similar and similarly situated to ABC. [VI. 18]

Then, since, as BC is to GH, so is GH to CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19, Por.] therefore, as BC is to CF, so is the triangle ABC to the triangle KGH. But, as BC is to CF, so also is the parallelogram BE to the parallelogram EF. [VI. 1] Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF; therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16] But the triangle ABC is equal to the parallelogram BE; therefore the triangle KGH is also equal to the parallelogram EF. But the parallelogram EF is equal to D; therefore KGH is also equal to D. And KGH is also similar to ABC.

Therefore one and the same figure KGH has been constructed similar to the given rectilineal figure ABC and equal to the other given figure D. Q. E. D.

PROPOSITION 26.

第二十六題

If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole

平行方形之內。減一平行方形。其減形與元形。相似而體勢等。又一角同。則減形必依元形之對角線。

For from the parallelogram ABCD let there be taken away the parallelogram AF similar and similarly situated to ABCD, and having the angle DAB common with it; I say that ABCD is about the same diameter with AF.

For suppose it is not, but, if possible, let AHC be the diameter , let GF be produced and carried through to H, and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31]

Since, then, ABCD is about the same diameter with KG, therefore, as DA is to AB, so is GA to AK. [VI. 24] But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE; therefore also, as GA is to AK, so is GA to AE. [V. 11] Therefore GA has the same ratio to each of the straight lines AK, AE. Therefore AE is equal to AK [V. 9], the less to the greater: which is impossible. Therefore ABCD cannot but be about the same diameter with AF; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF.

Therefore etc. Q. E. D.

PROPOSITION 27.

第二十七題

Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.

凡依直線之有闕平行方形。不滿線者。其闕形、與半線上之闕形、相似而體勢等。則半線上似闕形之有闕依形。必大于此有闕依形。

Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB; I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest. For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB; I say that AD is greater than AF.

For, since the parallelogram DB is similar to the parallelogram FB, they are about the same diameter. [VI. 26] Let their diameter DB be drawn, and let the figure be described.

Then, since CF is equal to FE, [I. 43] and FB is common, therefore the whole CH is equal to the whole KE. But CH is equal to CG, since AC is also equal to CB. [I. 36] Therefore GC is also equal to EK. Let CF be added to each; therefore the whole AF is equal to the gnomon LMN; so that the parallelogram DB, that is, AD, is greater than the parallelogram AF.

Therefore etc.

PROPOSITION 28.

第二十八題

To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.

一直線。求作依線之有闕平行方形。與所設直線形等。而其闕形、與所設平行方形相似。其所設直線形。不大于半線上所作平行方形。與所設平行方形相似者。

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D.

Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed.

If then AG is equal to C, that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D. But, if not, let HE be greater than C. Now HE is equal to GB; therefore GB is also greater than C. Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25] But D is similar to GB; therefore KM is also similar to GB. [VI. 21] Let, then, KL correspond to GE, and LM to GF. Now, since GB is equal to C, KM, therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM. Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM. Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26] Let GQB be their diameter, and let the figure be described.

Then, since BG is equal to C, KM, and in them GQ is equal to KM, therefore the remainder, the gnomon UWV, is equal to the remainder C. And, since PR is equal to OS, let QB be added to each; therefore the whole PB is equal to the whole OB. But OB is equal to TE, since the side AE is also equal to the side EB; [I. 36] therefore TE is also equal to PB. Let OS be added to each; therefore the whole TS is equal to the whole, the gnomon VWU. But the gnomon VWU was proved equal to C; therefore TS is also equal to C.

Therefore to the given straight line AB there has been applied the parallelogram ST equal to the given rectilineal figure C and deficient by a parallelogrammic figure QB which is similar to D. Q. E. F.

PROPOSITION 29.

第二十九題

To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.

一直線。求作依線之帶餘平行方形。與所設直線形等。而其餘形、與所設平行方形相似。

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar; thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D.

Let AB be bisected at E; let there be described on EB the parallelogram BF similar and similarly situated to D; and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25] Let KH correspond to FL and KG to FE. Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE. Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed; therefore MN is both equal and similar to GH. But GH is similar to EL; therefore MN is also similar to EL; [VI. 21] therefore EL is about the same diameter with MN. [VI. 26] Let their diameter FO be drawn, and let the figure be described.

Since GH is equal to EL, C, while GH is equal to MN, therefore MN is also equal to EL, C. Let EL be subtracted from each; therefore the remainder, the gnomon XWV, is equal to C. Now, since AE is equal to EB, AN is also equal to NB [I. 36], that is, to LP [I. 43]. Let EO be added to each; therefore the whole AO is equal to the gnomon VWX. But the gnomon VWX is equal to C; therefore AO is also equal to C.

Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24]. Q. E. F.

PROPOSITION 30.

第三十題

To cut a given finite straight line in extreme and mean ratio.

一直線。求作理分中末線。

Let AB be the given finite straight line; thus it is required to cut AB in extreme and mean ratio.

On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29]

Now BC is a square; therefore AD is also a square. And, since BC is equal to CD, let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD. But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, as FE is to ED, so is AE to EB. But FE is equal to AB, and ED to AE. Therefore, as BA is to AE, so is AE to EB. And AB is greater than AE; therefore AE is also greater than EB.

Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE. Q. E. F.

PROPOSITION 31.

第三十一題

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

三邊直角形之對直角邊上一形。與直角旁邊上兩形。若相似而體勢等。則一形與兩形幷、等。

Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

Let AD be drawn perpendicular.

Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8] And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1] And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [VI. 19, Por.] Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA. For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC. But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC.

Therefore etc. Q. E. D.

PROPOSITION 32.

第三十二題

If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.

兩三角形。此形之兩邊。與彼形之兩邊、相似。而平置兩形。成一外角。若备相似之各兩邊、各平行。則其餘各一邊、相聯為一直線。

Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE; I say that BC is in a straight line with CE.

For, since AB is parallel to DC, and the straight line AC has fallen upon them, the alternate angles BAC, ACD are equal to one another. [I. 29] For the same reason the angle CDE is also equal to the angle ACD; so that the angle BAC is equal to the angle CDE. And, since ABC, DCE are two triangles having one angle, the angle at A, equal to one angle, the angle at D, and the sides about the equal angles proportional, so that, as BA is to AC, so is CD to DE, therefore the triangle ABC is equiangular with the triangle DCE; [VI. 6] therefore the angle ABC is equal to the angle DCE. But the angle ACD was also proved equal to the angle BAC; therefore the whole angle ACE is equal to the two angles ABC, BAC. Let the angle ACB be added to each; therefore the angles ACE, ACB are equal to the angles BAC, ACB, CBA. But the angles BAC, ABC, ACB are equal to two right angles; [I. 32] therefore the angles ACE, ACB are also equal to two right angles. Therefore with a straight line AC, and at the point C on it, the two straight lines BC, CE not lying on the same side make the adjacent angles ACE, ACB equal to two right angles; therefore BC is in a straight line with CE. [I. 14]

Therefore etc. Q. E. D.

PROPOSITION 33.

THEOR. 23. PROPOS. 33.

第三十三題三支

In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences.

IN aequalibus circulis, anguli eandem habent rationem cum peripherijs, quibus insistunt, sive ad centra, sive ad peripherias constituti insistant: Insuper vero et sectores, quippe qui ad centra consistunt.

等圜之乘圜分角。或在心。或在界。其各相當兩乘圜角之比例。皆若所乘兩圜分之比例。而兩分圜形之比例。亦若所乘兩圜分之比例。

Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences; I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

SINT duo circuli aequales ABC, EFG, quorum centra D, H; sumanturque ex circulis duo arcus quicunque BC, FG, quibus ad centra quidem insistant anguli BDC, FHG; ad circumferentias vero anguli BAC, EFG. Dico esse ex sententia defin. 6. liber 5. ut arcum BC, ad arcum FG, ira angulum BDC, ad angulum FHG, et angulum BAC, ad angulum EFG; et sectorem insuper BDC, qui rectis BD, DC, et arcu BC, continetur ad sectorem FHG; quem comprehendunt rectae FH, HG, et arcus FG.

解曰甲乙丙、戊己庚、兩圜等。其心為丁、為辛。兩圜各任割一圜分為乙丙、為己庚。其乘圜角之在心者。為乙丁丙、己辛庚。在界者。為乙甲丙、己戊庚。題先言乙丙、與己庚、兩圜分之比例。若乙丁丙、與己辛庚、兩角。次言乙甲丙、與己戊庚、兩角之比例。若乙丙、與己庚、兩圜分。後言乙丁、丁丙、兩腰、偕乙丙圜分、內乙丁丙分圜形。與己辛、辛庚、兩腰、偕己庚圜分、內己辛庚分圜形、之比例。亦若乙丙、與己庚、兩圜分。

For let any number of consecutive circumferences CK, KL be made equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF; and let GK, GL, HM, HN be joined.

Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC. For the same reason also, whatever multiple the circumference NE is of EF, that multiple also is the angle NHE of the angle EHF. If then the circumference BL is equal to the circumference EN, the angle BGL is also equal to the angle EHN; [III. 27] if the circumference BL is greater than the circumference EN, the angle BGL is also greater than the angle EHN; and, if less, less. There being then four magnitudes, two circumferences BC, EF, and two angles BGC, EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN. And it has been proved that, if the circumference BL is in excess of the circumference EN, the angle BGL is also in excess of the angle EHN; if equal, equal; and if less, less. Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF. [V. Def. 5] But, as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF; for they are doubles respectively. Therefore also, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

Therefore etc. Q. E. D.

Permanent link

http://www2.hf.uio.no/common/apps/permlink/permlink.php?app=polyglotta&context=ctext&uid=6c7baf5e-3ed4-11e1-aecb-00215aecadea

Choose languages

Choose images, etc.

Choose languages

Choose display