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Euclid: Elementa
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Proposition 26. 
THEOR. 17. PROPOS. 26. 
第二十六題 
If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle. 
SI duo triangula duos angulos duobus angulis æquales habuerint, utrumque utrique, unumque latus uni lateri æquale, siue quod æqualibus adiacet angulis, seu quod uni æqualium angulorum subtenditur: & reliqua latera reliquis lateribus æqualia, utrumque utrique, & reliquum angulum reliquo angulo æqualem habebunt. 
二支
兩三角形。有相當之兩角等、及相當之一邊等。則餘兩邊必等。餘一角亦等。其一邊。不論在兩角之內、及一角之對。 
Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively,  namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD;  and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF;  I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF,  and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF. 
SINT duo anguli B, & C, trianguli A B C, æquales duobus angulis E, & E F D, trianguli D E F, uterque utrique,  hoc est, B, ipsi E, & C, ipsi E F D;  Sitque primo latus B C, quod angulis B, & C, adiacet, lateri E F, quod angulis E, & E F D, adiacet, æquale.  Dico, reliqua quoque latera A B, A C, reliquis lateribus D E, D F, æqualia esse, utrumque utrique, hoc est, A B, ipsi D E, & A C, ipsi D F, ea nimirum, quæ æqualibus angulis subtenduntur;  reliquumque angulum A, reliquo angulo D. 
.  .  .  .  . 
For, if AB is unequal to DE, one of them is greater.  Let AB be greater, and let BG be made equal to DE; and let GC be joined. 
Si enim latus A B, non est æquale lateri D E,  sit D E, maius, a quo abscindatur recta linea E G, æqualis rectæ lineæ A B, ducaturque recta G F. 
.  . 
Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively;  and the angle GBC is equal to the angle DEF;  therefore the base GC is equal to the base DF,  and the triangle GBC is equal to the triangle DEF,  and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]  therefore the angle GCB is equal to the angle DFE.  But the angle DFE is by hypothesis equal to the angle BCA;  therefore the angle BCG is equal to the angle BCA, the less to the greater: which is impossible.  Therefore AB is not unequal to DE, and is therefore equal to it.  But BC is also equal to EF; therefore the two sides AB, BC are equal to the two sides DE, EF respectively,  and the angle ABC is equal to the angle DEF;  therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I. 4] 
Quoniam igitur latera A B, B C, æqualia sunt lateribus G E, E F, utrumque utrique,  & anguli B, & E, æquales per hypothesin: Erit angulus C, æqualis angulo E F G.          Ponitur autem angulus C, æqualis angulo E F D.  Quare & angulus E F G, eidem angulo E F D, æqualis erit, pars toti; Quod est absurdum.  Non est igitur latus A B, inæquale lateri D E, sed æquale.  Quamobrem, cum latera A B, B C, æqualia sint lateribus D E, E F, utrumque utrique,  & anguli contenti B, & E, æquales;  erunt & bases A C, D F, & anguli reliqui A, & D, æquales. Quod est propositum. 
.  .  .  .  .  .  .  .  .  .  .  . 
Again, let sides subtending equal angles be equal, as AB to DE;  I say again that the remaining sides will be equal to the remaining sides,  namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF. 
SINT secundo latera A B, D E, subtendentia æquales angulos C, & E F D, inter se æqualia.  Dico rursus reliqua latera B C, C A, reliquis lateribus E F, F D, esse æqualia, utrumque utrique,  hoc est, B C, ipsi E F, & C A, ipsi F D; reliquumque angulum A, reliquo angulo D, æqualem. 
.  .  . 
For, if BC is unequal to EF, one of them is greater.  Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.  Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively,  and they contain equal angles;  therefore the base AH is equal to the base DF,  and the triangle ABH is equal to the triangle DEF,  and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]  therefore the angle BHA is equal to the angle EFD.  But the angle EFD is equal to the angle BCA;  therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA: which is impossible. [I. 16]  Therefore BC is not unequal to EF, and is therefore equal to it.  But AB is also equal to DE;  therefore the two sides AB, BC are equal to the two sides DE, EF respectively,  and they contain equal angles;  therefore the base AC is equal to the base DF, the triangle ABC equal to the triangle DEF, and the remaining angle BAC equal to the remaining angle EDF. [I. 4] 
Si enim latus B C, non est æquale lateri E F,  sit E F, maius; ex quo sumatur recta E G, æqualis ipsi B C, ducaturque recta D G.  Quoniam igitur latera A B, B C, æqualia sunt lateribus D E, E G, utrumque utrique,  & anguli contenti B, & E æquales, per hypothesin;        Erit angulus C, angulo E G D, æqualis:   Ponitur autem angulus C, angulo E F D, æqualis;  Igitur & angulus E G D, angulo eidem E F D, æqualis erit, externus interno, & opposito, quod est absurdum. Est enim maior.  Non ergo est latus B C, lateri E F, inæquale. Quocirca, ut prius, colligetur institutum ex 4. propos. huius libri.         
.  .  .  .  .  .  .  .  .  .  .  .  .  .  . 
Therefore etc.  Q. E. D. 
Si duo igitur triangula duos angulos duobus angulis æquales habuerint, &c.  Quod demonstrandum erat.

SCHOLION
PRIOR huius theorematis pars conuersa est 4. propositionis, quoad eam partem, in qua ex æqualitate laterum, & angulorum ipsis contenterum, collecta fuit æqualitas basium, & angulorum super bases. Nam in priori parte huius theorematis ex æqualitate basium B C, E F, & angulorum super has bases, demonstratum est, reliqua latera unius trianguli reliquis lateribus æqualia esse, reliquumque angulum reliquo angulo, &c. Quod quidem alia nos ratione iam demonstrauimus ad propositionem octauam huius liber quem modum eo loco monuimus. 
.  . 
 
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