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Euclid: Elementa
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Proposition 39. 
THEOR. 29. PROPOS. 39. 
第三十九題 
Equal triangles which are on the same base and on the same side are also in the same parallels. 
TRIANGVLA æqualia super eadem basi, & ad easdem partes constituta; & in eisdem sunt parallelis. 
兩三角形。其底同。其形等。必在兩平行線內。 
Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it;  [I say that they are also in the same parallels.] 
SINT duo triangula æqualia A B C, D B C, super eandem basin B C, & ad easdem partes.  Dico ipsa esse inter easdem parallelas constituta, hoc est, rectam ductam A D, parallelam esse ipsi B C. 
.  . 
And [For] let AD be joined; I say that AD is parallel to BC.  For, if not, let AE be drawn through the point A parallel to the straight line BC, [I. 31] and let EC be joined.  Therefore the triangle ABC is equal to the triangle EBC;  for it is on the same base BC with it and in the same parallels. [I. 37]  But ABC is equal to DBC;  therefore DBC is also equal to EBC, [C.N. 1] the greater to the less: which is impossible.  Therefore AE is not parallel to BC.  Similarly we can prove that neither is any other straight line except AD;  therefore AD is parallel to BC. 
  Si enim non est, ducatur ex A, parallela ipsi B C, quæ vel cadet supra A D, vel infra.  Cadat primum supra, qualis est A E, coeatque cum B D, protracta in E, & ducatur recta E C. Quoniam igitur parallelæ sunt A E, B C; erit triangulo A B C, triangulum E B C, æquale:    Est autem per hypothesin, triangulum quoque D B C, æquale eidem triangulo A B C.  Igitur erunt triangula D B C, E B C, æqualia, pars & totum, quod est absurdum.    Quod si parallela ducta per A, cadat infra A D, qualis est A F; ducta recta F C, erunt eadem ratiocinatione triangula B F C, B D C, æqualia, pars & totum; quod est absurdum.  Erit igitur A D, parallela ipsi B C. 
.  .  .  .  .  .  .  .  . 
Therefore etc.  Q. E. D. 
Quare triangula æqualia super eadem basi, &c.  Quod ostendendum erat.

SCHOLION
EX his infert Campanus sequens hoc theorema.

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

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

OMNE quadrilaterum, quod ab utraque diametro bifariam diuiditur, parallelogrammum est.
NAM quadrilaterum A B C D, diuidatur bifariam ab utraque diametro A C, B D. Dico ipsum esse parallelogrammum. Cum enim triangula A D C, B D C, dimidia sint eiusdem quadrilateri A B C D, ipsa inter se æqualia erunt. Quare cum eandem habeant basin D C, ad easdemque partes sint, ipsa in eisdem parallelis erunt; Atque idcirco rectæ A B, D C, parallelæ sunt. Non aliter ostendemus, parallelas esse A D, B C. Parallelogrammum igitur est A B C D, Quod est propositum. 
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