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Euclid: Elementa
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Proposition 14. 
THEOR. 7. PROPOS. 14. 
第十四題 
If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another. 
SI ad aliquam rectam lineam, atque ad eius punctum, duæ rectæ lineæ non ad easdem partes ductæ eos, qui sunt deinceps, angulos duobus rectis æquales fecerint; in directum erunt inter se ipsæ rectæ lineæ. 
一直線。於線上一點。出不同方兩直線。偕元線、每旁作兩角。若每旁兩角、與兩直角等卽後出兩線、為一直線。 
For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles;  I say that BD is in a straight line with CB. 
Ad punctum C, lineæ rectæ A B, in diuersas partes eductæ sint duæ rectæ C D, C E, facientes cum A B, duos angulos A C D, A C E, vel rectos, vel duobus rectis æquales.  Dico ipsas C D, C E, inter se esse constitutas in directum, ita ut D C E, sit una linea recta. 
.  . 
For, if BD is not in a straight line with BC, let BE be in a straight line with CB.  Then, since the straight line AB stands on the straight line CBE, the angles ABC, ABE are equal to two right angles. [I. 13]  But the angles ABC, ABD are also equal to two right angles;  therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1]  Let the angle CBA be subtracted from each;  therefore the remaining angle ABE is equal to the remaining angle ABD, [C.N. 3] the less to the greater:  which is impossible.  Therefore BE is not in a straight line with CB.  Similarly we can prove that neither is any other straight line except BD.  Therefore CB is in a straight line with BD. 
Si enim non est recta D C E; producta D C, ad partes C, in directum, & continuum cadet aut supra C E, ut sit recta D C F, aut infra C E, ut sit recta D C G.  Si cadit supra, cum A C, consistat super rectam D C F, fient duo anguli A C D, A C F, duobus rectis æquales;  Ponuntur autem & duo anguli A C D, A C E, æquales duobus rectis;  & omnes recti sunt inter se æquales. Quare duo anguli A C D, A C F, duobusangulis A C D, A C E, erunt æquales.   Ablato igitur cmmuni angulo A C D,   remanebunt anguli A C F, A C E, inter se æquales, pars & totum,  quod est absurdum.  Non igitur recta D C, producta cadet supra C E;   Sed neque infra cadet; Eadem enim ratione probarentur anguli A C E, A C G, æquales.  Igitur D C, producta eadem efficietur, quæ C E; 
.  .  .  .  .  .  .  .  .  . 
Therefore etc.  Q. E. D. 
proptereaque, si ad aliquam rectam lineam, atque ad eius punctum, &c.  Quod demonstrandum erat.

SCHOLION
EST hæc propositio præcedentis conuersa. In ea enim probatum fuit, si D C E, sit recta, angulos A C D, A C E, duobus esse rectis æquales. In hac vero demonstratum est, si dicti anguli sint duobus rectis æquales, rectas D C, E C, esse unam lineam rectam.

EX PROCLO
RECTE Euclides addidit in propositione hac (& non ad easdem partes.) Quoniam, ut ait Porphyrius, fieri potest, ut ad punctum aliquod lineæ datæ ad easdem partes duæ lineæ ducantur, facientes cum data duos angulos duobus rectis æquales, quæ tamen non constituant unam lineam, eo quod non ad diuersas sint ductæ partes. Sit enim punctum C, in linea A B, datum. Ducatur C D, perpendicularis ad A B, diuidaturque rectus angulus A C D, bifariam per rectam C E. Deinde ex D, quolibet puncto rectæ C D, ducatur D E, perpendicularis ad C D, secans rectam C D, in E. Producta autem E D, ad partes D, sumatur D F, æqualis rectæ D E, & ducatur recta F C. Quoniam igitur latera E D, D C, trianguli E D C, æqualia sunt lateribus F D, D C, trianguli F D C, utrumque utrique, & anguli D, ipsis contenti æquales, nempe recti; erit basis E C, basi C F, æqualis, & angulus E C D, angulo F C D. Sed angulus E C D, dimidium est recti. (Est enim rectus A C D, divisus bifariam.) Igitur & F C D, dimidium erit recti. Quare C F, cum A C, faciet angulum A C F, constantem ex recto, & dimidio recti; Facit autem C E, cum eadem A C, angulum A C E, dimidium etiam recti. Duo igitur anguli A C F, A C E, quos ad easdem partes faciunt rectæ C F, C E, cum A B; æquales sunt duobus rectis; Et tamen C F, C E, non sunt una linea recta, propterea quod non sunt ducta ad diuersas partes, sed ad easdem. 
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