Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5]
therefore the angle ADC is greater than the angle DCB;
therefore the angle CDB is much greater than the angle DCB.
Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB.
But it was also proved much greater than it: which is impossible.
Sint enim, si fieri potest, rectæ A C, A D, inter se, & rectæ B C, B D, inter se etiam æquales. Aut igitur punctum D, erit in alterutra rectarum A C, B C, ita vt recta A D, in ipsam rectam A C vel B D, in ipsam B C, cadat; aut intra triangulum A B C; aut extra. Sit primo punctum D, in altera rectarum A C, B C, nempe in A C, vt A D, sit pars ipsius A C. Quoniam igitur rectæ A C, A D, eundem terminum A, habentes dicuntur æquales, erit pars A D, toti A C, æqualis. Quod fieri non potest. Sit deinde punctum D, intra triangulum A B C, & ducta recta C D, producantur rectæ B C, B D, vsque ad E, & F. Quoniam igitur in triangulo A C D, ponuntur latera A C, A D, æqualia erunt anguli A C D, A D C, super basim C D, æquales;
Est autem angulus A C D, minor angulo D C E; nempe pars toto: Igitur & angulus A D C, minor erit eodem angulo D C E.
Quare angulus C D F, pars ipsius A D C, multo minor erit eodem angulo D C E.
Rursus, quia in triangulo B C D, latera B C, B D, ponuntur æqualia, erunt anguli C D F, D C E, sub basi C D, æquales.
Ostensum autem fuit, quod idem angulus C D F, multo sit minor angulo D C E. Idem ergo angulus C D F, & minor est angulo D C E, & eidem æqualis, quod est absurdum. Sit postremo punctum D, extra triangulum A B C. Aut igitur in tali erit loco, vt vna linea super alteram cadat, vt in priori figura, dummodo loco D, intelligas C, & loco C, ipsum D; ex quo rursus colligetur pars æqualis toti, quod est absurdum. Aut in tali erit loco, vt posteriores duæ lineæ ambiant priores duas, ceu in posteriori figura, si modo loco D, iterum intelligas C, & D, loco C. Quo posito, inidem absurdum incidemus, nempe angulum D C F, & minorem esse angulo C D E, & eidem æqualem, vt perspicuum est. Aut denique punctum D, ita erit extra triangulum A B C, vt altera linearum posteriorum, nempe A D, secet alteram priorum, vt ipsam B C. Ducta igitur recta C D, cum in triangulo A C D, latera A C, A D, ponantur æqualia, erunt anguli A C D, A D C, supra basim C D, æquales: Ac proinde cum angulus A D C, minor sit angulo B D C, pars toto, erit & angulus A C D, minor eodem angulo B D C. Quare multo minor erit angulus B C D, pars anguli A C D, angulo eodem B D C. Rursus, cum in triangulo B D C, latera B C, B D, ponantur æqualia, erunt anguli B C D, B D C, super basim C D, æquales: Est aut etiam ostensum, angulum B C D, multo esse minorem angulo B D C. Idem igitur angulus B C D, & minor est angulo B C D, & eidem æqualis, quod est absurdum. Non ergo æquales sunt inter se A C, A D, & inter se quoque B C, B D.