Bibliotheca Polyglotta

For let ABC be a triangle; I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA, AC greater than BC, AB, BC greater than AC, BC, CA greater than AB.

SIT triangulum A B C. Dico quælibet eius duo latera,¹ nempe A B, A C, simul maiora esse reliquo latere B C.

1. See also next record

. . .

For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined.

Producatur unum ex illis, ut C A, usque ad D, sitque recta A D, æqualis alteri lateri non producto A B, & ducatur recta D B.

Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5] therefore the angle BCD is greater than the angle ADC. [C.N. 5] And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19] therefore DB is greater than BC. But DA is equal to AC; therefore BA, AC are greater than BC. Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.

Quoniam igitur duo latera A B, A D, æqualia inter se sunt, per hypothesin, erunt anguli A B D, A D B, æquales inter se: Est autem angulo A B D, maior angulus C B D. Igitur & angulus C B D, maior erit angulo A D B. In triangulo ergo C B D, latus C D, oppositum maiori angulo C D B, maius erit latere B C, quod minori angulo C D B, opponitur. Cum igitur duo latera A B, A C, simul æqualia sint ipsi C D, (si enim æqualibus A B, A D, commune addatur A C, fient tota æqualia; nimirum linea composita ex A B, A C, & linea composita ex A D, A C) erunt quoque latera A B, A C, simul maiora latere B C. Eodem modo demonstrabitur, quælibet alia duo latera maiora esse reliquo.

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Therefore etc. Q. E. D.

Quare omnis trianguli duo latera reliquo sunt maiora, &c. Quod demonstrandum erat.

EX PROCLO
ALITER hoc theorema a familiaribus Heronis, & Porphyrii demonstratur, nullo latere producto, hac ratione. Sit probandum duo latera A B, A C, trianguli A B C, maiora esse latere B C. Diuidatur angulus B A C, illis lateribus contentus bifariam per rectam A D. Quoniam igitur trianguli C D A, latus C D, protractum est ad B, erit angulus externus B D A, maior interno & opposito C A D. Igitur & maior angulo B A D. Quare in triangulo A B D, latus A B, maiori angulo A D B, oppositum maius erit latere B D, quod minori angulo B A D, opponitur. Eadem ratione ostendetur, latus A C, maius esse quam C D, quia angulus C D A, maior est angulo B A D, hoc est, angulo C A D, &c. Quamobrem duo latera A B, A C, maiora erunt latere B C. Eademque est ratio quorumcunque duorum laterum, si angulus ipsis comprebensus bifariam secetur.

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