Bibliotheca Polyglotta

Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG; I say that the parallelogram ABCD is equal to EFGH.

SINT duo parallelogramma A C E F, G H D B, super æquales bases C E, H D, inter easdem parallelas A B, C D. Dico ea esse æqualia.

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For let BE, CH be joined. Then, since BC is equal to FG while FG is equal to EH, BC is also equal to EH. [C.N. 1] But they are also parallel. And EB, HC join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I. 33] () Therefore EBCH is a parallelogram. [I. 34] And it is equal to ABCD; for it has the same base BC with it, and is in the same parallels BC, AH with it. [I. 35] For the same reason also EFGH is equal to the same EBCH; [I. 35] so that the parallelogram ABCD is also equal to EFGH. [C.N. 1]

Connectantur enim extrema rectarum C E, G B, ad easdem partes lineis rectis C G, E B. Quoniam igitur recta C E, æqualis ponitur rectæ H D, & eidem H D, æqualis est G B, in parallelogrammo G H D B, opposita; erunt C E, G B, æquales inter se: Sunt autem & parallelæ, per hypothesin. Quare & C G, E B, ipsas coniungentes, parallelæ erunt, & æquales, ideoque C E B G, parallelogrammum erit. Itaque cum parallelogramma A C E F, G C E B, sint inter easdem parallelas, & super eandem basin C E, erit parallelogrammum A C E F, parallelogrammo G C E B, æquale. Rursus quia parallelogramma G C E B, G H D B, sunt inter easdem parallelas, & super eandem basin G B, erit quoque parallelogrammum G H D B, eidem parallelogrammo G C E B, æquale. Quare & parallelogramma A C E F, G H D B, inter se æqualia erunt.

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Therefore etc. Q. E. D.

Parallelogramma igitur super æqualibus basibus, & in eisdem parallelis constituta, & c. Quod ostendendum erat.

SCHOLION
CONVERSVM huius theorematis duplex est, ad hunc modum.
PARALLELOGRAMMA æqualia super bases æquales, & ad easdem partes constituta, inter easdem sunt parallelas: Et parallelogramma æqualia inter easdem parallelas, si non eandem habuerint basin, super æquales bases sunt constitura.
SINT primum duo parallelogramma æqualia A B C D, E F G H, super bases æquales B C, F G, & ad easdem partes constituta. Dico ea esse inter easdem parallelas, hoc est, A D, protractam coire in directum cum E H. Nam alias cadet aut infra E H, aut supra. Quo posito sequitur, totum & partem esse æqualia, quemadmodum in conuersa præcedentis propositionis est dictum, & figura facile commonstrat. Intelligendæ sunt autem bases æquales datæ in eadem linea recta B G.
SINT secundo eadem parallelogramma æqualia inter easdem parallelas A H, B G. Dico bases B C, F G, esse æquales. Si enim altera, nempe B C, dicatur maior, abscindatur B I, æqualis rectæ F G, & ducatur I K, parallela ipsi C D. Erit ergo parallelogrammum A B I K, æquale parallelogrammo E F G H; & ideo parallelogrammo A B C D, pars toti, quod est absurdum. Non ergo B C, maior est, quam F G. Eadem ratione neque minor erit. Quare bases B C, F G, æquales sunt.

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