Bibliotheca Polyglotta

Let ABCD be a parallelogram, and AC its diameter; and about AC let EH, FG be parallelograms, and BK, KD the so-called complements; I say that the complement BK is equal to the complement KD.

IN parallelogrammo A B C D, sint circa diametrum A C, parallelogramma A E G H, C F G I, & complementa D F G H, E B I G, ut in 36. defin. diximus. Dico complementa hæc inter se esse æqualia.

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For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD. [I. 34] Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK. For the same reason the triangle KFC is also equal to KGC. Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC, the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2] And the whole triangle ABC is also equal to the whole ADC; therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3]

Cum enim triangula A B C, C D A, æqualia sint; Itemque triangula A E G, G H A; si hæc ab illis demantur, remanebunt trapezia C B E G, C D H G, æqualia: Sunt autem & triangula C G I, C G F, æqualia. Quare si detrahantur ex trapeziis, remanebunt æqualia complementa D F G H, E B I G.

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Therefore etc. Q. E. D.

In omni igitur parallelogrammo, complementa, &c. Quod ostendendum erat.

SCHOLION
EODEM modo hoc theorema demonstratur a Proclo, etiam si duo parallelogramma circa diametrum non coniunguntur in puncto G, sed vel unum ab altero sit semotum, vel ambo se mutuo intersecent. Sit enim prius unum ab altero distans, ita ut complementa sint figuræ quinquangulæ. Vt in parallelogrammo A B C D, circa diametrum A C, consistant parallelogramma A E F G, C H I K. Dico complementa D E F I H, B K I F G, esse æqualia. Cum enim triangula A B C, C D A, æqualia inter se sint: Item triangula A E F, G H I, æqualia triangulis A G F, C K I: erunt reliqua complementa D E F I H, B K I F G, æqualia. Quod est propositum.
SECENT se iam mutuo parallelogramma A E F G, C H I K, circa diametrum consistentia, ita ut communem partem habeant I L F M. Dico adhuc complementa D E L H, B G M K, esse æqualia. Cum enim æqualia sint triangula A B C, C D A: Item triangula A F G, A F E: erunt reliqua quadrilatera B C F G, D C F E, æqualia: Sunt autem rursus æqualia triangula I F M, I F L. Igitur si hæc addantur dictis quadrilateris, erunt figura B C I M G, D C I L E, æquales. Cum igitur & æqualia sint triangula C I K, C I H: erunt reliqua complementa B G M K, D E L H, etiam æqualia. Quod est propositum.
CONVERSVM quoque huius theorematis cum Peletario demonstrabimus, hoc modo.

SI parallelogrammum diuisum fuerit in quatuor parallelogramma, ita ut ex illis duo aduersa sint æqualia: consistent reliqua duo circa diametrum.
DVCTIS duabus rectis E F, G H, quæ sint parallelæ rectis B C, C D, seque secent in I, diuisum sit parallelogrammum A B C D, in quatuor parallelogramma, quorum aduersa duo B E I H, D F I G, sint æqualia. Dico reliqua duo A E I G, C F I H, circa diametrum consistere, hoc est, diametrum a puncto C, ad punctum A, ductam transire per punctum I. Si enim non transit, secet diameter C K A, rectam G H, in K, si fieri potest, & per K, ducatur L M, parallela ipsi B C. Erunt igitur complementa B H K L, D G K M, æqualia: Est autem D G K M, maius quam D G I F. Quare & maius erit B H K L, quam D G I F. Cum ergo D G I F, æquale ponatur ipsi B E I H; erit etiam B H K L, maius quam B E I H, pars quam totum. Quod est absurdum. Non ergo diameter A C, rectam G H, in K, secat, sed per punctum I, transit. Quodest propositum.

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