Let BC be bisected at E, and let AE be joined;
on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23]
through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF.
Then FECG is a parallelogram.
And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC,
for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38]
therefore the triangle ABC is double of the triangle AEC.
But the parallelogram FECG is also double of the triangle AEC,
for it has the same base with it and is in the same parallels with it; [I. 41]
therefore the parallelogram FECG is equal to the triangle ABC.
And it has the angle CEF equal to the given angle D.
Diuidatur latus unum trianguli, nempe B C, bifariam in E,
& fiat angulus C E F, æqualis angulo D,
Ducatur item per A, recta A F, parallela ipsi B C, quæ secet E F, in F. Rursus per C, vel B, ducatur ipsi E F, parallela C G, occurrens rectæ A F, productæ in G.
Eritque constitutum parallelogrammum C E F G, quod dico esse æquale triangulo A B C.
Ducta enim recta E A; quoniam parallelogrammum C E F G, duplum est trianguli A E C;
& triangulum A B C, duplum eiusdem trianguli A E C, quod triangula A E C, A B E, super æquales bases E C, B E, & in eisdem parallelis, sint æqualia.
Erunt parallelogrammum C E F G, & triangulum A B C, æqualia inter se.
Cum igitur angulus C E F, factus sit æqualis angulo D, constat propositum.
Complete text
Book I