Let the triangles ABC, DEF have their sides proportional, so that AB is to BC as DE is to EF; and BC to CA as EF is to FD; and, consequently, ex aequali, BA to AC as ED is to DF: the triangle ABC shall be equian^ilar to the triangle DEF, and they shall have those angles equal which are opposite to the homologous sides, namely, the angle ABC equal to the angle DEF, and the angle BCA equal to the angle EFD, and the angle BAC equal to the angle EDF.
At the point E, in the straight line EF, make the angle FEG equal to the angle ABC; and at the point F, in the straight line EF, make the angle EFG equal to the angle BCA; [I. 23.
therefore the remaining angle EGF is equal to the remaining angle BAC.
Therefore the triangle ABC is equiangular to the triangle GEF;
and therefore they have their sides opposite to the equal angles proportionals; [VI. 4.
therefore AB is to BC as GE is to EF.
But AB is to BC as DE is to EF: [Hypothesis.
therefore DE is to EF as GE is to EF; [V. 11.
therefore DE is equal to GF. [V. 9.
For the same reason, DF is equal to GF.
Then, because in the two triangles DEF, GEF, DE is equal to GE, and EF is common;
the two sides DE, EF are equal to the two sides GE, EF, each to each;
and the base DF is equal to the base GF;
therefore the angle DEF is equal to the angle GEF, [I. 8.
and the other angles to the other angles, each to each, to which the equal sides are opposite. [I. 4,
therefore the angle DFE is equal to the angle GFE, and the angle EDF is equal to the angle EGF.
