Proposition I-26

This is impossible because angle DEF equals the angle ABC, and the whole is greater than the part.
(CN-5)

So, AC cannot be less than DF. Therefore AC equals DF. Correspondingly, AB equals DE and angle BAC equals angle EDF, which was to be proved.

The other part of the theorem is proved similarly.

QED

Let ABC be the given triangle.

Let ABC and DEF be two triangles having the two angles ABC and BCA equal to the two angles DEF and EFD respectively.

Also, let BC equal EF.

We will show that AB equals DE and that AC equals DF.

Since GF equals AC, BC equals EF, and the angles at C and F are equal, the triangles ABC and GEF are congruent.
(Prop I-4)

Therefore the angles ABC and GEF are equal.

Next, let sides opposite equal angles be equal, as AB equals DE. I say again that the remaining sides equal the remaining sides, namely AC equals DF and BC equals EF, and further the remaining angle BAC equals the remaining angle EDF. If BC is unequal to EF, then one of them is greater. Let BC be greater, if possible. Make BH equal to EF, and join AH. I.3 Post.1 Since BH equals EF, and AB equals DE, the two sides AB and BH equal the two sides DE and EF respectively, and they contain equal angles, therefore the base AH equals the base DF, the triangle ABH equals the triangle DEF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle BHA equals the angle EFD. I.4 But the angle EFD equals the angle BCA, therefore, in the triangle AHC, the exterior angle BHA equals the interior and opposite angle BCA, which is impossible. C.N.1 I.16 Therefore BC is not unequal to EF, and therefore equals it. But AB also equals DE. Therefore the two sides AB and BC equal the two sides DE and EF respectively, and they contain equal angles.

Therefore the base AC equals the base DF, the triangle ABC equals the triangle DEF, and the remaining angle BAC equals the remaining angle EDF.

QED