Proposition I-9

Proposition I-8. If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.
Proof.

Suppose that the point A does not coincide with E.

It is then at point G.

Let ABC and EDF be the given triangles.

Sides AB and ED are equal. Sides AC and EF are equal. Also, the bases BC and DF are equal.

We now show that angles BAC and DEF are equal.

Apply the triangle ABC to the triangle EDF so that the point B coincides with the point D and the point C coincides with the point F.

(This follows because BC is equal to DF)

Of course, the sides GD and ED are equal because GD is equal to AB. Similarly sides GF and EF are equal.

Therefore the triangles EDF and GEF have all equal sides. But this is impossible. (Prop I-7)

...unless

Thus G coincides with E and the triangles ABC adn EDF are congruent.

Therefore all the respective angles are equal, in particular angles BAC and DEF, which was to be proved.

QED