Prove that the angles at the base of an isosceles triangle are equal.

Consider an isosceles triangle ABC, since it is isosceles, then AC = BC.
Now, to prove equality of the angles at the base of this triangle, draw the bisector CD from the apex of the triangle, therefore the angle ACB will be divided into two equal angles: ∠ACD = ∠BCD.
Based on the sign of equality of triangles, which are considered equal if they have at least two sides equal and the angles between these sides are equal, then in the end we get two equal triangles ACD and BCD. Hence it follows that their angles ∠A and ∠B are also equal.