Prove that isosceles triangles are similar if the angles at their vertices opposite to their bases are equal.

Let there be given two isosceles triangles ABC and A1B1C1 in which the angle ABC = A1B1C1 = α0.

Since the triangles are isosceles, their angles at the base of AC and A1C1 are equal.

In the triangle ABC, the angle BAC = (180 – ABC) / 2 = (180 – α) / 2.

In triangle A1B1C1, angle B1A1C1 = (180 – ABC) / 2 = (180 – α) / 2.

Then in triangles ABC and A1B1C1 the angle ABC = A1B1C1 and BAC = B1A1C1, and therefore triangles ABC and A1B1C1 are similar in two angles, the first sign of the similarity of triangles, which was required to be proved.