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

October 9, 2021 | education

| 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.

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