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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