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<AEF = 90^o if AB=AC, PQ// AB, F circumcener of PQC

Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2011 Seniors p2

May 21, 2022
geometryright angleisoscelesChampions Tournament

Problem Statement

Let ABCABC be an isosceles triangle in which AB=ACAB = AC. On its sides BCBC and ACAC respectively are marked points PP and QQ so that PQ∥ABPQ\parallel AB. Let FF be the center of the circle circumscribed about the triangle PQCPQC, and EE the midpoint of the segment BQBQ. Prove that ∠AEF=90o\angle AEF = 90^o .
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