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Brocard point in isosceles triangle

Source: 9.5 of XX Geometrical Olympiad in honour of I.F.Sharygin

August 6, 2024
geometrygeo

Problem Statement

Let ABCABC be an isosceles triangle (AC=BC)(AC = BC), OO be its circumcenter, HH be the orthocenter, and PP be a point inside the triangle such that ∠APH=∠BPO=π/2\angle APH = \angle BPO = \pi /2. Prove that ∠PAC=∠PBA=∠PCB\angle PAC = \angle PBA = \angle PCB.
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