Let $ABC$ be an isosceles triangle $(AC = BC)$, $O$ be its circumcenter, $H$ be the orthocenter, and $P$ be a point inside the triangle such that $\angle APH = \angle BPO = \pi /2$. Prove that $\angle PAC = \angle PBA = \angle PCB$.
Source: 9.5 of XX Geometrical Olympiad in honour of I.F.Sharygin
Tags: geometry, geo
Let $ABC$ be an isosceles triangle $(AC = BC)$, $O$ be its circumcenter, $H$ be the orthocenter, and $P$ be a point inside the triangle such that $\angle APH = \angle BPO = \pi /2$. Prove that $\angle PAC = \angle PBA = \angle PCB$.