Given is an acute triangle $ABC$. Point $P$ lies inside this triangle and lies on the bisector of angle $\angle BAC$. Suppose that the point of intersection of the altitudes $H$ of triangle $ABP$ lies inside triangle $ABC$. Let $Q$ be the intersection of the line $AP$ and the line perpendicular to $AC$ passing through $H$. Prove that $Q$ is the point symmetrical to $P$ wrt the line $BH$.
Problem
Source: 2023 Czech-Polish-Slovak Match Junior, individual p3 CPSJ
Tags: geometry, orthocenter
