Question 24
Let $ABC$ be a triangle such that angle $BAC = 45^\circ$. Let $P$ and $Q$ be interior points of the triangle $ABC$ such that $ABQ = QBP = PBC$ and $ACQ = QCP = PCB$. Let $D$ and $E$ be the feet of the perpendiculars drawn from $P$ to the sides $CA$ and $AB$, respectively. Show that $Q$ is the orthocenter of the triangle $ADE$.