Question 20
Let $ABC$ be a triangle whose angles $\angle CAB$ and $\angle ABC$ are greater than $45^\circ$. $PQRS$ is a square such that $P$ and $Q$ are inside the side $AB$ in the order $APQB$, $R$ is inside the side $BC$, $S$ is inside the side $CA$. Let $M$ and $N$ be the feet of the perpendiculars drawn from $P$ to the side $CB$ and from $Q$ to the side $CA$, respectively. If $H$ is the intersection of $PM$ and $QN$, show that $CH$ is perpendicular to $AB$.