Question 156
Consider a triangle $A B C$ with $\angle A C B=90^\circ$. Let $F$ be the foot of the altitude from $C$. Circle $\omega$ touches the line segment $F B$ at point $P$, the altitude $C F$ at point $Q$ and the circumcircle of $A B C$ at point $R$. Prove that points $A, Q, R$ are collinear and $A P=A C$.
