USAJMO 2013 Problem 5

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$ . Segments $AY$ and $BX$ meet at $P$ . Point $Z$ is the foot of the perpendicular from $P$ to line $XY$ . Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$ . Let $Q$ be the intersection of segments $AY$ and $XC$ . Prove that $$\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.$$