TH-TMO 2012 Problem 1

Let $\vartriangle ABC$ be a right triangle with $\angle B = 90^o$. Let $P$ be a point on side $BC$, and let $\omega$ be the circle with diameter $CP$. Suppose that $\omega$ intersects $AC$ a nd $AP$ again at $Q$ and $R$, respectively. Show that $CP^2 = AC \cdot CQ - AP \cdot P R$.