Let $A, B,C$ be colinear points in this order, $\omega$ an arbitrary circle passing through $B$ and $C$, and $l$ an arbitrary line different from $BC$, passing through A and intersecting $\omega$ at $M$ and $N$. The bisectors of the angles $\angle CMB$ and $\angle CNB$ intersect $BC$ at $P$ and $Q$. Prove that $AP\cdot AQ = AB \cdot AC$.