Question 184
Let $A$ be a point outside the circle $\omega$. Two points $B, C$ lie on $\omega$ such that $A B, A C$ are tangent to $\omega$. Let $D$ be any point on $\omega$ ($D$ is neither $B$ nor $C$) and $M$ the foot of perpendicular from $B$ to $C D$. The line through $D$ and the midpoint of $B M$ meets $\omega$ again at $P$. Prove that $A P \perp C P$.