Question 106
Let $\omega_1$ and $\omega_2$ be circles with no common points. Points $M$ and $N$ are chosen on the circles $\omega_1$ and $\omega_2$, respectively, such that the tangent to the circle $\omega_1$ at $M$ and the tangent to the circle $\omega_2$ at $N$ intersect at $P$ and $\triangle PMN$ is an isosceles triangle with apex $P$. The circles $\omega_1$ and $\omega_2$ meet the segment $MN$ again at $A$ and $B$, respectively. The line $PA$ meets the circle $\omega_1$ again at $C$ and the line $PB$ meets the circle $\omega_2$ again at $D$. Prove that $\angle BCN = \angle ADM$.