Question 122
Let $ABC$ be an acute triangle with circumcircle $\omega$ centered at $O$. A point $P$ is chosen on the extension of the segment $BC$. The line $AP$ meets $\omega$ again at $L$. Let $K$ be the reflection of $L$ over the line $OP$ and $M$ be the point of intersection of lines $AK$ and $OP$. Prove that $\angle OMB + \angle OMC = 180^\circ$.