Question 156
Let $\triangle ABC$ be a triangle with incenter $I$. The incircle of triangle $\triangle ABC$ touches $\overline{BC}$ at $D$. Let $M$ be the midpoint of $\overline{BC}$, and let line $AI$ meet the circumcircle of triangle $\triangle ABC$ again at $L \neq A$. Let $\omega$ be the circle centered at $L$ tangent to $AB$ and $AC$. If $\omega$ intersects segment $\overline{AD}$ at point $P$, prove that $\angle IPM = 90^{\circ}$.