Question 113
Let $ABC$ be a triangle, $\omega$ its incircle and $D$, $E$ and $F$ the points of tangency of $\omega$ with the sides $BC$, $CA$ and $AB$ respectively. The perpendicular to $(BC)$ at $C$ meets the line $(EF)$ at $M$, and the perpendicular to $(BC)$ at $B$ meets the line $(EF)$ at $N$. The line $(DM)$ meets $\omega$ again at $P$ and the line $(DN)$ meets $\omega$ again at $Q$. Show that $DP = DQ$.