Question 105
Let $ABC$ be an acute triangle with $AB \neq AC$. The incircle $\omega$ of the triangle touches the sides $BC$, $CA$ and $AB$ in $D$, $E$ and $F$, respectively. The perpendicular line erected from $C$ to $BC$ intersects $EF$ at $M$, and, similarly, the perpendicular line erected at $B$ to $BC$ intersects $EF$ at $N$. The line $DM$ meets $\omega$ again in $P$, and the line $DN$ meets $\omega$ again in $Q$. Prove that $DP = DQ$.
Rubén Dario, Perú, and Leonard Giugliuc, Romania
Rubén Dario, Perú, and Leonard Giugliuc, Romania