Question 119
Let $ABC$ be an acute triangle with circumcircle $\Gamma$ and $D$ the foot of the altitude from $A$. Suppose that $AD = BC$. Point $M$ is the midpoint of $DC$, and the bisector of $\angle ADC$ meets $AC$ at $N$. Point $P$ lies on $\Gamma$ such that lines $BP$ and $AC$ are parallel. Lines $DN$ and $AM$ meet at $F$, and line $PF$ meets $\Gamma$ again at $Q$. Line $AC$ meets the circumcircle of $\triangle PNQ$ again at $E$. Prove that $\angle DQE = 90^\circ$.