PMO 2020 Problem 4

Let $\triangle ABC$ 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 $\angle DQE = 90^{\circ}$ .