Question 169
In triangle $A B C$ with $\angle A B C=60^{\circ}$ and $5 A B=4 B C$, points $D$ and $E$ are the feet of the altitudes from $B$ and $C$, respectively. $M$ is the midpoint of $B D$ and the circumcircle of triangle $B M C$ meets line $A C$ again at $N$. Lines $B N$ and $C M$ meet at $P$. Prove that $\angle E D P=90^{\circ}$.