IT-CESENATICO 2014 Problem 2

Let $ABC$ be a triangle. Let $H$ be the foot of the altitude from $C$ on $AB$ . Suppose that $AH = 3HB$ . Suppose in addition we are given that

(a) $M$ is the midpoint of $AB$ ; (b) $N$ is the midpoint of $AC$ ; (c) $P$ is a point on the opposite side of $B$ with respect to the line $AC$ such that $NP = NC$ and $PC = CB$ .

Prove that $\angle APM = \angle PBA$ .