Question 157
Let $\triangle ABC$ be an acute triangle with orthocenter $H$. The point $H'$ is symmetric to $H$ with respect to the line $AB$. Let $N$ be the intersection point of $HH'$ and $AB$. The circle passing through the points $A$, $N$ and $H'$ intersects the line $AC$ again at $M$, and the circle passing through the points $B$, $N$ and $H'$ intersects the line $BC$ again at $P$. Prove that the points $M$, $N$ and $P$ are collinear.