Given is an acute $\triangle ABC$ with orthocenter $H$. The point $H'$ is symmetric to $H$ over the side $AB$. Let $N$ be the intersection point of $HH'$ and $AB$. The circle passing through $A$, $N$ and $H'$ intersects $AC$ for the second time in $M$, and the circle passing through $B$, $N$ and $H'$ intersects $BC$ for the second time in $P$. Prove that $M$, $N$ and $P$ are collinear. Proposed by Petar Filipovski