Let $ABC$ be a triangle and let $N$ and $M$ be the midpoints of $AB$ and $CA$, respectively. Let $H$ be the foot of altitude from $A$. The circumcircle of $ABH$ intersects $MN$ at $P$, with $P$ and $M$ on the same side relative to $N$, and the circumcircle of $ACH$ intersects $MN$ at $Q$, with $Q$ and $N$ on the same side relative to $M$. $BP$ and $CQ$ intersect at $X$. Prove that $AX$ is the angle bisector of $\angle CAB$.