First of all, $\angle{QCP} = 90^{\circ} - \frac{\angle{A}}{2} + 90^{\circ} - \frac{\angle{B}}{2} - 90^{\circ} = 45^{\circ} \implies \triangle{POQ}$ has angles of $45^{\circ}, 45^{\circ}, 90^{\circ}$. Furthermore, $AO, BO$ bisect $\angle{A}, \angle{B} \implies O$ is the incenter of $\triangle{ABC}$. Thus, if $M_C$ is the midpoint of $\widehat{AB}$ not containing $C$, then $O$ lies on the circle $(M_C)$ centered at $M_C$ passing through $A$ and $B$, and $O = M_CC \cap (M_C)$. Also, if $D$ is the foot from $O$ to $AB$ then it is the midpoint of $PQ$, so $OH$ passes through the centroid $G$ of $\triangle{CPQ}$, which is $\frac{2}{3}$ of the way from $C$ to $D$. Let $R$ be the intersection of $GO$ and the perpendicular bisector of $AB$; then it suffices to show that $RM_C$ is constant. By the Ratio Lemma on $\triangle{OCD}$, $$\frac{1}{2} \div \tan 45^{\circ} = \frac{1}{2} \div \frac{OD}{OC} = \frac{\sin \angle{DOG}}{\sin \angle{COG}} = \frac{\sin \angle{ORM_C}}{\sin \angle{ROM_C}},$$which by LoS in $\triangle{ROM_C}$ equals $\frac{OM_C}{RM_C}$. $O_MC$