Let $AH,BH$ hit $BC, AC$ at $E,F$ respectively, note that $(BEFA)$ has center $M$ and thus $M$ lies in the perendicular bisector of $EF$, if you let $N$ midpoint of $AH$ then it also lies in the perpendicular bisector of $EF$ because its the center of $(CEHF)$, but to finish whether $\angle C$ is acute or obtuse $P$ is either midpoint of arc $ECF$ or $EHF$ in $(CEHF)$ both of which lie in the perpendicular bisector of $EF$ thus we are done .

Let $N$ be the midpoint of $CH$. It is well known that $CH=2OM$ so $CN=OM$. Since $CN\perp AB$ and $OM\perp AB$ we have that $CN\parallel OM$ and thus $CNOM$ is a parallelogram. To finish $$\measuredangle HNP=2\measuredangle HCP=\measuredangle HCO=\measuredangle HNM$$

Took more time than I’d like to admit, but, yeah, simple solution. Let $O$ be the circumcentre of $\triangle ABC$. Let $N$ be mid point of $AH$. Let $MN \cap (N,AN)=P’$. We show that $P’ \equiv P$. It is well known that $MN \parallel AO$ and that $H$ and $O$ are isogonal conjugates. We get $\angle HAO=\angle HNM=\frac{\angle HAP’}{2} \implies \angle HAP=\angle PAO \implies AP$ is the angle bisector, and we are done.$\blacksquare$