Bariz - Problem

I found a very simple solution different from the official one. Let $H$ be the orthocenter, $O$ be the circumcenter. Let $\angle A=60^{\circ}$ and $HO\cap AB=M$; $HO\cap AC=N$. Since $H$ is the orthocenter, we have$\angle ABH=90^{\circ}-\angle A=60^{\circ}$, analogously $\angle ACH=60^{\circ}$. We wish to prove that $\angle BHM=30^{\circ}=\angle CHN$. This the same as saying that $HO$ is the external angle bisector of $\angle BHC$. By simple angle chasing, $\angle BHC=\angle BOC=120^{\circ}$. This means that points $B,H,O,C$ are concyclic. Also, $O$ is the midpoint of arc $BC$ containing the point $H$. THis implies that $HO$ is the angle bisector of $\angle BHC$. From this follows the result