Let $\triangle ABC$ be a acute-angled triangle so that $AB>AC$ and $\angle BAC = 60^\circ$. Let $O$ be the circumcenter and $H$ the orthocenter. Let $OH$ intersect $AB$ and $AC$ in $P$ and $Q$ respectively. Show that $AQ=AP$. Or alternatively, show that $\angle AOP$ and/or $\angle AHQ$ equal $\angle ACB+90-\angle BAC$
I know $BAO=CAH=90-\angle ACB$, but that's all I've got. I need one of two above. An elementary solution is preferred.