Let $O$ and $H$ denote the circumcenter and the orthocenter of an acute triangle $\triangle ABC$ respectively, Prove that $\angle BAH = \angle CAO$.
The only thing that I've managed to get is $$\angle CAO+\angle OBA+\angle BCO= 90^{\circ}$$ That’s because $OB, OA$ and $OC$ are all radiuses of $(C)$.