If AO=AH. prove that angle A=60 degree.

Question

$ABC$ is a triangle with the circumcenter $O$ and orthocenter $H$. If $AO=AH$, prove that $m(\hat A)= 60^\circ$.

Also, if $H$ lies on the circumcircle of $BOC$, prove that $m(\hat A)= 60^\circ$.

Answer 1

$AH=2OM_1$ and $AH=AO=CO=R$ $ \Rightarrow OM_1=\frac12CO \Rightarrow \angle OCM_1 =30^{\circ}=\angle BOC$ Then in triangle $BOC:$ $$\angle BOC =180^{\circ}-2\cdot 30^{\circ}=120^{\circ} \Rightarrow \angle BAC =\frac12 \angle BOC=60^{\circ}$$