Question 134
In a non-isosceles triangle $\triangle A B C$ we have $\angle B A C=60^{\circ}$. Let $D$ be the intersection of the bisector of $\angle B A C$ with the side $B C$, $O$ the centre of the circumcircle of $\triangle A B C$ and $E$ the intersection of $A O$ with $B C$. Prove that $\angle A E D+\angle A D O=90^{\circ}$.
