NL-BXMO-TST 2018 Problem 4

In a non-isosceles triangle $\vartriangle ABC$ we have $\angle BAC = 60^o$ . Let $D$ be the intersection of the angular bisector of $\angle BAC$ with side $BC, O$ the centre of the circumcircle of $\vartriangle ABC$ and $E$ the intersection of $AO$ and $BC$ . Prove that $\angle AED + \angle ADO = 90^o$ .