CONO-SUR 2018 Problem 5

Let $ABC$ be an acute-angled triangle with $\angle BAC = 60^{\circ}$ and with incenter $I$ and circumcenter $O$ . Let $H$ be the point diametrically opposite(antipode) to $O$ in the circumcircle of $\triangle BOC$ . Prove that $IH=BI+IC$ .