KMO 2001 Problem 5

In a triangle $ABC$ with $\angle B < 45^{\circ}$ , $D$ is a point on $BC$ such that the incenter of $\triangle ABD$ coincides with the circumcenter $O$ of $\triangle ABC$ . Let $P$ be the intersection point of the tangent lines to the circumcircle $\omega$ of $\triangle AOC$ at points $A$ and $C$ . The lines $AD$ and $CO$ meet at $Q$ . The tangent to $\omega$ at $O$ meets $PQ$ at $X$ . Prove that $XO=XD$ .