ELMO 1999 Problem 1

In nonisosceles triangle $ABC$ the excenters of the triangle opposite $B$ and $C$ be $X_B$ and $X_C$ , respectively. Let the external angle bisector of $A$ intersect the circumcircle of $\triangle ABC$ again at $Q$ . Prove that $QX_B = QB = QC = QX_C$ .