Belarus2022

Denote by $O$ the center of the circumcircle of the triangle $XYZ$. Then $$\angle YOZ=2\angle YXZ=2\angle ABC=180^{\circ }-\angle BAC=180^{\circ }-\angle ZAY,$$ whence the quadrilateral $ZAYO$ is cyclic. The chords $OY$ and $OZ$ are equal so $\angle ZAO=\angle OAY$. Thus $O$ lies on the bisector of the angle $BAC$, which is the perpendicular bisector of the segment $XW$, therefore $OW=OX=OY=OZ$ and $W$ lies on the circumcircle of the triangle $XYZ$.