Question 165
Let $ABC$ be a triangle, and let $X, Y$, and $Z$ be collinear points such that $AY = AZ$, $BZ = BX$, and $CX = CY$. Points $X'$, $Y'$, and $Z'$ are the reflections of $X$, $Y$, and $Z$ over $BC, CA$, and $AB$, respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$.