China National Team Selection Test

Proof Refer to the figure, join points $O$ and $M$. Then $OM$ is the perpendicular bisector of $AB$. So $△XES\sim △OMS$, and thus $SX=XE$. Now draw a circle with center $X$ whose radius is $XE$. Then the circle $X$ is tangent to chord $AB$ and line $CS$. Draw the circumcircle of $△PQR$, line $MA$ and line $MC$. It is easy to see ($△AMR\sim △PMA$ etc) $$MR\cdot MP=M{A}^2=ME\cdot MS.\stackrel{◯}{1}$$ By the Power of a Point theorem, $$CQ\cdot CP=C{S}^2.\stackrel{◯}{2}$$ So $M$, $C$ are on the radical axis of circle $Z$ and circle $X$. Thus $$ZX\perp MC.$$ Similarly, we have $ZY\perp MC$. So $X$, $Y$, $Z$ are collinear.