Question 42
Let $ABC$ be a triangle and let $\omega$ be its circumcircle. Let $E$ be the midpoint of the minor arc $BC$ of $\omega$, and $M$ the midpoint of $BC$. Let $V$ be the other point of intersection of $AM$ with $\omega$, $F$ the point of intersection of $AE$ with $BC$, $X$ the other point of intersection of the circumcircle of $FEM$ with $\omega$, $X'$ the reflection of $V$ with respect to $M$, $A'$ the foot of the perpendicular from $A$ to $BC$ and $S$ the other point of intersection of $XA'$ with $\omega$. If $Z \in \omega$ with $Z \neq X$ is such that $AX = AZ$, then prove that $S$, $X'$ and $Z$ are collinear.