Suppose we have a fixed triangle $ABC$. $M$ is the midpoint of $BC$ and $D$ is a variable point on segment $BC$. $D'$ is the reflection of $D$ across $M$. The circumcircles of $\triangle ADM$ and $\triangle AD'M$ intersects $AB$ at $X,Y$ respectively. Prove that the midpoint of $XY$ is fixed as $D$ varies.
So naturally, I considered $W,X$, the intersection of the two circles on $AC$. I found that circles $AYW$ and $AXZ$ intersect on circle $ABC$, at the point $G$ that makes $ABCG$ an isosceles trapezium. However, I can't seem to prove those claims... I suspect that there are some Miquel properties and Spiral similarities at work here... Help!
By power of a point, $DB \cdot MB = XB \cdot AB$, and $D'B \cdot MB = YB \cdot AB$. So adding the two equations, $MB(DB+D'B) = AB(XB+YB)$. But note that $DB + D'B = 2MB$, so the LHS is fixed, and so the RHS is fixed too. As $AB$ is fixed, we obtain $XB+YB$ is fixed, and so $EB = \frac{XB+YB}{2}$ is fixed too.