In my class we are learning geometry and the instructor gave us this problem: Let $ABC$ be a scalene triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$.Prove that $X, Y, H$ are collinear.
I already solved this trivial problem using motivated methods of radical axis but my instructor proposed another solution where you reflect $W$ over the midpoint of $BC$. Suppose this point is $V$. I noticed that if I can prove $VY$ is perpendicular to $AC$, then the problem is immediately proven by Pappus. Does anyone know how to prove this claim?
Through D, a line is drawn parallel to BM cutting WY at T and CM at K. After this setup, CM is perpendicularly bisected by DTK and, in addition, T is the center of $\omega_2$ (because it is the intersection of the diameter WY and the perpendicular bisector of the chord CM).
This makes DT the line of centers of $\omega_2$ and $\omega_3$ (centered at D, radius = DV).
Let $\omega_2$ and $\omega_3$ intersect at Z. Then, TD extended cuts WZ (the common chord) at right angles. Note that ZVY is a straight line because $\angle WZV$ (wrt $\omega_3$) $= \angle WZY$ (wrt $\omega_2$) $= 90^0$. We then conclude that $BM$, $DTK$ and $VY$ are all parallel. Result follows.
Remark: Point X plays no role in this question.