Let $K=BA\cap CD,O,O'$ the circumcenter of $BCX$ and $ADY$ .Since $YA\perp BX,YD\perp CX $ then \angle BXC =\angle AYD$ $ thus $K $ is the ex-similicenter of $(ADY),(XBC)$ moreover if $Y'$ is the symmetric of $Y$ in $O'$ by angle chase we get $XBC \sim Y'AD $ hence $Y'AD$ is the image of $XBC$ ; by the the homothety of center $K$ hence the locus of $Y'$ is line say $\ell '$ but $O'$ is on the bisector of $AD$ thus $Y$ is on the symmetric of $\ell '$ in the $AD$-bisector . RH HAS