Let $O$ be the center of $\omega$. $\angle{AED} = \angle{AEO} = 90^{\circ}$, so $O, D, E$ are collinear. By Power of a Point, $BC \cap AE$ lies on $XY$, and by radical axis $XY$ is perpendicular to the line through $O$ and the center of $(AD)$, so by Brocard on $MDEA$, $AD \cap EM \in XY$. $\square$

I'll add one more. As above $O$, $D$, and $E$ are collinear and let $S$ be the intersection of $BC$, $AE$, and $XY$. Let $(EMO)$ intersect $\omega$ at $Z$ and $W$. $\angle OZS=\angle OWS=\angle OMS=90^{\circ}$ so $OZ$ and $OW$ are tangents to $\omega$. Note that since $BC$ is the polar of $A$ wrt $\omega$. Since $S$ lies on $BC$, by LeHire's Theorem $A$ lies on the polar of $S$ wrt $\omega$. So $A$ lies in $WZ$. Also since $DB\cdot DC=DO\cdot DE$ we have that $D$ lies in the radical axis of $\omega$ and $(EMO)$. This gives us that $AD$ is the radical axis of $\omega$ and $(EMO)$. From here, apply Radical Axes Theorem to the circles $(EMO)$, $\omega$, and $(AMD)$ and get that $AD$, $EM$, and $XY$ are collinear.

Another aproach that seems to work quite nicely is inversion at circle omega which insta kills if u know where the points go.