It is claimed that $XY$ is the perpendicular bisector of $EF$. To show it, first a lemma is proved.

Lemma. Let $D$ be a point on the extension of side $AB$ of triangle $ABC$ such that $A$ is between $B$ and $D$, and $AD = BC$. Let $O$ be the intersection point of external angle bisector of vertex $B$ in triangle $ABC$ and the perpendicular bisector of $DC$. Then $O$ lies on the perpendicular bisector of $AB$.

Proof. Let $E$ be a point on the extension of side $AB$ of triangle $ABC$ such that $B$ is between $E$ and $A$ and $BE = BC$. Since triangle $EBC$ is isosceles, the external angle bisector of $B$ is the perpendicular bisector of $EC$. Therefore, $O$ is the intersection point of perpendicular bisectors of $AC$ and $CE$, which means $O$ is the circumcenter of triangle $DEC$, so $O$ lies on the perpendicular bisector of $DE$. Since $BE = AD$, this line is the perpendicular bisector of $AB$ as well, which completes the proof.

Note that $AE$ and $AF$ are the external angle bisectors of triangle $DEF$. Using the lemma for triangle $DEF$ and point $Q$ on the extension of $EF$ implies that $X$ lies on the perpendicular bisector of $EF$. By a similar argument, $Y$ lies on it too, which means that $XY$ is the perpendicular bisector of $EF$. On the other hand, since points $E$, $F$, $C$ and $B$ lie on a circle with diameter $BC$ ($\angle BEC = \angle BFC = 90^\circ$), the midpoint of $BC$ is the center of this circle and lies on the perpendicular bisector of $EF$. Therefore, the midpoint of $BC$ lies on line $XY$.