smc

Because $\overline{XY}$ is a diameter of the circle and $\overline{AB}\perp \overline{XY}$, we know that $\overline{XY}$ bisects $\overline{AB}$, so $AQ=BQ$. Thus, $M$ is on the perpendicular bisector of $\overline{AB}$, and so $AM=BM$. Furthermore, by Thales' Theorem, $\measuredangle AMB=90^{\circ }$. Thus, because $△AMB$ is a right isosceles triangle with $AM=BM$, it is a 45-45-90 triangle. Thus, $\measuredangle ABM=45^{\circ }$. Now, draw $\overline{OA}$ and $\overline{OD}$. Because $\angle ABD\cong \angle ABM$ is an inscribed angle which intercepts minor arc $AD$, the measure of central angle $\angle AOD$ must be $2\measuredangle ABD=2\cdot 45^{\circ }=90^{\circ }$. Because $OA=OD=r$, $△AOD$ is also a $45$-$45$-$90$ triangle, so $AD=\boxed{r\sqrt{2}}$.