1957 AMC 12 Problem 47

Since $\overline{XY}$ perpendicularly bisects $\overline{AB},$ compare $MA$ and $MB$

Use the semicircle to find $\angle AMB,$ then use that $B,$ $M,$ and $D$ are collinear

Line $XY$ is the perpendicular bisector of $AB,$ so $MA=MB.$ Since $M$ lies on the semicircle with diameter $AB,$ $\angle AMB=90^\circ.$ Thus triangle $AMB$ is an isosceles right triangle and $\angle ABM=45^\circ.$ Because $B,M,D$ are collinear, the inscribed angle $\angle ABD=45^\circ,$ so its intercepted arc $AD$ measures $90^\circ.$ Therefore chord $AD$ is a side of an inscribed square and has length $$AD=r\sqrt2.$$

Thus, the correct answer is A.