In circle $ O$, the midpoint of radius $ OX$ is $ Q$; at $ Q$, $ \overline{AB} \perp \overline{XY}$. The semi-circle with $ \overline{AB}$ as diameter intersects $ \overline{XY}$ in $ M$. Line $ \overline{AM}$ intersects circle $ O$ in $ C$, and line $ \overline{BM}$ intersects circle $ O$ in $ D$. Line $ \overline{AD}$ is drawn. Then, if the radius of circle $ O$ is $ r$, $ AD$ is:
[asy]defaultpen(linewidth(.8pt));
unitsize(2.5cm);

real m = 0;
real b = 0;

pair O = origin;
pair X = (-1,0);
pair Y = (1,0);
pair Q = midpoint(O--X);
pair A = (Q.x, -1*sqrt(3)/2);
pair B = (Q.x, -1*A.y);
pair M = (Q.x + sqrt(3)/2,0);

m = (B.y - M.y)/(B.x - M.x);
b = (B.y - m*B.x);

pair D = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b));

m = (A.y - M.y)/(A.x - M.x);
b = (A.y - m*A.x);

pair C = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b));

draw(Circle(O,1));
draw(Arc(Q,sqrt(3)/2,-90,90));
draw(A--B);
draw(X--Y);
draw(B--D);
draw(A--C);
draw(A--D);
dot(O);dot(M);

label("$B$",B,NW);
label("$C$",C,NE);
label("$Y$",Y,E);
label("$D$",D,SE);
label("$A$",A,SW);
label("$X$",X,W);
label("$Q$",Q,SW);
label("$O$",O,SW);
label("$M$",M,NE+2N);[/asy]$ \textbf{(A)}\ r\sqrt {2} \qquad \textbf{(B)}\ r\qquad \textbf{(C)}\ \text{not a side of an inscribed regular polygon}\qquad \textbf{(D)}\ \frac {r\sqrt {3}}{2}\qquad \textbf{(E)}\ r\sqrt {3}$
1. **Identify Key Points and Relationships:**
   - Given: Circle \( O \) with radius \( r \).
   - \( Q \) is the midpoint of radius \( OX \), so \( OQ = \frac{r}{2} \).
   - \( \overline{AB} \perp \overline{XY} \) at \( Q \).
   - \( \overline{AB} \) is the diameter of a semicircle intersecting \( \overline{XY} \) at \( M \).
   - Lines \( \overline{AM} \) and \( \overline{BM} \) intersect circle \( O \) at points \( C \) and \( D \) respectively.
   - We need to find the length of \( \overline{AD} \).

2. **Analyze the Geometry:**
   - Since \( Q \) is the midpoint of \( OX \), \( OQ = \frac{r}{2} \).
   - \( \triangle BQO \) is a \( 30^\circ-60^\circ-90^\circ \) triangle because \( OB = r \) and \( OQ = \frac{r}{2} \).
   - In a \( 30^\circ-60^\circ-90^\circ \) triangle, the side opposite the \( 30^\circ \) angle is half the hypotenuse, and the side opposite the \( 60^\circ \) angle is \( \frac{\sqrt{3}}{2} \) times the hypotenuse.
   - Therefore, \( BQ = \frac{\sqrt{3}}{2}r \).

3. **Determine \( QM \) and \( AQ \):**
   - Since \( \overline{AB} \) is perpendicular to \( \overline{XY} \) at \( Q \), and \( \overline{AB} \) is the diameter of the semicircle, \( AQ = BQ = \frac{\sqrt{3}}{2}r \).
   - \( QM = BQ = \frac{\sqrt{3}}{2}r \).

4. **Calculate \( OM \):**
   - \( OM = QM - OQ = \frac{\sqrt{3}}{2}r - \frac{1}{2}r = \frac{\sqrt{3} - 1}{2}r \).

5. **Apply Power of a Point Theorem:**
   - By the Power of a Point theorem, \( (MB)(MD) = (MX)(MY) \).
   - \( MB = \frac{\sqrt{6}}{2}r \) (since \( \triangle BQM \) is a \( 45^\circ-45^\circ-90^\circ \) triangle).
   - \( OX = r \), so \( MX = OX + OM = r + \frac{\sqrt{3} - 1}{2}r = \frac{2 + \sqrt{3} - 1}{2}r = \frac{1 + \sqrt{3}}{2}r \).
   - \( MY = OX - OM = r - \frac{\sqrt{3} - 1}{2}r = \frac{2 - \sqrt{3} + 1}{2}r = \frac{3 - \sqrt{3}}{2}r \).

6. **Solve for \( MD \):**
   - \( MD \cdot \frac{\sqrt{6}}{2}r = \left(\frac{1 + \sqrt{3}}{2}r\right) \left(\frac{3 - \sqrt{3}}{2}r\right) \).
   - Simplify the right-hand side: 
     \[
     \left(\frac{1 + \sqrt{3}}{2}r\right) \left(\frac{3 - \sqrt{3}}{2}r\right) = \frac{(1 + \sqrt{3})(3 - \sqrt{3})}{4}r^2 = \frac{3 + 3\sqrt{3} - \sqrt{3} - 3}{4}r^2 = \frac{2\sqrt{3}}{4}r^2 = \frac{\sqrt{3}}{2}r^2
     \]
   - Therefore, \( MD = \frac{\sqrt{3}}{2}r \).

7. **Use Pythagorean Theorem in \( \triangle AMD \):**
   - \( \triangle AMD \) is a right triangle because \( \angle BMA \) is an inscribed angle in a semicircle.
   - By the Pythagorean Theorem:
     \[
     AD = \sqrt{AM^2 + MD^2} = \sqrt{\left(\frac{\sqrt{6}}{2}r\right)^2 + \left(\frac{\sqrt{3}}{2}r\right)^2} = \sqrt{\frac{6r^2}{4} + \frac{3r^2}{4}} = \sqrt{\frac{9r^2}{4}} = \frac{3r}{2}
     \]

The final answer is \( \boxed{r\sqrt{2}} \).