jmc

We begin by drawing a diagram: We draw diagonals $\overline{AC}$ and $\overline{BD}$ and let the intersection point be $H$. Since $\angle ADC=60^{\circ }$ and $AD=CD$, $△ACD$ is equilateral, so $AC=24$. Since $ABCD$ has two pairs of equal sides, it is a kite, and so its diagonals are perpendicular and $\overline{BD}$ bisects $\overline{AC}$. Thus, $$AH=HC=24/2=12.$$Applying the Pythagorean Theorem on $△BHC$ and $△CHD$ gives $$BH=\sqrt{B{C}^2-H{C}^2}=\sqrt{13^2-12^2}=5$$and $$HD=\sqrt{C{D}^2-H{C}^2}=\sqrt{24^2-12^2}=12\sqrt{3}.$$

Let $Y^{\prime }$ be the midpoint of $\overline{CD}$. We look at triangle $BCD$. Since segment $\overline{X{Y}^{\prime }}$ connects midpoints $X$ and $Y^{\prime }$, it is parallel to $\overline{BD}$ and has half the length of $\overline{BD}$. Thus, $$X{Y}^{\prime }=\frac{1}{2}(BH+HD)=\frac{1}{2}(5+12\sqrt{3}).$$Now, we look at triangle $ACD$. Similarly, since $Y$ and $Y^{\prime }$ are midpoints, $\overline{Y{Y}^{\prime }}$ is parallel to $\overline{AC}$ and has half the length of $\overline{AC}$, so $$Y{Y}^{\prime }=24/2=12.$$Since $\overline{BD}\perp \overline{AC}$, we have $\overline{X{Y}^{\prime }}\perp \overline{Y{Y}^{\prime }}$, so $\angle X{Y}^{\prime }Y=90^{\circ }$. Finally, we use the Pythagorean theorem on $△X{Y}^{\prime }Y$ to compute $$\begin{array}{rl}X{Y}^2=Y{Y}^{\prime 2}+X{Y}^{\prime 2}& =12^2+{\left(\frac{1}{2}(5+12\sqrt{3})\right)}^2\\ & =144+\frac{1}{4}(25+120\sqrt{3}+144\cdot 3)\\ & =\boxed{\frac{1033}{4}+30\sqrt{3}}.\end{array}$$