In a circle with center $O$, $AD$ is a [diameter](https://artofproblemsolving.com/wiki/index.php/Diameter), $ABC$ is a [chord](https://artofproblemsolving.com/wiki/index.php/Chord), $BO=5$ and $\angle ABO= \ \stackrel{\frown}{CD} \ =60^\circ$. Then the length of $BC$ is

$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{3} \qquad \mathrm{(C) \  } 5-\frac{\sqrt{3}}{2} \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of the above}$
Since $\angle CAD$ is inscribed and intersects an arc of length $60^\circ$, $\angle CAD=30^\circ$. Thus, $\triangle ABO$ is a $30-60-90$ right triangle. Thus, $AO=BO\sqrt{3}=5\sqrt{3}$ and $AB=2BO=10$. Since $AO$ and $DO$ are both radii, $DO=AO=5\sqrt{3}$ and $AD=10\sqrt{3}$. Since $\angle ACD$ is inscribed in a semicircle, it's a right angle, and $\triangle ACD$ is also a $30-60-90$ right triangle. Thus, $CD=\frac{AD}{2}=5\sqrt{3}$ and $AC=CD\sqrt{3}=15$. Finally, $BC=AC-AB=15-10=5, \boxed{\text{D}}$.