In a circle with center $ O$, $ AD$ is a diameter, $ ABC$ is a chord, $ BO \equal{} 5$, and $ \angle ABO = \stackrel{\frown}{CD} = 60^{\circ}$.  Then the length of $ BC$ is:

[asy]size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C);
draw(C--A--D^^B--O^^Circle(O,1));
pair point=O;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$O$", O, dir(305));

label("$5$", B--O, dir(O--B)*dir(90));
label("$60^\circ$", dir(185), dir(185));
label("$60^\circ$", B+0.05*dir(-25), dir(-25));[/asy]

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 3 \plus{} \sqrt3 \qquad \textbf{(C)}\ 5 \minus{} \frac{\sqrt3}{2} \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of the above}$
1. Given that \(AD\) is a diameter of the circle with center \(O\), and \(BO = 5\). Also, \(\angle ABO = 60^\circ\) and \(\stackrel{\frown}{CD} = 60^\circ\).

2. Since \(\stackrel{\frown}{CD} = 60^\circ\), the central angle \(\angle COD = 60^\circ\). Because \(AD\) is a diameter, \(\angle CAD = 30^\circ\) (since the angle subtended by an arc at the center is twice the angle subtended at the circumference).

3. \(\angle ABO = 60^\circ\) implies that \(\angle BOA = 90^\circ\) (since \(\angle BOA = 180^\circ - \angle ABO - \angle OAB\) and \(\angle OAB = 30^\circ\)).

4. In \(\triangle BOA\), which is a \(30^\circ-60^\circ-90^\circ\) triangle:
   - \(BO = 5\)
   - \(AO = 5\sqrt{3}\) (since \(AO = BO \cdot \sqrt{3}\))
   - \(BA = 10\) (since \(BA = 2 \cdot BO\))

5. Since \(AO\) is a radius of the circle, \(AD = 2 \cdot AO = 10\sqrt{3}\).

6. Draw a segment to connect \(C\) and \(D\) to form \(\triangle ACD\). Since \(AD\) is a diameter, \(\angle ACD = 90^\circ\) (by the Inscribed Angle Theorem).

7. \(\triangle ACD\) is a \(30^\circ-60^\circ-90^\circ\) triangle:
   - \(AC = 15\) (since \(AC = AD \cdot \frac{1}{2} = 10\sqrt{3} \cdot \frac{1}{2} = 5\sqrt{3}\))

8. Since \(AB = 10\) and \(AC = AB + BC\), we have:
   \[
   AC = AB + BC \implies 15 = 10 + BC \implies BC = 5
   \]

Conclusion:
\[
\boxed{5}
\]