smc

Since $\angle CAD$ is an angle inscribed in a $60{}^{\circ }$ arc, we obtain $\angle CAD=\frac{60^{\circ }}{2}=30^{\circ }$, so $△ABO$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ right triangle. This gives $AO=BO\sqrt{3}=5\sqrt{3}$ and $AB=2BO=10$, and now since $AD$ is a diameter, $AD=2AO=10\sqrt{3}$. The fact that $AD$ is a diameter also means that $\angle ACD=90^{\circ }$, so $△ACD$ is again a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ right triangle, yielding $CD=\frac{1}{2}AD=5\sqrt{3}$ and $AC=\frac{\sqrt{3}}{2}AD=15$, which finally gives $BC=AC-AB=15-10=\boxed{5}$.