jmc

Since arc $CD$ is $60^{\circ }$, $\angle CAD=60^{\circ }/2=30^{\circ }$. Since triangle $AOC$ is isosceles with $AO=CO$, $\angle OCA=\angle OAC=30^{\circ }$.



Since $\angle ABO=60^{\circ }$, and this angle is external to triangle $BCO$, $\angle BOC=\angle ABO-\angle BCO=60^{\circ }-30^{\circ }=30^{\circ }$. Hence, triangle $BCO$ is isosceles, and $BC=BO=\boxed{5}$.