imc

As $△ABC$ is equilateral, we have $\angle BAC=\angle BCA=60^{\circ }$, hence $\angle OAC=\angle OCA=30^{\circ }$. Then $\angle AOC=120^{\circ }$, and from symmetry we have $\angle AOB=\angle COB=60^{\circ }$. Thus, this gives us $\angle ABO=\angle CBO=30^{\circ }$. We know that $DO=AO$, as $D$ lies on the circle. From $△ABO$ we also have $AO=BO\sin 30^{\circ }=\frac{BO}{2}$, Hence $DO=\frac{BO}{2}$, therefore $BD=BO-DO=\frac{BO}{2}$, and $\frac{BD}{BO}=\boxed{\frac{1}{2}}$.