Points $A$ and $C$ lie on a circle centered at $O$, and $\overline{BA}$, $\overline{BC}$ are tangent to the circle, with $\angle ABC=120^\circ$. The circle intersects $\overline{BO}$ at $D$. What is $\dfrac{BD}{BO}$?

1. Since $\overline{BA},\overline{BC}$ are tangent to the circle, $BA=BC$ (equal tangent lengths), so $\overline{BO}$ bisects $\angle ABC$, $\angle OBA=60^\circ$.
2. In right triangle $OAB$ (right angle at $A$, since $OA\perp BA$), $\sin\angle OBA=\dfrac{OA}{BO}$. With $OA=1$: $BO=\dfrac1{\sin60^\circ}=\dfrac2{\sqrt3}=\dfrac{2\sqrt3}3$.
3. $D$ is where the circle meets $\overline{BO}$, so $OD=1$, and $BD=BO-OD=\dfrac{2\sqrt3}3-1$.
4. $\dfrac{BD}{BO}=1-\dfrac{OD}{BO}=1-\dfrac1{2\sqrt3/3}=1-\dfrac{\sqrt3}2=\boxed{\dfrac{2-\sqrt3}2}$.