Points $A$ and $C$ lie on a circle centered at $O$, each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$. What is $\frac{BD}{BO}$?
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$
Solution 1

As $\triangle ABC$ 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 $\triangle ABO$ we also have $AO = BO \sin 30^\circ = \frac{BO}2$, Hence $DO = \frac{BO}2$, therefore $BD = BO - DO = \frac{BO}2$, and $\frac{BD}{BO} = \boxed{\frac 12 \Longrightarrow B}$.

Solution 2

As in the previous solution, we find out that $\angle AOB = \angle COB = 60^\circ$. Hence $\triangle AOD$ and $\triangle COD$ are both equilateral.
We then have $\angle SCD = \angle SAD = 30^\circ$, hence $D$ is the incenter of $\triangle ABC$, and as $\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \cdot SD = BD$, and as $SD = SO$, we have $2\cdot SD = SD + SO = OD$, therefore $BD=OD$, and as before we conclude that $\frac{BD}{BO} = \boxed{\frac 12}$.