2009 AMC 10B Problem 16

Tangent lines always come with right angles, so draw the radius to a point of tangency and look for a right triangle.

1. Draw the radius $OA$ and call its length $r$. A radius drawn to the point where a tangent touches is perpendicular to the tangent, so $\angle OAB = 90^\circ$.
2. The picture is symmetric across the line $BO$: the tangent segments $BA$ and $BC$ are mirror images. So $BO$ splits $\angle ABC$ into two equal halves.
3. Triangle $ABC$ is equilateral, so $\angle ABC = 60^\circ$ and each half is $\angle ABO = 30^\circ$.
4. Now triangle $OAB$ has angles $90^\circ$ at $A$ and $30^\circ$ at $B$, so it is a $30$-$60$-$90$ triangle. In such a triangle, the side across from the $30^\circ$ angle is half the longest side. The side across from $B$ is $OA = r$, so the longest side is $BO = 2r$.
5. The point $D$ is where the circle crosses $BO$, so $OD$ is a radius: $OD = r$. That leaves $BD = BO - OD = 2r - r = r$.
6. So
$$\frac{BD}{BO} = \frac{r}{2r} = \frac12.$$

The answer is $\boxed{\textbf{(B)}\ \frac12}$.

Two facts about tangents do all the work. The radius to the point of tangency makes a right angle with the tangent, and the two tangents from one outside point are mirror images across the line to the center. Together they give a right triangle with a known angle, and then $D$ is found just by subtracting one radius.

Draw the radius $OA$. Because $BA$ is tangent at $A$, $OA \perp BA$, so triangle $OAB$ has a right angle at $A$.

The two tangent segments from $B$ are symmetric about the line $BO$, so $BO$ bisects $\angle ABC = 60^\circ$ and $\angle ABO = 30^\circ$. Triangle $OAB$ is therefore a $30$-$60$-$90$ triangle with the short leg $OA = r$ opposite the $30^\circ$ angle, which makes the hypotenuse $BO = 2r$.

Point $D$ lies on the circle and on segment $BO$, so $OD = r$ and $BD = BO - OD = 2r - r = r$. Hence
$$\frac{BD}{BO} = \frac{r}{2r} = \frac12.$$

The answer is $\boxed{\textbf{(B)}\ \frac12}$.