[https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles](https://artofproblemsolving.comhttps://artofproblemsolving.com/community/c4h256249s1_lines_and_circles)
The straight line $\overline{AB}$ is divided at $C$ so that $AC=3CB$.  Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $AB$ produced at $D$.  Then $BD$ equals:
$\textbf{(A)}\ \text{diameter of the smaller circle} \\ \textbf{(B)}\ \text{radius of the smaller circle} \\ \textbf{(C)}\ \text{radius of the larger circle} \\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii}$
Let $x=\overline{BD}$ and let $r$ be the radius of the small circle. Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. By similar triangles, $\frac{x+r}{r}=\frac{x+5r}{3r} \implies x=r$, or $\boxed{\textbf{(B)}}$.
==Video Solution
[https://youtu.be/cbU3DxJKB8U](https://artofproblemsolving.comhttps://youtu.be/cbU3DxJKB8U)
~Lucas