1954 AHSME Problem 26

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:

\begin{array}{l} \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} \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.