AHSME 1967 Problem 29

$\overline{AB}$ is a diameter of a circle. Tangents $\overline{AD}$ and $\overline{BC}$ are drawn so that $\overline{AC}$ and $\overline{BD}$ intersect in a point on the circle. If $\overline{AD}=a$ and $\overline{BC}=b$, $a \not= b$, the diameter of the circle is: $\textbf{(A)}\ |a-b| \textbf{(B)}\ \frac{1}{2}(a+b) \textbf{(C)}\ \sqrt{ab} \textbf{(D)}\ \frac{ab}{a+b} \textbf{(E)}\ \frac{1}{2}\frac{ab}{a+b}$