1960 AHSME Problem 40

Given right \triangle ABC with legs BC=3, AC=4. Find the length of the shorter angle trisector from C to the hypotenuse:

\textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad

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