AHSME 1960 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: $\textbf{(A)}\ \frac{32\sqrt{3}-24}{13} \textbf{(B)}\ \frac{12\sqrt{3}-9}{13} \textbf{(C)}\ 6\sqrt{3}-8 \textbf{(D)}\ \frac{5\sqrt{10}}{6} \textbf{(E)}\ \frac{25}{12}$