AHSME 1973 Problem 9

In $\triangle ABC$ with right angle at $C$, altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$, then the area of $\triangle ABC$ is $\textbf{(A)}\ 6K \textbf{(B)}\ 4\sqrt3\ K \textbf{(C)}\ 3\sqrt3\ K \textbf{(D)}\ 3K \textbf{(E)}\ 4K$