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 
 $\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$
Draw diagram as shown (note that $A$ and $B$ can be interchanged, but it doesn’t change the solution).
Note that because $CM$ is a median, $AM = BM$.  Also, by ASA Congruency, $\triangle CHA = \triangle CHM$, so $AH = HM$.  That means $HM = \tfrac{1}{4} \cdot AB$, and since $\triangle CHM$ and $\triangle ABC$ share an altitude, $[ABC] = \boxed{\textbf{(E)}\ 4K}$.