smc

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, $△CHA=△CHM$, so $AH=HM$. That means $HM=\frac{1}{4}\cdot AB$, and since $△CHM$ and $△ABC$ share an altitude, $[ABC]=\boxed{4K}$.