$17 \cdot 141$ In a right triangle $\triangle ABC$ with $\angle C$ being the right angle, the altitude $CH$ and the median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $k$, then the area of $\triangle ABC$ is
(A) $6 k$.
(B) $4 \sqrt{3} k$.
(C) $3 \sqrt{3} k$.
(D) $3 k$.
(E) $4 k$.
(24th American High School Mathematics Examination, 1973)
[ Sol ] $\triangle A B C, \triangle A M C, \triangle M B C$ and $\triangle H B C$ all have the same height $H C$, so their areas are proportional to their bases.
Since $A M=B M$, we have
$$
S_{\triangle A M C}=S_{\triangle M B C} .
$$

Similarly, $S_{\triangle M B C}=S_{\triangle M H C}+S_{\triangle H B C}=k+k=2 k$, noting that $\triangle M H C$ and $\triangle H B C$ are congruent $30^{\circ}-60^{\circ}$ right triangles.
$$
\therefore S_{\triangle A B C}=S_{\triangle A M C}+S_{\triangle M B C}=2 k+2 k=4 k \text {. }
$$

Therefore, the answer is $(E)$.