$18 \cdot 61$ As shown in the figure, in rectangle $ABCD$, $F$ is the midpoint of side $CD$, and the length of side $BC$ is three times the length of $BE$. Then the area of this rectangle is a multiple of the shaded area.
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] Let $C F=a, B E=b$,
then $C D=2 a, B C=3 b$, and $C E=2 b$.
Then $S_{\text {shaded }}=S_{\triangle B C D}-S_{\triangle E C F}$
$$
\begin{array}{l}
=\frac{1}{2} \cdot 2 a \cdot 3 b-\frac{1}{2} \cdot a \cdot 2 b \\
=2 a b .
\end{array}
$$

Also, $S_{A B C D}=2 a \cdot 3 b=6 a b=3 S_{\text {shaded }}$.
Therefore, the answer is $(B)$.