18・118 As shown in the figure, in trapezoid $ABCD$, $AB // CD$, the ratio of the area of $\triangle DCE$ to the area of $\triangle DCB$ is $1: 3$. Then, the ratio of the area of $\triangle DEC$ to the area of $\triangle ABD$ is
(A) $1: 5$.
(B) $1: 6$.
(C) $1: 7$.
(D) $1: 9$.
[Solution 1] Given $S_{\triangle D C E}: S_{\triangle D C B}=1: 3$, by the property of the ratio of areas of triangles with equal (or same) heights, we know: $D E: D B=1: 3$, or $D E: E B=1: 2$.
By $\triangle D E C \backsim \triangle B E A$, we know $C E: A E=1: 2$.
Thus, $S_{\triangle D E C}: S_{\triangle A B D}=1: 6$.
Therefore, the answer is $(B)$.
[Solution 2] Let $S_{\triangle A B E}=n, S_{\triangle C D E}=m$,

Then
$$
\begin{array}{r}
S_{\triangle A D E}=S_{\triangle B C E}=\sqrt{m n}, \\
\therefore \quad \frac{S_{\triangle C D E}}{S_{\triangle B C D}}=\frac{m}{m+\sqrt{m n}}=\frac{1}{3},
\end{array}
$$

This implies $\sqrt{m n}=2 m$ or $n=4 m$,
$$
\therefore \quad \frac{S_{\triangle C D E}}{S_{\triangle A B D}}=\frac{m}{n+\sqrt{m n}}=\frac{m}{4 m+2 m}=\frac{1}{6} \text {. }
$$

Therefore, the answer is (B).