$18 \cdot 24$ In triangle $ABC$, $\angle CBA=72^{\circ}$, $E$ is the midpoint of side $AC$, $D$ is on $BC$ such that $2BD=DC$, and $AD$ intersects $BE$ at $F$. Then the ratio of the area of $\triangle BDF$ to the area of quadrilateral $FDEC$ is
(A) $\frac{1}{5}$.
(B) $\frac{1}{4}$.
(C) $\frac{1}{3}$.
(D) $\frac{2}{5}$.
(E) None of these.
(31st American High School Mathematics Examination, 1980)
[Solution] As shown in the figure, draw a line from $E$ to the midpoint $G$ of $DC$.
Since $EG$ is the line connecting the midpoints of sides $AC$ and $DC$ in $\triangle ACD$, $EG \parallel AD$, then
$$
S_{\triangle EBG}=\frac{2}{3} S_{\triangle EBC},
$$

and $S_{\triangle BDF}=\frac{1}{4} S_{\triangle EBG}=\frac{1}{6} S_{\triangle EBC}$.
Therefore, $S_{FDC E}=\frac{5}{6} S_{\triangle EBC}$.
Thus, $\frac{S_{\triangle BDF}}{S_{FDC E}}=\frac{1}{5}$.
Hence, the answer is $(A)$.