$17 \cdot 174$ On the side $AB$ of $\triangle ABC$, take points $D, E$, and on the side $CA$, take points $F, G$. Connect $BF, FE, EG, GD$. Divide $\triangle ABC$ into five smaller triangles of equal area, that is,
$$
S_{\triangle CBF}=S_{\triangle FBE}=S_{\triangle EFG}=S_{\triangle GED}=S_{\triangle DCA}.
$$

Then the ratio $\frac{FG}{CA}$ equals
(A) $\frac{1}{3}$.
(B) $\frac{4}{15}$.
(C) $\frac{2}{5}$.
(D) $\frac{7}{30}$.
(China Jiangxi Province Nanchang City Junior High School Mathematics Competition, 1990)
[Solution] According to the problem, $S_{\triangle C F B}=\frac{1}{5} S_{\triangle C A B}, \therefore C F=\frac{1}{5} C A$.
Also, $S_{\triangle A G E}=2 S_{\triangle G F F}, \therefore A G=2 G F$.
$\because \quad C A=A G+G F+F C=2 G F+G F+\frac{1}{5} C A$.
$\therefore \frac{4}{5} C A=3 G F$, thus $\frac{F G}{C A}=\frac{4}{15}$. Therefore, the answer is $(B)$.