$17 \cdot 99$ In triangle $ABC$, point $F$ divides $AC$ in the ratio 1:2. Let $E$ be the intersection of side $BC$ and line $AG$, where $G$ is the midpoint of $BF$. Then the ratio in which $E$ divides side $BC$ is 
(A) $1: 4$
(B) $1: 3$.
(C) $2: 5$.
(D) $4: 11$.
(E) $3: 8$.
[Solution] Draw $F H / / A E$. Since $B G=G F$, we have
$$
B E=E H \text {. }
$$

At the same time, from $2 A F=F C$, we get $2 E H=H C$.
$$
\therefore 3 B E=E H+H C=E C \text {. }
$$

Thus, $E$ divides $B C$ in the ratio $1: 3$. Therefore, the answer is $(B)$.