$17 \cdot 107$ In $\triangle A B C$, $D$ and $E$ are points on sides $B C$ and $A C$ respectively, and $\frac{B D}{D C}=\frac{2}{3}, \frac{A E}{E C}=\frac{3}{4}$. Then the value of $\frac{A F}{F D} \cdot \frac{B F}{F E}$ is
(A) $\frac{7}{3}$.
(B) $\frac{14}{9}$.
(C) $\frac{35}{12}$.
(D) $\frac{56}{15}$.
(Chinese Junior High School Mathematics Competition, Taiyuan City, Shanxi Province, 1997)
[Solution] Draw $E G / / B C$ intersecting $A D$ at $G$. By the given conditions,
$$
\frac{G E}{B D}=\frac{G E / D C}{B D / D C}=\frac{A E / A C}{B D / D C}=\frac{3 / 7}{2 / 3}=\frac{9}{14} \text {. }
$$

Thus, $\frac{F G}{D F}=\frac{F E}{B F}=\frac{G E}{B D}=\frac{9}{14}$,
By the Componendo Theorem, we have
$$
\frac{D G}{D F}=\frac{23}{14},
$$

Also, $\frac{A G}{A D}=\frac{3}{7}$, so $\frac{G D}{A D}=\frac{4}{7}$, which means
$$
G D=\frac{4}{7} A D .
$$

Therefore,
$$
\frac{A D}{D F}=\frac{\frac{7}{4} G D}{D F}=\frac{7}{4} \cdot \frac{23}{14}=\frac{23}{8} \text {. }
$$

By the Dividendo Theorem, we have
$$
\frac{A F}{D F}=\frac{15}{8} \text {. }
$$

Thus,
$$
\frac{A F}{F D} \cdot \frac{B F}{F E}=\frac{15}{8} \cdot \frac{14}{9}=\frac{35}{12} \text {. }
$$

Therefore, the answer is $(C)$.