$17 \cdot 104$ On side $AB$ of $\triangle ABC$, take a point $E$ such that $AE: EB=1: 3$. On another side $BC$, take a point $D$ such that $CD: DB=1: 2$. If $AD$ and $CE$ intersect at point $F$, then $\frac{EF}{FC}+\frac{AF}{FD}$ is
(A) $\frac{4}{5}$.
(B) $\frac{5}{4}$.
(C) $\frac{3}{2}$.
(D) 2.
(E) $\frac{5}{2}$.
(16th American High School Mathematics Examination, 1965)
[Solution] Draw $D H / / A B$ intersecting $E C$ at $G$, and $A C$ at $H$, and let $A E=a$, $D C=b$.
Then $\frac{D G}{3 a}=\frac{b}{3 b}, D G=a=E A$,
so $E F=F G$, and $A F=F D$.
Thus $\frac{A F}{F D}=1$. Also, since $D G / / B E$,
we have $G C=\frac{1}{3} E C, E G=\frac{2}{3} E C$.
Furthermore, since $E F=F G, F C=\frac{2}{3} E C$, hence $\frac{E F}{F C}=\frac{1}{2}$.
Therefore,
$$
\frac{E F}{F C}+\frac{A F}{F D}=\frac{1}{2}+1=\frac{3}{2} \text {. }
$$

Thus, the answer is $(C)$.
Note In $\triangle A B D$ and $\triangle E B C$, by Menelaus' theorem, the result can be directly obtained.