$17 \cdot 103$ As shown in the figure, in $\triangle A B C$, $M$ is the midpoint of side $B C$, $A B=12, A C$ $=16 . E$ and $F$ are points on $A C$ and $A B$ respectively, line $E F$ intersects $A M$ at $G$. If $A E$ $=2 A F$, then $\frac{E G}{G F}$ equals
(A) $\frac{3}{2}$.
(B) $\frac{4}{3}$.
(C) $\frac{5}{4}$.
(D) $\frac{6}{5}$.
(E) The given conditions are insufficient to solve this problem.
[Solution 1] As shown in Figure (1), extend $BC$ and $EF$ to intersect at $H$. In $\triangle FBH$ and $\triangle HCE$, by Menelaus' theorem, we respectively get
$$
\frac{GH}{GF} \cdot \frac{AF}{AB} \cdot \frac{MB}{MH}=1, \quad \frac{GH}{GE} \cdot \frac{AE}{AC} \cdot \frac{MC}{MH}=1,
$$

Since $MC=MB, AE=2AF$, dividing the two equations above, we get
$$
\frac{GE}{GF}=2 \frac{AB}{AC}=2 \cdot \frac{12}{16}=\frac{3}{2}.
$$

Therefore, the answer is $(A)$.
[Solution 2] Draw a line $CP / / EF$ intersecting $AB$ at $P$. According to the problem, $AP=8$. Let $a$, $x, y, \alpha, \beta, \delta$ and $\theta$ be as shown in Figure (2). Then,
$$
\frac{a}{\sin \alpha}=\frac{12}{\sin \theta} ; \quad \frac{a}{\sin \beta}=\frac{16}{\sin \left(180^{\circ}-\theta\right)} ; \quad \frac{\sin \beta}{\sin \alpha}=\frac{3}{4}.
$$

So, $\frac{x}{\sin \alpha}=\frac{8}{\sin \delta} ; \quad \frac{y}{\sin \beta}=\frac{16}{\sin \left(180^{\circ}-\delta\right)}$;
and $\quad \frac{EG}{GF}=\frac{y}{x}=2 \frac{\sin \beta}{\sin \alpha}=\frac{3}{2}$.
Therefore, the answer is $(A)$.