$17 \cdot 164$ In $\triangle A B C$, $A C: A B=1: 2 . \angle A$'s internal and external angle bisectors are $A E$ and $A F$, respectively, then the area $S_{\triangle A B C}$ : $S_{\triangle A B E}: S_{\triangle A B F}$ equals
(A) $3: 2: 4$.
(B) $3: 2: 6$.
(C) $3: 1: 4$.
(D) $2: 1: 4$.
[Solution] Given that $A E$ is the angle bisector of $\angle A$, and $A C: A B=1: 2$, then similarly
$$
\begin{array}{l}
C E: B E=1: 2 . \\
F C: F B=1: 2 .
\end{array}
$$

Let $S_{\triangle A B C}=S$, then $S_{\triangle A B E}=\frac{2}{3} S, S_{\triangle A B F}=2 S$.
$$
S_{\triangle A B C}: S_{\triangle A B E}: S_{\triangle A B F}=S: \frac{2}{3} S: 2 S=3: 2: 6 \text {. }
$$

Therefore, the answer is $(B)$.