$17 \cdot 137$ As shown in the figure, $S_{\triangle A B C}=10$, and $A D=2, B D=3$, and $S_{\triangle A B E}=S_{\text {quadrilateral } D B E F}$, then $S_{\triangle A B E}$ equals
(A) 5.
(B) 6.
(C) 7.
(D) 8.
(Singapore Mathematical Olympiad, 1988)
［Solution］Connect $D E$, as shown in the figure, it is clear that
$$
\begin{array}{l}
S_{\triangle A B E}=S_{\triangle A D E}+S_{\triangle B D E}, \\
\text { Also, } S_{\text {quadrilateral } D B E F}=S_{\triangle D E F}+S_{\triangle B D E}, \\
\therefore S_{\triangle A D E}=S_{\triangle D E F}, \text { hence } A F \parallel D E. \\
\therefore \quad B E: B C=B D: A B=3: 5, \\
\text { Therefore, } \quad S_{\triangle A B E}=\frac{B E}{B C} \quad S_{\triangle A B C}=\frac{3}{5} \cdot 10=6. \\
\end{array}
$$

Thus, the answer is $(B)$.
Note The original problem was a fill-in-the-blank question, here it is changed to a multiple-choice question.