$17 \cdot 149$ As shown in the figure, in the obtuse triangle $A B C$, $A M=M B, M D \perp B C, E C$ $\perp B C$. If the area of $\triangle A B C$ is 24, then the area of $\triangle B E D$ is
(A) 9.
(B) 12.
(C) 15.
(D) 18.
(E) cannot be uniquely determined.
(35th American High School Mathematics Examination, 1984)
[Solution] Connect $M C$. In $\triangle D M C$ and $\triangle D M E$, they share the common base $M D$, and the heights on $M D$ are equal, so these two triangles have equal areas.
And
$$
\begin{array}{l}
S_{\triangle B M C}=S_{\triangle B M D}+S_{\triangle D M C}, \\
S_{\triangle B E D}=S_{\triangle B M D}+S_{\triangle D M E} .
\end{array}
$$

Therefore, $S_{\triangle B M C}=S_{\triangle B E D}$.
There are two methods to prove that
$$
S_{\triangle B M C}=\frac{1}{2} S_{\triangle B A C} .
$$

Thus, $S_{\triangle B E D}=S_{\triangle B M C}=\frac{1}{2} S_{\triangle B A C}=12$.
First method: Draw the altitude from $A$ to $B C$, intersecting the extension of $B C$ at $F$, so $M D / / A F$.
Since $M$ is the midpoint of $B A$, therefore $M D=\frac{1}{2} A F$.
Furthermore, $\triangle B M C$ and $\triangle B A C$ share the common base $B C$, so
$$
S_{\triangle B M C}=\frac{1}{2} S_{\triangle B A C} \text {. }
$$

Second method: $\quad S_{\triangle B M C}=\frac{1}{2} B M \cdot B C \sin B=\frac{1}{2}\left[\frac{1}{2} B A \cdot B C \sin B\right]$
$$
=\frac{1}{2} S_{\triangle B A C} \text {. }
$$

Therefore, the answer is $(B)$.