$20 \cdot 70$ As shown, $AB$ is the diameter of $\odot O$, $AB=4$, chord $BC=3$, the angle bisector of $\angle ABC$ intersects the circle at $D$, and the extensions of $AD$ and $BC$ meet at $E$. Then the area of quadrilateral $ABCD$ is the area of $\triangle DCE$ 
(A) 7 times.
(B) 8 times.
(C) 3 times.
(D) 4 times.
(Anhui Province Hefei City Junior High School Mathematics Competition, 1990)
[Solution 1] Connect $D C$. It is easy to prove that $\triangle E D C \backsim \triangle E B A$, with the similarity ratio being $\frac{E D}{E B}$. Also, $\cos B=\frac{3}{4}, \quad \therefore \quad 1-2 \sin ^{2} \frac{B}{2}=\frac{3}{4}$,
thus $\sin ^{2} \frac{B}{2}=\frac{1}{8}$. Also, $\sin \frac{B}{2}=\frac{E D}{E B}$,
$\therefore S_{\triangle E D C}: S_{\triangle E R A}=1: 8$.
Therefore, the area of quadrilateral $A B C D$ is 7 times that of $\triangle D C E$.
Hence, the answer is $(A)$.
[Solution 2] Connect $A C, O D, D C$.
$$
\begin{array}{lc}
\because & B D \text { bisects } \angle A B C, \text { then } \overparen{A D}=\overparen{D C}, \\
\therefore & A D=D C .
\end{array}
$$

Also, $O D=O B$,
then $\angle A O D=\angle O D B+\angle O B D$
$$
\begin{aligned}
= & 2 \angle O B D=\angle A B C, \\
\therefore \quad \cos \angle A O D & =\cos \angle A B C=\frac{B C}{A B}=\frac{3}{4},
\end{aligned}
$$

In $\triangle A D O$, by the cosine rule, we can find $A D^{2}=2$, i.e., $C D^{2}=2$.
$$
\because
$$
$$
\triangle E D C \backsim \triangle E B A,
$$

then $S_{\triangle D C E}: S_{\triangle E A B}=C D^{2}: A B^{2}=1: 8$,
$$
\therefore \quad S_{\triangle D C E}=\frac{1}{7} S_{A B C D} \text {. }
$$

Hence, the answer is $(A)$.