20. (12 points) As shown in Figure 3, in $\triangle A B C$, $\sin \frac{\angle A B C}{2} = \frac{\sqrt{3}}{3}$, $A B = 2$, point $D$ is on segment $A C$, and $A D = 2 D C$, $B D = \frac{4 \sqrt{3}}{3}$. Find
(1) the length of $B C$;
(2) the area of $\triangle D B C$.
20. (1) From $\sin \frac{\angle A B C}{2}=\frac{\sqrt{3}}{3}$, we know
$$
\cos B=1-2 \times \frac{1}{3}=\frac{1}{3} \text {. }
$$

In $\triangle A B C$, let $B C=a, A C=3 b$.
By the cosine rule, we have
$$
9 b^{2}=a^{2}+4-\frac{4}{3} a \text {. }
$$

In $\triangle A B D$ and $\triangle D B C$, by the cosine rule and
$$
\begin{array}{l}
\cos \angle A D B=-\cos \angle B D C, \\
\frac{4 b^{2}+\frac{16}{3}-4}{\frac{16 \sqrt{3}}{3} b}=\frac{b^{2}+\frac{16}{3}-a^{2}}{\frac{8 \sqrt{3}}{3} b} \\
\Rightarrow 3 b^{2}-a^{2}=-6 .
\end{array}
$$

From equations (1) and (2), we get $a=3, b=1$, i.e., $B C=3$.
(2) From (1), we have
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \times 2 \times 3 \times \frac{2 \sqrt{2}}{3}=2 \sqrt{2} \\
\Rightarrow S_{\triangle D B C}=\frac{2 \sqrt{2}}{3} .
\end{array}
$$