20. In $\triangle A B C$, $\sin \frac{\angle A B C}{2}=\frac{\sqrt{3}}{3}, A B=2, D$ is a point on line segment $A C$, and $A D=2 D C, B D=\frac{4 \sqrt{3}}{3}$.
(1) Find the length of $B C$;
(2) Find the area of $\triangle D B C$.
$$
\begin{array}{l}
\text { (1) } \cos \angle A B C=1-2 \sin ^{2} \frac{\angle A B C}{2}=\frac{1}{3}, \\
\overrightarrow{A D}=\frac{2}{3} \overrightarrow{A C} \Rightarrow \overrightarrow{B D}=\frac{2}{3} \overrightarrow{B C}+\frac{1}{3} \overrightarrow{B A} \text {, let } B C=x, \\
\text { then } \frac{16}{3}=\frac{1}{9}\left(4 x^{2}+4+2 \cdot 2 \cdot 2 x \cdot \frac{1}{3}\right) \\
\Rightarrow 3 x^{2}+2 x-33=(3 x+11)(x-3)=0 \Rightarrow x=3,
\end{array}
$$

So $B C=3$.
(2) From (1), we know $\sin \angle A B C=\frac{2 \sqrt{2}}{3} \Rightarrow S_{\triangle A B C}=\frac{1}{2} \cdot 2 \cdot 3 \cdot \frac{2 \sqrt{2}}{3}=2 \sqrt{2}$, so $S_{\triangle D B C}=\frac{1}{3} S_{\triangle A B C}=\frac{2 \sqrt{2}}{3}$.