Three, (Full marks 16 points) In the cyclic quadrilateral $ABCD$, $\angle A = 60^{\circ}, \angle B = 90^{\circ}, AD = 3, CD = 2$.
(1) Find the length of $BC$;
(2) Find the area of quadrilateral $ABCD$;
(3) Compare $AB + BC$ with $AD + CD$.

保留源文本的换行和格式，直接输出翻译结果。
Three 1$) \because \angle B$ is a right angle, we know $AC$ is the diameter,
$$
\therefore \angle D=90^{\circ} \text {. }
$$

As shown in Figure 5, extend $AB$ and
$DC$ to intersect at $E$, with $\angle E=$ $30^{\circ}$, then
$$
\begin{array}{l}
AE=2AD=6, \\
DE=3\sqrt{3}, \\
EC=3\sqrt{3}-2 .
\end{array}
$$

Thus, $BC=\frac{1}{2}EC=\frac{3}{2}\sqrt{3}-1$.
(2) $BE=BC \cdot \sqrt{3}=\frac{9}{2}-\sqrt{3}$.

Then $S_{\text{quadrilateral } AIXC}=S_{\triangle ADE}-S_{\triangle HXC}=\frac{5}{8}\sqrt{3}+\frac{9}{2}$.
(3) $\because AB=AE-BE=\sqrt{3}+\frac{3}{2}$,
$$
\begin{aligned}
\therefore AB+BC & =\frac{5}{2}\sqrt{3}+\frac{1}{2}<\frac{5}{2} \times 1.8+\frac{1}{2} \\
& =5=AD+CD .
\end{aligned}
$$