Three, (20 points) As shown in Figure 2, in trapezoid $ABCD$, $AD$ $/ / BC, AC \perp DB, AC=5, \angle DBC=30^{\circ}$.
(1) Find the length of diagonal $BD$;
(2) Find the area of trapezoid $ABCD$.
(1) As shown in Figure 5, draw $A E / /$ $D B$ intersecting the extension of $C B$ at $E$.
Since $A C \perp D B$,
$A E / / D B$
$\Rightarrow A C \perp A E$,
$$
\angle A E C=\angle D B C=30^{\circ}
$$
$\Rightarrow \angle E A C=90^{\circ}$, which means $\triangle E A C$ is a right triangle
$\Rightarrow E C=2 A C=10$
$\Rightarrow A E=\sqrt{E C^{2}-A C^{2}}=\sqrt{10^{2}-5^{2}}=5 \sqrt{3}$.
Since $A D / / B C$ and $A E / / D B$, quadrilateral $A E B D$ is a parallelogram.
Thus, $D B=A E=5 \sqrt{3}$.
(2) Let the area of trapezoid $A B C D$ be $S$. Draw $A F$
$\perp B C$ at $F$, then $\triangle A F E$ is a right triangle.
Since $\angle A E F=30^{\circ}$, we have,
$$
A F=\frac{1}{2} A E=\frac{5 \sqrt{3}}{2} \text {, }
$$

which means the height of trapezoid $A B C D$ $A F=\frac{5 \sqrt{3}}{2}$.
Since quadrilateral $A E B D$ is a parallelogram, therefore,
$A D=E B$.
Thus, $S=\frac{1}{2}(A D+B C) A F=\frac{1}{2} E C \cdot A F$
$$
=\frac{1}{2} \times 10 \times \frac{5 \sqrt{3}}{2}=\frac{25 \sqrt{3}}{2} .
$$