II. (25 points) As shown in Figure 2, $BC$ is the diameter of semicircle $O$, and $D$ is the midpoint of $\overparen{AC}$. The diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $E$.
(1) Prove that $AC \cdot BC = 2 BD \cdot CD$;
(2) If $AE = 3$ and $CD = 2\sqrt{5}$, find the lengths of chord $AB$ and diameter $BC$.
(1) As shown in Figure 8, connect $O D$, intersecting $A C$ at point $F$.
Since $D$ is the midpoint of $\overparen{A C}$, we have
$$
\begin{array}{l}
\overparen{A D}=\overparen{C D}, \\
\angle 1=\angle 2=\angle 3, \\
O D \perp A C, \\
A F=F C=\frac{1}{2} A C .
\end{array}
$$

Also, $B C$ is the diameter of the semicircle $O$, so,
$$
\angle B D C=\angle B A C=90^{\circ}, \angle C F D=\angle B D C=90^{\circ} \text {. }
$$

Thus, $\triangle C D F \backsim \triangle B C D$.
Therefore, $\frac{C F}{B D}=\frac{C D}{B C}$.
Hence, $C F \cdot B C=B D \cdot C D$, which means
$$
A C \cdot B C=2 B D \cdot C D \text {. }
$$
(2) As shown in Figure 8, extend $B A$ and $C D$ to intersect at point $G$. From (1), we know $\angle B D C=\angle B A C=90^{\circ}, \angle 2=\angle 3$,
so, $\triangle B C G$ is an isosceles triangle.
Thus, $A D=D G=C D=2 \sqrt{5}$.
Also, $\triangle C D E \backsim \triangle C A G$,
therefore, $\frac{C E}{C G}=\frac{C D}{A C}$, which means $\frac{C E}{4 \sqrt{5}}=\frac{2 \sqrt{5}}{C E+3}$.
Solving this, we get $C E=5$ or -8 (discard -8).
In $\mathrm{Rt} \triangle A C G$, by the Pythagorean theorem,
$$
A G=\sqrt{C G^{2}-A C^{2}}=\sqrt{(4 \sqrt{5})^{2}-(3+5)^{2}}=4 \text {. }
$$

By the secant theorem, $A G \cdot B G=C G \cdot D G$, which means
$$
4(A B+4)=(2 \sqrt{5}+2 \sqrt{5}) \times 2 \sqrt{5} \text {. }
$$

Solving this, we get $A B=6$.
In $\mathrm{Rt} \triangle A B C$, by the Pythagorean theorem,
$$
B C=\sqrt{A B^{2}+A C^{2}}=\sqrt{6^{2}+(3+5)^{2}}=10 .
$$