13. In the figure, $BC$ is the diameter of the semicircle $\odot O$, $D$ is the midpoint of $\overparen{AC}$, and the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $E$.
(1) Prove: $AC \cdot BC = 2 BD \cdot CD$;
(2) If $AE = 3$, $CD = 2\sqrt{5}$, find the lengths of chord $AB$ and diameter $BC$.
13. (1) Proof 1: As shown in Figure 9, extend $BA$ and $CD$ to intersect at point $G$. Since $BC$ is the diameter of the semicircle $\odot O$, we have
$$
\angle BAC = \angle BDC = 90^{\circ}.
$$

Thus, $\triangle CAG$ and $\triangle BDC$ are both right triangles.

Since $D$ is the midpoint of $\overparen{AC}$, then $\overparen{AD} = \overparen{CD}$. Therefore,
$$
\angle 1 = \angle 2 = \angle 3.
$$

Hence, $\mathrm{Rt} \triangle CAG \sim \mathrm{Rt} \triangle BDC$, and we have
$$
\frac{AC}{BD} = \frac{CG}{BC},
$$

which implies $AC \cdot BC = BD \cdot CG$.
Since $BD \perp CD$ and $\angle 2 = \angle 3$, $\triangle BCG$ is an isosceles triangle. Therefore,
$$
DG = CD = \frac{1}{2} CG.
$$

Thus, $AC \cdot BC = 2BD \cdot CD$.
Proof 2: As shown in Figure 9, connect $OD$, intersecting $AC$ at point $F$.
Since $D$ is the midpoint of $\overparen{AC}$, we have $\overparen{AD} = \overparen{CD}$.
Therefore, $\angle 1 = \angle 2 = \angle 3$, $OD \perp AC$, and
$$
AF = FC = \frac{1}{2} AC.
$$

It is easy to see that $\angle BDC = 90^{\circ}$.
Since $\angle CFD = 90^{\circ}$, we have $\mathrm{Rt} \triangle CDF \sim \mathrm{Rt} \triangle BCD$.
Thus, $\frac{CF}{BD} = \frac{CD}{BC}$.
Hence, $CF \cdot BC = BD \cdot CD$, which means
$$
AC \cdot BC = 2BD \cdot CD.
$$
(2) In $\mathrm{Rt} \triangle CDE$ and $\mathrm{Rt} \triangle CAG$, since $\angle 1$ is a common angle, we have
$\mathrm{Rt} \triangle CDE \sim \mathrm{Rt} \triangle CAG$.
Thus, $\frac{CE}{CG} = \frac{CD}{AC}$, which implies
$$
\frac{CE}{4\sqrt{5}} = \frac{2\sqrt{5}}{CE + 3}.
$$

Solving this, we get $CE = 5$ or $CE = -8$ (discard the negative value).
In $\mathrm{Rt} \triangle ACG$, by the Pythagorean theorem, we have
$$
AG = \sqrt{CG^2 - AC^2} = \sqrt{(4\sqrt{5})^2 - (3 + 5)^2} = 4.
$$

By the secant theorem, we have $GA \cdot GB = GD \cdot GC$, which implies
$$
4(AB + 4) = 2\sqrt{5} \times 4\sqrt{5}.
$$

Solving this, we get $AB = 6$.
In $\mathrm{Rt} \triangle ABC$, by the Pythagorean theorem, we have
$$
BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + (3 + 5)^2} = 10.
$$