Connect $AC, BD$. In the right-angled triangle, by the Pythagorean theorem, we get
$$BD=\sqrt{AB^{2}-AD^{2}}=\sqrt{8^{2}-2^{2}}=2\sqrt{15}$$

By the acute angle trigonometric functions, we get
$$\sin \angle DAB=\frac{BD}{AB}=\frac{2\sqrt{15}}{8}=\frac{\sqrt{15}}{4}$$

In the right-angled $\triangle ABC$, by the acute angle trigonometric functions, we get
$$\sin \angle ABC=\frac{AC}{AB}=\frac{AC}{8}$$

Since $S_{\text{quadrilateral } ABCD}=S_{\triangle ABD}+S_{\triangle BCD}$
$$=\frac{1}{2} AD \cdot AB \cdot \sin \angle DAB+\frac{1}{2} CD$$
- $CB \cdot \sin \angle DCB$
$=\frac{1}{2} \cdot 2 \cdot 8 \cdot \sin \angle DAB+\frac{1}{2} \cdot 2 \cdot CB \cdot \sin \angle DAB$
$=(8+BC) \sin \angle DAB=\frac{\sqrt{15}(8+BC)}{4}$.
Similarly, $S_{\text{quadrilateral } ABCD}=S_{\triangle ACD}+S_{\triangle ABC}=\frac{AC(2BC+1)}{4}$.
Therefore, $\frac{\sqrt{15}(8+BC)}{4}=\frac{AC(2BC+1)}{4}$.
In the right-angled $\triangle ABC$, by the Pythagorean theorem, we get
$AC^{2}+BC^{2}=AB^{2}$.
That is, $\left[\frac{\sqrt{15}(8+BC)}{2BC+1}\right]^{2}+BC^{2}=8^{2}$,
Simplifying and organizing, we get $BC^{4}+BC^{3}-60BC^{2}+224=0$.
Solving this higher-degree equation, we get
$$BC=7, BC=-8, BC= \pm 2$$

By the problem's conditions, only $BC=7$ is valid.
Therefore, $BC=7(\mathrm{~cm})$.
The solution above is a bit complicated. In fact, the solution can be simplified:
In Figure 3, connect $OC, OD$.
Let $\angle COD=\angle DOA=\theta$, in $\triangle AOD$, $|OA|=|OD|=4,|AD|=2$, by the cosine theorem, we get
$\cos \theta=\frac{|OA|^{2}+|OD|^{2}-|AD|^{2}}{2|OA| \cdot|OD|}=\frac{7}{8}$,
Thus, $\cos 2\theta=2\cos^{2}\theta-1=\frac{17}{32}$.
In $\triangle ABC$, $\angle BOC=\pi-2\theta,|OB|=|OC|$ $=4$, by the cosine theorem, we get
$$|BC|^{2}=|OB|^{2}+|OC|^{2}-2|OB| \cdot|OC|$$
- $\cos (\pi-2\theta)$
$$\begin{array}{l}
=4^{2}+4^{2}+2 \cdot 4 \cdot 4 \cdot \frac{17}{32}=49 \\
BC=7(\mathrm{~cm})
\end{array}$$
Question 3 As shown in Figure 3, given the diameter $A B=8 \mathrm{~cm}$ of circle $O$, chord $A D=C D$ $=2 \mathrm{~cm}$, find the length of $B C$.