Example 6 As shown in Figure 5, in $\triangle A B C$, side $B C=a=14$, height $h_{a}$ $=12$, inradius $r$ $=4$. Find $b, c$.
Given
$B$
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \times 14 \times 12 \\
=p \cdot 4=\frac{1}{2} b c \sin \dot{A},
\end{array}
$$

we get
$$
p=21, b+c=2 p-a=28 \text {. }
$$

Notice that
$$
\tan \frac{A}{2}=\frac{O N}{A N}=\frac{r}{p-a}=\frac{4}{7} ,
$$

thus, $\sin A=\frac{2 \tan \frac{\mathrm{A}}{2}}{1+\tan ^{2} \frac{A}{2}}=\frac{56}{65}$.
Therefore, $b c=\frac{14 \times 12}{\sin A}=195$.
Combining this with $b+c=28$, we find that $b$ and $c$ are 13 and 15, or 15 and 13.