Example 9 As shown in Figure 8, let $I$ be the incenter of $\triangle ABC$, $B_{1}$ and $C_{1}$ be the midpoints of sides $AB$ and $AC$ respectively, the extension of $C_{1}I$ intersects $AB$ at point $B_{2}$, and the extension of $B_{1}I$ intersects the extension of $AC$ at point $C_{2}$. Given that $S_{\triangle AB_{2}C_{2}} = S_{\triangle ABC}$, find $\angle A$.
Solution: As shown in Figure 8, connect $A I$. Let $A C=b, A B=c$, $B C=a, A B_{2}=c^{\prime}, A C_{2}=b^{\prime}$, and the inradius of the incircle be $r$.
From $\frac{b c}{2} \sin A=S_{\triangle A B C}=S_{\triangle A B_{2} C_{2}}=\frac{b^{\prime} c^{\prime}}{2} \sin A$, we get $b c=b^{\prime} c^{\prime}$.
Notice that
$$
\begin{array}{l}
S_{\triangle B_{1} C_{2} A}=\frac{A B_{1} \cdot A C_{2}}{A B \cdot A C} S_{\triangle A B C} \\
=\frac{b^{\prime}}{2 b} \cdot \frac{r}{2}(a+b+c) \\
=S_{\triangle A R C_{2}}+S_{\triangle A B B_{1}}=\frac{r}{2} A C_{2}+\frac{r}{2} A B_{1} \\
=\frac{r}{2} \cdot b^{\prime}+\frac{r}{2} \cdot \frac{c}{2} .
\end{array}
$$

From equations (1) and (2), we get
$$
b^{\prime}+\frac{c}{2}=(a+b+c) \frac{b^{\prime}}{2 b},
$$

which simplifies to $b c=(a-b+c) b^{\prime}$.
For $\triangle A C_{1} B_{2}$, similarly, we get
$$
b c=(a+b-c) c^{\prime} \text {. }
$$
Multiplying (3) and (4), and using $b c=b^{\prime} c^{\prime}$, we get
$$
b c=a^{2}-b^{2}-c^{2}+2 a b,
$$

which simplifies to $a^{2}=b^{2}+c^{2}-b c=b^{2}+c^{2}-2 b c \cos A$.
Therefore, $\cos A=\frac{1}{2}, \angle A=60^{\circ}$.
The area method is a widely used and important method in geometry, often serving as a transitional step in the solution process. This idea should be carefully appreciated.