Property 4 Let $I$ be a point inside $\triangle ABC$, $I$ is the incenter of $\triangle ABC$ if and only if: $\angle BIC=90^{\circ}+\frac{1}{2} \angle A, \angle AIC=90^{\circ}+\frac{1}{2} \angle B, \angle AIB=90^{\circ}+\frac{1}{2} \angle C$.
Solution From Property 4, we know
$$
\begin{aligned}
\angle BFE & =90^{\circ}+\frac{1}{2} \angle BDE=90^{\circ}+\frac{1}{4} \angle BDA \\
& =90^{\circ}+\frac{1}{4}\left(90^{\circ}+\frac{1}{2} \angle ACB\right) \\
& =112^{\circ}+\frac{1}{8}\left(4^{\circ}+\angle ACB\right) .
\end{aligned}
$$

Therefore, when $\angle ACB=4^{\circ}$, the minimum degree of $\angle BFE$ is $113^{\circ}$.
Example 1 As shown in Figure $12-5, D$ is the incenter of $\triangle A B C$, $E$ is the incenter of $\triangle A B D$, $F$ is the incenter of $\triangle B D E$. If the degree of $\angle B F E$ is an integer, find the minimum degree of $\angle B F E$.