$-、 \triangle A B C$ has side lengths $a, b, c$ with $b < c$, and $A D$ is the angle bisector of $\angle A$, with point $D$ on $B C$.
(1) Find the necessary and sufficient condition for the existence of points $E, F$ (not endpoints) on segments $A B, A C$ respectively, such that $B E = C F$ and $\angle B D E = \angle C D F$ (expressed using $\angle A, \angle B, \angle C$);
(2) Given the existence of points $E$ and $F$, express the length of $B E$ in terms of $a, b, c$.
（Xiong Bin, problem contributor）
(1) As shown in Figure 1, if there exist points $E$ and $F$ on segments $AB$ and $AC$ respectively, such that $BE = CF$ and $\angle BDE = \angle CDF$, then because the distances from point $D$ to $AB$ and $AC$ are equal, we have
$S_{\triangle BDE} = S_{\triangle CDF}$.
Thus, we have
$BD \cdot DE = DC \cdot DF$.
By the Law of Cosines, we have
$BD^2 + DE^2 - 2BD \cdot DE \cos \angle BDE = BE^2 = CF^2$
$= CD^2 + DF^2 - 2CD \cdot DF \cos \angle CDF$.
Since $\angle BDE = \angle CDF$, from (1) and (2) we can get
$BD^2 + DE^2 = CD^2 + DF^2$,
$BD + DE = CD + DF$.
From (1) and (4), one of the following two cases must hold:
$\left\{\begin{array}{l}BD = CD, \\ DE = DF ;\end{array}\right.$
$\left\{\begin{array}{l}BD = DF, \\ DE = CD .\end{array}\right.$
Since $\angle DAF = \frac{1}{2} \angle A$,

i.e., $2 \angle B > \angle A$.
This is the necessary condition for the existence of points $E$ and $F$.
Conversely, if $2 \angle B > \angle A$, then
$\angle ADC = \angle B + \frac{1}{2} \angle A > \angle A$,
$\angle ADB = \angle C + \frac{1}{2} \angle A > \angle A$.
Thus, we can construct $\angle BDE = \angle A$ and $\angle CDF = \angle A$ inside $\triangle ABC$, with point $E$ on $AB$ and point $F$ on $AC$. This leads to $A, C, D, E$ and $A, B, D, F$ being concyclic.
Therefore, $\angle B = \angle DFC$.
Thus, $\triangle BDE \sim \triangle FDC$.
Since the heights from point $D$ to the bases of these two triangles are equal, we have
$\triangle BDE \cong \triangle FDC$.
Hence, $BE = CF$.
In summary, the necessary and sufficient condition is $2 \angle B > \angle A$.
(2) From the discussion in (1), we know that $A, C, D, E$ and $A, B, D, F$ are concyclic. Therefore,
$$
BE \cdot BA = BD \cdot BC, \quad CF \cdot CA = CD \cdot CB .
$$

Since $BE = CF$, we have
$$
\begin{array}{l}
BE(A B + A C) = B C(B D + D C) = B C^2 . \\
BE = \frac{a^2}{b + c} .
\end{array}
$$