9. 10 (CMO 17) In $\triangle ABC$, the lengths of the three sides are $a$, $b$, and $c$, with $b < c$, and $AD$ is the angle bisector of $\angle A$, with point $D$ on $BC$. (1) Find the necessary and sufficient condition for the existence of points $E$ and $F$ (not endpoints) on segments $AB$ and $AC$ respectively, such that $BE = CF$ and $\angle BDE = \angle CDF$ (expressed in terms of $\angle A$, $\angle B$, and $\angle C$);
(2) When points $E$ and $F$ exist, express the length of $BE$ in terms of $a$, $b$, and $c$.
9.10 (1) First find the necessary condition, then prove its sufficiency. If there exist points $E$ and $F$, then $S_{BDE} = S_{CDF}$, thus $BD \cdot DE = CD \cdot DF$. Combining the application of the cosine theorem to $\triangle BDE$, $\triangle CDF$, and $\angle BDE = \angle CDF$, we can derive $BD^2 + DE^2 = CD^2 + DF^2$, and $BD + DE = CD + DF$. After discussion, it is known that only $BD = DF, DE = CD$ holds.

Thus, $\triangle DEB \cong \triangle DCF, \angle B = \angle DFC, \angle C = \angle BED, \angle DFC > \angle DAF$, which implies $2 \angle B > \angle A$, which is the necessary condition.

Conversely, if $2 \angle B > \angle A$, then $\angle ADC > \angle A, \angle ADB > \angle A$. At this time, using the construction method, we can construct $\angle BDE = \angle A, \angle CDF = \angle A$ inside $\triangle ABC$, obtaining points $E$ and $F$ on $AB$ and $AC$, respectively, and thus prove that $BE = CF$, i.e., there exist points $E$ and $F$.

(2) From (1), we know that $A, C, D, E$ and $A, B, D, F$ are each cyclic quadrilaterals, $BE \cdot BA = BD \cdot BC, CF \cdot CA = CD \cdot CB$. Therefore, we can calculate $BE = \frac{a^2}{b+c}$.