the edges of triangle $ABC$ are $a,b,c$ respectively,$b<c$,$AD$ is the bisector of $\angle A$,point $D$ is on segment $BC$.
(1)find the property $\angle A$,$\angle B$,$\angle C$ have,so that there exists point $E,F$ on $AB,AC$ satisfy $BE=CF$,and $\angle NDE=\angle CDF$
(2)when such $E,F$ exist,express $BE$ with $a,b,c$
1. **Given:**
   - Triangle \(ABC\) with sides \(a, b, c\) opposite to vertices \(A, B, C\) respectively.
   - \(b < c\).
   - \(AD\) is the angle bisector of \(\angle A\), and \(D\) is on segment \(BC\).
   - Points \(E\) and \(F\) on \(AB\) and \(AC\) respectively such that \(BE = CF\) and \(\angle NDE = \angle CDF\).

2. **To find:**
   - The property of \(\angle A\), \(\angle B\), \(\angle C\) such that points \(E\) and \(F\) exist.
   - Express \(BE\) in terms of \(a, b, c\) when such \(E\) and \(F\) exist.

3. **Using the Law of Sines in triangles \(CYZ\) and \(CXB\):**
   \[
   \frac{CF}{\sin x} = \frac{CZ}{\sin(C + x)}
   \]
   \[
   \frac{BE}{\sin x} = \frac{BZ}{\sin(B + x)}
   \]
   Since \(CF = BE\), we have:
   \[
   \frac{CZ}{BZ} = \frac{\sin(C + x)}{\sin(B + x)}
   \]

4. **Using the Angle Bisector Theorem:**
   Since \(AZ\) is the angle bisector of \(\angle BAC\), we have:
   \[
   \frac{CZ}{BZ} = \frac{AC}{AB} = \frac{\sin B}{\sin C}
   \]

5. **Equating the two expressions:**
   \[
   \frac{\sin(C + x)}{\sin(B + x)} = \frac{\sin B}{\sin C}
   \]
   This implies:
   \[
   \sin B \cdot \sin(B + x) = \sin C \cdot \sin(C + x)
   \]

6. **Using the product-to-sum identities:**
   \[
   \sin B \cdot \sin(B + x) = \frac{1}{2} [\cos(x) - \cos(2B + x)]
   \]
   \[
   \sin C \cdot \sin(C + x) = \frac{1}{2} [\cos(x) - \cos(2C + x)]
   \]
   Equating these, we get:
   \[
   \cos(2B + x) = \cos(2C + x)
   \]

7. **Solving the cosine equation:**
   Since \(\cos \theta = \cos \phi\) implies \(\theta = 2k\pi \pm \phi\) for integer \(k\), we have:
   \[
   2B + x = 2C + x \quad \text{or} \quad 2B + x = 360^\circ - (2C + x)
   \]
   The first equation simplifies to:
   \[
   2B = 2C \implies B = C \quad \text{(not possible since \(b < c\))}
   \]
   The second equation simplifies to:
   \[
   2B + x + 2C + x = 360^\circ \implies 2B + 2C + 2x = 360^\circ \implies B + C + x = 180^\circ
   \]
   Thus:
   \[
   x = 180^\circ - (B + C) = A
   \]

8. **Condition for \(F\) on \(AC\):**
   Since \(F\) lies on \(AC\), we must have:
   \[
   x \leq \angle AZC = \frac{A}{2} + B
   \]
   Hence:
   \[
   A \leq \frac{A}{2} + B \implies \frac{A}{2} \leq B
   \]
   This is the sufficient condition.

9. **Expressing \(BE\) in terms of \(a, b, c\):**
   When such \(E\) and \(F\) exist, we use the Angle Bisector Theorem:
   \[
   \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b}
   \]
   Let \(BD = x\) and \(DC = y\). Then:
   \[
   \frac{x}{y} = \frac{c}{b} \implies x = \frac{c}{b} y
   \]
   Since \(x + y = a\):
   \[
   \frac{c}{b} y + y = a \implies y \left( \frac{c}{b} + 1 \right) = a \implies y = \frac{ab}{b + c}
   \]
   Thus:
   \[
   x = \frac{c}{b} \cdot \frac{ab}{b + c} = \frac{ac}{b + c}
   \]
   Therefore:
   \[
   BE = BD = \frac{ac}{b + c}
   \]

The final answer is \(\boxed{\frac{ac}{b + c}}\)