It is given isosceles triangle $ABC$ ($AB=AC$) such that $\angle BAC=108^{\circ}$. Angle bisector of angle $\angle ABC$ intersects side $AC$ in point $D$, and point $E$ is on side $BC$ such that $BE=AE$. If $AE=m$, find $ED$
1. Given an isosceles triangle \(ABC\) with \(AB = AC\) and \(\angle BAC = 108^\circ\). The angle bisector of \(\angle ABC\) intersects side \(AC\) at point \(D\), and point \(E\) is on side \(BC\) such that \(BE = AE\). We need to find \(ED\) if \(AE = m\).

2. Since \(AB = AC\), \(\triangle ABC\) is isosceles, and \(\angle BAC = 108^\circ\). Therefore, the remaining angles \(\angle ABC\) and \(\angle ACB\) are equal. Let \(\angle ABC = \angle ACB = x\).

3. Using the angle sum property of a triangle:
   \[
   \angle BAC + \angle ABC + \angle ACB = 180^\circ
   \]
   \[
   108^\circ + x + x = 180^\circ
   \]
   \[
   108^\circ + 2x = 180^\circ
   \]
   \[
   2x = 72^\circ
   \]
   \[
   x = 36^\circ
   \]
   Therefore, \(\angle ABC = \angle ACB = 36^\circ\).

4. The angle bisector of \(\angle ABC\) intersects \(AC\) at \(D\). By the Angle Bisector Theorem:
   \[
   \frac{AD}{DC} = \frac{AB}{BC}
   \]
   Since \(AB = AC\), we have:
   \[
   \frac{AD}{DC} = \frac{AB}{BC}
   \]

5. Since \(BE = AE\), \(\triangle AEB\) is isosceles with \(AE = BE\). Therefore, \(\angle AEB = \angle ABE\).

6. In \(\triangle AEB\), \(\angle AEB = 180^\circ - \angle ABE - \angle BAE\). Since \(\angle BAE = 36^\circ\) (half of \(\angle BAC\)):
   \[
   \angle AEB = 180^\circ - 36^\circ - 36^\circ = 108^\circ
   \]

7. Since \(\angle AEB = 108^\circ\) and \(\angle AEC = 72^\circ\) (as \(\angle BAC = 108^\circ\) and \(\angle AEB = 108^\circ\)), we have:
   \[
   \angle AEC = \angle EAC = 72^\circ
   \]

8. Since \(EC = AC = AB\), we have:
   \[
   \frac{AD}{DC} = \frac{BE}{EC}
   \]
   Since \(BE = AE\) and \(EC = AC\), we have:
   \[
   DE \parallel AB
   \]

9. Therefore, \(\angle BDE = 18^\circ = \angle DBE\). Since \(\angle DBE = 18^\circ\) and \(DE = BE = AE = m\), we have:
   \[
   DE = BE = AE = m
   \]

The final answer is \(\boxed{m}\)