The bisectors of the angles $A$ and $B$ of the triangle $ABC$ intersect the sides $BC$ and $AC$ at points $D$ and $E$. It is known that $AE + BD = AB$. Find the angle $\angle C$.
1. **Identify the given information and setup the problem:**
   - We have a triangle \(ABC\) with angle bisectors of \(\angle A\) and \(\angle B\) intersecting sides \(BC\) and \(AC\) at points \(D\) and \(E\) respectively.
   - It is given that \(AE + BD = AB\).

2. **Use the Angle Bisector Theorem:**
   - The Angle Bisector Theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides.
   - For \(\angle A\), the bisector intersects \(BC\) at \(D\), so:
     \[
     \frac{BD}{DC} = \frac{AB}{AC}
     \]
   - For \(\angle B\), the bisector intersects \(AC\) at \(E\), so:
     \[
     \frac{AE}{EC} = \frac{AB}{BC}
     \]

3. **Reflect points \(D\) and \(E\) over the angle bisectors:**
   - Reflecting \(D\) and \(E\) over the angle bisectors of \(\angle A\) and \(\angle B\) respectively, we find that the reflections coincide on \(AB\).
   - This implies that the incenter \(I\) of \(\triangle ABC\) is equidistant from \(D\) and \(E\), i.e., \(ID = IE\).

4. **Use the property of the incenter and the circumcircle:**
   - Since \(ID = IE\), the incenter \(I\) lies on the circumcircle of \(\triangle CDE\).
   - This implies that \(\angle C = \angle BID\).

5. **Calculate \(\angle BID\):**
   - The angle \(\angle BID\) can be expressed in terms of the angles of the triangle. Specifically, \(\angle BID = \frac{\angle AID}{3}\).
   - Since \(I\) is the incenter, \(\angle AID = 180^\circ - \frac{\angle A}{2} - \frac{\angle B}{2}\).

6. **Determine the value of \(\angle C\):**
   - Given that \(AE + BD = AB\), we can infer that the triangle is equilateral. This is because the sum of the segments created by the angle bisectors equals the side of the triangle, which is a property of equilateral triangles.
   - In an equilateral triangle, all angles are \(60^\circ\).

Therefore, \(\angle C = 60^\circ\).

The final answer is \(\boxed{60^\circ}\).