Points $K$ and $L$ are chosen on the sides $AB$ and $BC$ of the isosceles $\triangle ABC$ ($AB = BC$) so that $AK +LC = KL$. A line parallel to $BC$ is drawn through midpoint $M$ of the segment $KL$, intersecting side $AC$ at point $N$. Find the value of $\angle KNL$.
1. **Identify the given information and setup the problem:**
   - We have an isosceles triangle $\triangle ABC$ with $AB = BC$.
   - Points $K$ and $L$ are chosen on sides $AB$ and $BC$ respectively such that $AK + LC = KL$.
   - A line parallel to $BC$ is drawn through the midpoint $M$ of segment $KL$, intersecting side $AC$ at point $N$.
   - We need to find the value of $\angle KNL$.

2. **Consider the circle with center $E$:**
   - The circle is tangent to $KL$ and tangent to $BA$ and $BC$ at points $A$ and $C$ respectively.
   - This implies that $E$ is the excenter of $\triangle KLC$ opposite to vertex $K$.

3. **Intersection of $EL$ with $AC$:**
   - Let $EL$ intersect $AC$ at point $N'$.
   - By the properties of the excenter, $E$ lies on the angle bisector of $\angle KLC$.

4. **Angle chasing to show cyclic quadrilateral:**
   - Since $E$ is the excenter, $\angle KAN' = \angle KEN'$.
   - This implies that quadrilateral $KAEN'$ is cyclic.
   - Therefore, $\angle KN'E = \angle KAE = 90^\circ$.

5. **Midpoint properties and parallel lines:**
   - Since $M$ is the midpoint of $KL$, we have $MK = ML$.
   - Given that $MN'$ is parallel to $BC$, and $M$ is the midpoint, it follows that $MN' = MK = ML$.
   - This implies that $N \equiv N'$.

6. **Conclusion:**
   - Since $N \equiv N'$, we have $\angle KNL = \angle KN'L = 90^\circ$.

The final answer is $\boxed{90^\circ}$