In a triangle $ABC, L$ and $K$ are the points of intersections of the angle bisectors of $\angle ABC$ and $\angle BAC$ with the segments $AC$ and $BC$, respectively. The segment $KL$ is angle bisector of $\angle AKC$, determine $\angle BAC$.
1. Let $\alpha = \angle CAK = \angle KAB$. We need to determine $\angle BAC$.

2. Note that $L$ is the $B$-excenter of triangle $\triangle ABK$. This means that $L$ is the center of the excircle opposite to vertex $B$ in $\triangle ABK$.

3. Since $L$ is the $B$-excenter, it lies on the angle bisector of $\angle AKC$. Therefore, $\angle LAK$ is half of the exterior angle at $A$ in $\triangle ABK$.

4. The exterior angle at $A$ in $\triangle ABK$ is $180^\circ - \angle KAB$. Since $L$ is the $B$-excenter, $\angle LAK$ is half of this exterior angle:
   \[
   \angle LAK = \frac{180^\circ - \angle KAB}{2}
   \]

5. Given that $\alpha = \angle KAB$, we substitute $\alpha$ into the equation:
   \[
   \angle LAK = \frac{180^\circ - \alpha}{2}
   \]

6. Since $L$ is the $B$-excenter, $\angle LAK$ is also equal to $\alpha$:
   \[
   \alpha = \frac{180^\circ - \alpha}{2}
   \]

7. Solving for $\alpha$, we multiply both sides by 2:
   \[
   2\alpha = 180^\circ - \alpha
   \]

8. Adding $\alpha$ to both sides, we get:
   \[
   3\alpha = 180^\circ
   \]

9. Dividing both sides by 3, we find:
   \[
   \alpha = 60^\circ
   \]

10. Since $\alpha = \angle CAK = \angle KAB$, the angle $\angle BAC$ is:
    \[
    \angle BAC = 2\alpha = 2 \times 60^\circ = 120^\circ
    \]

The final answer is $\boxed{120^\circ}$