On the bisector $AL$ of triangle $ABC$ a point $K$ is chosen such that $\angle BKL=\angle KBL=30^{\circ}$. Lines $AB$ and $CK$ intersect at point $M$, lines $AC$ and $BK$ intersect at point $N$. FInd the measure of angle $\angle AMN$


[I]Proposed by D. Shiryaev, S. Berlov[/i]
1. **Define Points and Angles:**
   Let $E, F, G$ be the intersections of a circle with center $L$ and radius $LB$ with $AB$, $AC$ (between $A$ and $C$), and $BC$, respectively. Since $\angle BKL = \angle KBL = 30^\circ$, we know that $K$ lies on the circle centered at $L$ with radius $LB$.

2. **Angle Relationships:**
   Since $\angle EAK = \angle FAK = \alpha$ and $AK$ passes through the center of $(BEF)$, we have $\angle EBK = \angle FBK = 30^\circ - \alpha$. This implies $\angle FBG = \alpha$ and $\angle FLG = 2\alpha$.

3. **Equilateral Triangle:**
   It is easy to see that $\triangle LKG$ is an equilateral triangle because $\angle LKG = \angle KGL = \angle GLK = 60^\circ$. Therefore, $\angle AEK = \angle KGB = 60^\circ$.

4. **Reflection and Cyclic Quadrilateral:**
   Since $F$ is the reflection of $E$ with respect to the bisector $AK$, we have $\angle AFK = 60^\circ$. This implies that $KFCL$ is a cyclic quadrilateral. Therefore, $\angle FKC = \angle FLC = 2\alpha$.

5. **Cyclic Quadrilateral and Equilateral Triangle:**
   Since $AMKF$ is cyclic, $\angle MAF = 2\alpha$. Therefore, $\angle EMK = 60^\circ$, making $\triangle MKE$ an equilateral triangle.

6. **Kite Formation:**
   We see that $KG$ is a bisector of $\angle FKC = 2\alpha$. Therefore, $\angle BKF = \angle BKM = 90^\circ + \alpha$. Since $MK = KF$, $\triangle MBFN$ is a kite. Therefore, $\angle KMN = \angle KFN = 60^\circ$.

7. **Conclusion:**
   Since $\angle KMN = 60^\circ$, we conclude that $\angle AMN = 60^\circ$.

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