Let $K$ be a point on the angle bisector, such that $\angle BKL=\angle KBL=30^\circ$. The lines $AB$ and $CK$ intersect in point $M$ and lines $AC$ and $BK$ intersect in point $N$. Determine  $\angle AMN$.
1. **Understanding the Problem:**
   - We are given a point \( K \) on the angle bisector of \( \angle BAC \) such that \( \angle BKL = \angle KBL = 30^\circ \).
   - We need to determine \( \angle AMN \) where \( M \) is the intersection of lines \( AB \) and \( CK \), and \( N \) is the intersection of lines \( AC \) and \( BK \).

2. **Constructing the Triangle:**
   - Consider an equilateral triangle \( \triangle ABC \). In an equilateral triangle, all internal angles are \( 60^\circ \).
   - The angle bisector of \( \angle BAC \) will also be the altitude, median, and perpendicular bisector of \( BC \).

3. **Placing Point \( K \):**
   - Since \( K \) is on the angle bisector of \( \angle BAC \), it lies on the line segment from \( A \) to the midpoint of \( BC \).
   - Given \( \angle BKL = \angle KBL = 30^\circ \), \( K \) must be such that these angles are formed.

4. **Identifying Points \( M \) and \( N \):**
   - \( M \) is the intersection of \( AB \) and \( CK \).
   - \( N \) is the intersection of \( AC \) and \( BK \).

5. **Analyzing the Geometry:**
   - In an equilateral triangle, the altitude from \( A \) to \( BC \) (which is also the angle bisector) will intersect \( BC \) at its midpoint, say \( D \).
   - Since \( \angle BKL = \angle KBL = 30^\circ \), \( K \) must be the midpoint of \( AD \).

6. **Determining \( \angle AMN \):**
   - Since \( K \) is the midpoint of \( AD \), \( CK \) is the median of \( \triangle ABC \).
   - \( BK \) is also the median of \( \triangle ABC \).
   - Therefore, \( M \) and \( N \) are the same point, which is the centroid of \( \triangle ABC \).

7. **Calculating \( \angle AMN \):**
   - In an equilateral triangle, the centroid divides each median in a 2:1 ratio.
   - The centroid is the intersection of the medians, and it forms \( 90^\circ \) angles with the sides of the triangle.

Thus, \( \angle AMN = 90^\circ \).

The final answer is \( \boxed{90^\circ} \).