4. On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $L$ and $K$ are marked, respectively, and $M$ is the intersection point of segments $A K$ and $C L$. It is known that the area of triangle $A M C$ is equal to the area of quadrilateral $L B K M$. Find the angle $A M C$.
Solution. From the equality of the areas of triangle $A M C$ and quadrilateral $L B K M$, it follows that the areas of triangles $A C K$ and $C B L$ are equal (see the figure), because in this case, the same area of triangle $C M K$ is added to the equal areas. We have $S_{A C K}=\frac{1}{2} A C \cdot C K \cdot \sin \angle A C K$, and $S_{C B L}=\frac{1}{2} B C \cdot B L \cdot \sin \angle C B L$, considering the equality of sides and angles in the equilateral triangle, we get that $C K=L B$, i.e., $\triangle A C K=\triangle C B L$ by two sides and the included angle. Therefore, $\angle C K A=\angle B L C$. Considering that $\angle B L C+\angle B C L=180^{\circ}-\angle L B C=120^{\circ}$, we obtain

$$
\angle A M C=\angle M C K+\angle M K C=\angle L C B+\angle B L C=120^{\circ} .
$$

Answer: $120^{\circ}$.