On sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $D$ and $K$ are marked, respectively, and on side $A C$, points $E$ and $M$ are marked such that $D A + A E = K C + C M = A B$. Segments $D M$ and $K E$ intersect. Find the angle between them.
Consider triangles $A D M$ and $C E K$ (see the figure): $D A = A B - A E = A C - A E = C E$; $A M = A C - C M = A B - C M = K C$; $\angle D A M = 60^{\circ} = \angle E C K$. Therefore, these triangles are congruent (by two sides and the included angle). Then $\angle A M D = \angle C K E$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae6ec2c916c776dd1759g-24.jpg?height=455&width=575&top_left_y=1&top_left_x=753)

Let the lines $D M$ and $K E$ intersect at point $P$. From triangle $P E M: \angle M P E = 180^{\circ} - (\angle P E M + \angle P M E) = 180^{\circ} - (\angle K E C + \angle C K E) = \angle K C E = 60^{\circ}$.

## Answer

$60^{\circ}$.