28.3. On the sides $B C$ and $C D$ of the square $A B C D$, points $K$ and $M$ are chosen such that the perimeter of triangle $K M C$ is equal to twice the side length of the square $A B C D$. Find the angle $K A M$.

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(7-10 \text { grades })
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28.3. Rotate triangle $A M D$ around point $A$ by an angle of $90^{\circ}$ until segment $A D$ coincides with segment $A B$ (Fig. 25). Let 

![](https://cdn.mathpix.com/cropped/2024_05_21_f0d089db3d527e92a992g-40.jpg?height=465&width=611&top_left_y=744&top_left_x=720)

Fig. 25

point $M$ move to point $P$. We obtain triangle $A P K$, which is equal to triangle $A M K$ by the three sides: side $A K$ is common, $A P=A M, K M=P K$ (since by the condition of the problem $K M=B K+M D=$ $=B K+P B=P K$). Since angle $M A P$ is a right angle, angle $M A K$ is $45^{\circ}$.