Problem 4. Square $A B C D$ is inscribed in circle $\omega$. On the smaller arc $C D$ of circle $\omega$, an arbitrary point $M$ is chosen. Inside the square, points $K$ and $L$ are marked such that $K L M D$ is a square. Find $\angle A K D$.
Answer: $135^{\circ}$.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_90b2af0281b2e8bdc481g-3.jpg?height=595&width=652&top_left_y=211&top_left_x=397)

We will prove the equality of triangles $D A K$ and $D C M$. Let's check that the conditions of the first criterion for the equality of triangles are met. Segments $D A$ and $D C$ are equal as sides of the square $A B C D$, and segments $D K$ and $D M$ are equal as sides of the square $K L M D$. Furthermore, since angles $A D C$ and $K D M$ of the original squares are right angles, we can write

$$
\angle A D K=90^{\circ}-\angle K D C=\angle C D M
$$

Thus, triangles $D A K$ and $D C M$ are indeed equal. Therefore, $\angle A K D=\angle C M D$. But $\angle C M D=135^{\circ}$, since it is an inscribed angle subtending the arc $D A B C$ of the circle $\omega$, and the measure of this arc is $270^{\circ}$. Then $\angle A K D=135^{\circ}$, which is what we needed to find.

## Criteria

0 points. Any incomplete or incorrect solution.

1 point. Only the correct answer without proof.

1 point. The possible equality of triangles $D A K$ and $D C M$ is mentioned in the work, but not proven or incorrectly proven; no further progress.

2 points. The possible equality of triangles $D A K$ and $D C M$ is mentioned in the work, but not proven or incorrectly proven; the solution is completed and the correct answer is obtained.

2 points. The equality of triangles $D A K$ and $D C M$ is proven, but the angle $A K D$ is not found or calculated incorrectly.

4 points. Any complete and correct solution.