Problem 5. On the side $A D$ of the square $A B C D$, point $K$ is marked, and on the extension of ray $A B$ beyond point $B$ - point $L$. It is known that $\angle L K C=45^{\circ}, A K=1, K D=2$. Find $L B$.
Answer: $L B=2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ea70eb8ee17be29735a7g-2.jpg?height=395&width=608&top_left_y=649&top_left_x=420)

Fig. 1: to the solution of problem 5

Solution. Note that $\angle L A C=45^{\circ}=\angle L K C$, which implies that quadrilateral $L A K C$ is cyclic. Then $\angle K C L=90^{\circ}$ (Fig. 1). Therefore, triangle $L C K$ is an isosceles right triangle, i.e., $L C=K C$. Right triangles $B L C$ and $D K C$ are congruent by hypotenuse and leg, so $B L=K D=2$.

Criteria

4 6. A complete and justified solution is provided.

In the absence of a correct solution, the highest applicable criterion from those listed below is used

3 6. The cyclic nature of quadrilateral $L A K C$ is proven, but there is no further progress.

1 6. The correct answer is provided, but there is no justification.

06. The problem is not solved or is solved incorrectly.