Problem 5. An isosceles triangle $ABC (AB = BC)$ and a square $AKLM$ are positioned as shown in the figure. Point $S$ on $AB$ is such that $AS = SL$. Find the measure of angle $SLB$.

[8 points] (L. Ponov)
Answer: $90^{\circ}$.

Solution. Consider triangles $A K S$ and $L K S$. They are equal by three sides.

![](https://cdn.mathpix.com/cropped/2024_05_06_ef48f16512240b91bbc6g-3.jpg?height=383&width=293&top_left_y=416&top_left_x=984)
Therefore, angles $K A S$ and $K L S$ are equal.

In the isosceles triangle $A B C$, angles $B A C$ and $A C B$ are equal. Lines $K L$ and $A C$ are parallel, so angles $A C B$ and $K L B$ are equal as corresponding angles. Then $\angle S L B=\angle S L K + \angle K L B=\angle K A S + \angle A C B=\angle K A S + \angle B A C=\angle K A M=90^{\circ}$.