Answer: $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-10.jpg?height=454&width=457&top_left_y=1806&top_left_x=822)

Solution. Let $\angle A B L=\varphi$. Then $\angle K B L=\varphi$ and $\angle K B C=2 \varphi$. Since $B K=B C$, we get $\angle B K C=\angle B C K=90^{\circ}-\varphi$. Moreover,

$$
\angle L C K+\angle B K C=\angle L B K+\angle B K C=\varphi+\left(90^{\circ}-\varphi\right)=90^{\circ},
$$

which means that lines $B K$ and $C L$ are perpendicular. On the other hand,

$$
\angle B L C+\angle A C L=\angle B K C+\angle A B L=\left(90^{\circ}-\varphi\right)+\varphi=90^{\circ},
$$

so lines $B L$ and $A C$ are also perpendicular. Therefore, point $P$ is the orthocenter of triangle $B C L$, from which it follows that $L P \perp B C$.
3. A circle is circumscribed around an acute-angled triangle $A B C$. Point $K$ is the midpoint of the smaller arc $A C$ of this circle, and point $L$ is the midpoint of the smaller arc $A K$ of this circle. Segments $B K$ and $A C$ intersect at point $P$. Find the angle between the lines $B C$ and $L P$, given that $B K = B C$.