3. Given an isosceles triangle \(ABC\) with base \(BC\). On the extension of side \(AC\) beyond point \(C\), point \(K\) is marked, and a circle is inscribed in triangle \(ABK\) with center at point \(I\). A circle passing through points \(B\) and \(I\) is tangent to line \(AB\) at point \(B\). This circle intersects segment \(BK\) again at point \(L\). Find the angle between lines \(IK\) and \(CL\). Answer: \(90^{\circ}\).

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-02.jpg?height=496&width=820&top_left_y=483&top_left_x=638)
Solution. Let $I M$ and $I N$ be the perpendiculars dropped from point $I$ to $A K$ and $B K$ respectively. Note that $\angle A B I = \angle B L I$, since line $A B$ is tangent to the circumcircle of triangle $B L I$. Line $A I$ is the angle bisector of angle $A$, and $A B = A C$ by the given condition. Therefore, triangles $A B I$ and $A C I$ are congruent by two sides and the included angle. Hence,

$$
\angle A C I = \angle A B I = \angle B L I \quad \text{and} \quad I M = I N
$$

which means that right triangles $C M I$ and $L N I$ are congruent. Consequently,

$$
K C = K M - C M = K N - L N = K L
$$

Line $K I$ is the angle bisector of isosceles triangle $C L K$, from which it follows that $K I \perp C L$.