3. In triangle $ABC$, the angle bisectors $BK$ and $CL$ are drawn. A point $N$ is marked on segment $BK$ such that $LN \parallel AC$. It turns out that $NK = LN$. Find the measure of angle $ABC$. (A. Kuznetsov)
Answer: $120^{\circ}$. Solution. In the isosceles triangle $L N K$, $\angle K L N = \angle L K N$. Moreover, the angles $\angle K L N$ and $\angle L K A$ are equal as alternate interior angles when the lines $L N$ and $A C$ are parallel. Thus, $\angle K L N = \angle L K A$, which means that the ray $K L$ is the bisector of angle $A K B$. Therefore, the point $L$ lying on it is equidistant from the lines $K A$ and $K B$. Additionally, it is equidistant from the lines $C A = K A$ and $C B$, as it lies on the bisector of angle $A C B$. This means that the point $L$ is equidistant from the lines $C B$ and $K B$, and therefore must lie on the bisector of one of the angles formed by these lines, in which it is contained. This is the angle $K B C_{1}$, where $C_{1}$ is a point on the extension of the segment $C B$ beyond point $B$, and its bisector must be the ray $B L = B A$. From this, we get that $\angle A B C_{1} = \angle A B K = \angle C B K$. Since these three angles together form a straight angle, each of them is $60^{\circ}$, from which $\angle A B C = \angle A B K + \angle C B K = 120^{\circ}$.