8.2. Given an isosceles triangle $A B C$ with base $A C$. Points $K$ and $N$ are marked on the side $B C$ ( $K$ lies between $B$ and $N$ ). It turns out that $K N=A N$ and $\angle B A K=\angle N A C$. Find $\angle B A N$.
Answer: $60^{\circ}$. Solution. Let $\angle B A K=\angle N A C=x, \angle K A N=y$. In the isosceles triangle $A K N$, the angle $A K N$ (at the base) is also equal to $y$. The angles at the base of triangle $A B C$ are equal to $2 x+y$, and the angle at the vertex $B$ is therefore equal to $180^{\circ}-2(2 x+y)$. The exterior angle $\angle A K N=y$ in triangle $A K B$ is equal to $x+180^{\circ}-2(2 x+y)=180^{\circ}-3 x-2 y$. Then from the equation $y=180^{\circ}-3 x-2 y$ we get $x+y=60^{\circ}$, i.e., $\angle B A N=60^{\circ}$.