3. (17 points) In triangle $A B C$, a point $N$ is taken on side $A C$ such that $A N=N C$. Side $A B$ is twice as long as $B N$, and the angle between $A B$ and $B N$ is $50^{\circ}$. Find the angle $A B C$.
Answer: 115.

Solution. Complete triangle $A B C$ to parallelogram $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9c8efc6111fa3479c256g-1.jpg?height=597&width=277&top_left_y=2220&top_left_x=524)

Then, $D B=2 N B=A B$. Therefore, triangle $A B D$ is isosceles and $\angle A D B=\frac{180^{\circ}-50^{\circ}}{2}=65^{\circ}$. $\angle A D B=\angle D B A$ (alternate interior angles when lines are parallel). In the end, we get $\angle A B C=\angle A B D+\angle D B C=50^{\circ}+65^{\circ}=115^{\circ}$.