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 $40^{\circ}$. Find the angle $A B C$.
Answer: 110.

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_9c8efc6111fa3479c256g-3.jpg?height=594&width=340&top_left_y=2204&top_left_x=524)

Extend triangle $ABC$ to parallelogram $ABCD$. Then, $DB=2NB=AB$. Therefore, triangle $ABD$ is isosceles and $\angle ADB=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ}$. $\angle ADB=\angle DBA$ (alternate interior angles when lines are parallel). In the end, we get $\angle ABC=\angle ABD+\angle DBC=40^{\circ}+70^{\circ}=110^{\circ}$.