2. In triangle $A B C$, the median $B K$ is twice as small as side $A B$ and forms an angle of $32^{\circ}$ with it. Find the angle $A B C$.
Answer: $106^{\circ}$.

Solution. Let $K$ be the midpoint of segment $B D$. Then $A B C D$ is a parallelogram. In triangle $A B D$, we have the equality of sides $A B$ and $B D$. Therefore,

$$
\angle B D A=\frac{1}{2}\left(180^{\circ}-32^{\circ}\right)=74^{\circ}
$$

Angles $A D B$ and $C B D$ are equal as alternate interior angles. Thus,

$$
\angle A B C=\angle A B D+\angle D B C=32^{\circ}+74^{\circ}=106^{\circ} .
$$

Evaluation. 13 points for a correct solution.