2. In triangle $A B C$, the median $B M$ is twice as short as side $A B$ and forms an angle of $40^{\circ}$ with it. Find the angle $A B C$.
Solution. Extend median $B M$ beyond point $M$ by the same length to get point $D$ (see figure). Since $A B = B D$, triangle $A B D$ is isosceles. Therefore, $\angle B A D = \angle B D A = (180^{\circ} - 40^{\circ}) : 2 = 70^{\circ}$.

Quadrilateral $A B C D$ is a parallelogram because its diagonals bisect each other. Thus, $\angle C B D = 2 B M = 70^{\circ}$. Then $\angle A B C = \angle A B D + \angle C B D = 110^{\circ}$.

Answer $110^{\circ}$.