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

![](https://cdn.mathpix.com/cropped/2024_05_06_1b324abf723c207e3278g-15.jpg?height=420&width=583&top_left_y=1322&top_left_x=746)

Quadrilateral $A B C D$ is a parallelogram because its diagonals are bisected by their point of intersection. Thus, $\angle C B D=\angle A D B=70^{\circ}, \mathrm{a}$

$\angle B=\angle A B D+\angle C B D=110^{\circ}$.

## Answer

$110^{\circ}$.