4. In triangle $A B C$, the median $B M$ was drawn. It turned out that $A B=2 B M$ and $\angle M B A=40^{\circ}$. Find $\angle C B A$.

#
# Answer: $110^{\circ}$.

## Solution:

Extend the median $B M$ beyond point $M$ by its length and obtain point $D$. Since $A B=2 B M$, then $A B=B D$, which means triangle $A B D$ is isosceles. Therefore, angles $B A D$ and $B D A$ are each equal to $\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$. $A B C D$ is a parallelogram because its diagonals are bisected by their point of intersection. Thus, angle $C B D$, like $A D B$, is $70^{\circ}$, and angle $A B C$, which is the sum of $C B D$ and $A B D$, is $110^{\circ}$.

## Criteria:

Correct answer without correct justification - 1 point.

The idea of extending the triangle to form a parallelogram without further substantial progress - 2 points (not combined with the point for the correct answer).