3. In triangle $ABC$, the median $BM$ is twice as short as side $AB$ and forms an angle of 40 degrees with it. Find angle $ABC$.


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 $110^{\circ}$.

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Grading Guidelines. Correct answer without justification - 0 points. The idea of extending the triangle to a parallelogram without further substantial progress - 2 points.