3. In triangle ABC, the median BM is drawn. It is known that $\angle A B M=40^{\circ}$, and $\angle C B M=70^{\circ}$. Find the ratio $A B: B M$.
Answer: 2. First solution. Complete the triangle $A B C$ to a parallelogram $A B C D$. Since the diagonals of a parallelogram bisect each other, point $M$ is the point of their intersection and $B D=2 B M$. On the other hand, $\angle B D A=\angle C B D=70^{\circ}$, and $\angle B A D=180^{\circ}-\angle B D A-\angle A B D=70^{\circ}=\angle B D A$, from which $A B=B D=2 B M$ and $A B: B M=2 B M: B M=2$. Second solution. Note that $B C$ is the bisector of the external angle at vertex $B$ of triangle $A B M$. Therefore, $A B: B M=A C: C M=2$.