9.2. In triangle $ABC$, the median $BM$ was drawn. It turned out that $\angle ABM=40^{\circ}$, $\angle MBC=70^{\circ}$. Find the ratio $AB: BM$. Justify your answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_94f1ab38259b8995cf4bg-4.jpg?height=494&width=708&top_left_y=1763&top_left_x=137)

To solve problem 9.2 Solution: Extend the median $BM$ beyond point $M$ and lay off a segment $MD = BM$ (see the figure). In quadrilateral $ABCD$, the diagonals bisect each other, so this quadrilateral is a parallelogram. Then $\angle ABC + \angle BCD = 180^{\circ}$, from which $\angle BCD = 70^{\circ} = \angle MBC = \angle DBC$. This means that triangle $BCD$ is isosceles, as is the congruent triangle $BAD$. Therefore,

$$
AB: BM = AB: \frac{BD}{2} = AB: \frac{AB}{2} = 2
$$
Answer: $A B: B M=2$.

| in progress | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| An example of a triangle satisfying the conditions of the problem is provided | 0 points |
| Incorrect solution or its absence | 0 points |