3. Let $D$ be the midpoint of the hypotenuse $B C$ of the right triangle $A B C$. A point $M$ is chosen on the leg $A C$ such that $\angle A M B = \angle C M D$. Find the ratio $\frac{B M}{M D}$.
Answer: $2: 1$.

Solution. Let point $E$ be the intersection of rays $B A$ and $D M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9c0cded10981f14efe49g-07.jpg?height=610&width=588&top_left_y=278&top_left_x=820)

Then angles $A M E$ and $C M D$ are equal as vertical angles. Therefore, angles $A M B$ and $A M E$ are also equal. Hence, triangles $A M B$ and $A M E$ are congruent (by the common side $A M$ and the adjacent angles). Thus, $B A = E A$. Consider triangle $E B C$. In this triangle, point $M$ is the intersection of medians $C A$ and $E D$. By the property of the intersection of medians, $E M: M D = 2: 1$. But $B M = E M$ (from the congruence of triangles $A M B$ and $A M E$). Hence the answer.

Evaluation. 14 points for a complete solution.