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

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

![](https://cdn.mathpix.com/cropped/2024_05_06_9c0cded10981f14efe49g-02.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). This means $B A = E A$. 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 centroid, $A M: M C = 1: 2$.

Evaluation. Full solution: 14 points.