Problem 8-3. $\quad$ In a right triangle $A B C$, angle $B$ is a right angle. On the leg $A B$, a point $M$ is chosen such that $A M=B C$, and on the leg $B C$, a point $N$ is chosen such that $C N=M B$. Find the acute angle between the lines $A N$ and $C M$.

Answer: $45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0eb6495bef493c26fee2g-2.jpg?height=694&width=468&top_left_y=1489&top_left_x=797)
Solution. Complete the triangle $A B C$ to a rectangle $A B C D$ and choose a point $P$ on its side $C D$ such that $A P$ is parallel to $C M$. Then $P C = A D$, $D P = C N$, and the right triangles $A D P$ and $C P N$ are equal, with $\angle D A P = \angle C P N$. Therefore, $\angle A P D + \angle C P N = 90^{\circ}$ and $\angle A P N = 90^{\circ}$, meaning that $A P N$ is an isosceles right triangle. Thus, $\angle P A N = 45^{\circ}$ and, due to the parallelism of lines $A P$ and $C M$, the acute angle between the lines $A N$ and $C M$ also measures 45 degrees.

| Rating | Score | Content Criterion |
| :---: | :---: | :--- |
| + | 20 | Full solution |
| - | 0 | Solution is completely incorrect/only answer |