9.4. The distance between the midpoints of sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ is equal to the distance between the midpoints of its diagonals. Find the angle formed by the lines $AD$ and $BC$ at their intersection. Justify your answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_974cd9d9a997a850d94fg-6.jpg?height=471&width=762&top_left_y=918&top_left_x=133)

$\mathrm{K}$ to the solution of problem 9.4
Solution: We will apply the theorem of the midline of a triangle four times and obtain that 1) $E M=A D / 2=F N ; 2) F M=B C / 2=E N$ (points $M, N, E, F-$ are the midpoints of segments $A B, C D, D B, A C$, respectively). Therefore, quadrilateral $E M F N$ is a parallelogram. According to the condition, its diagonals are equal, which means this parallelogram is actually a rectangle. Since the angle between lines $A D$ and $B C$ is equal to the angle $E M F$ of the rectangle, it is equal to $90^{\circ}$.

Answer: $90^{\circ}$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Proven that quadrilateral $E M F N$ is a parallelogram | 4 points |
| Theorem of the midline of a triangle applied (at least once) | 2 points |
| Answer without justification and/or incorrect answer | 0 points |