10.2. In parallelogram $A B C D$, the angle $\angle A=\alpha$ and diagonal $B D=d$ are known. Let $M$ and $N$ be the feet of the perpendiculars dropped from vertex $B$ to lines $C D$ and $A D$, respectively. Find $M N$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c26acbdc585bc94b69dg-3.jpg?height=365&width=708&top_left_y=1779&top_left_x=160)

To the solution of problem 10.2
Solution: 1) Note that the equality $M N=d \sin \alpha$ holds. $\angle M B N=\angle B A D=\alpha$, since the sides of angle $M B N$ are perpendicular to the sides of angle $B A D$.

2) Consider the circle with diameter $B D=d$. Points $M$ and $N$ lie on this circle, as $\angle B M D=\angle B N D=90^{\circ}$.
3) By the Law of Sines for triangle $B M N$, we have $\frac{M N}{\sin \angle M B N}=d$, from which we get the answer: $M N=d \sin \alpha$.

Note: The statement of the problem and the provided solution do not depend on whether the feet of the perpendiculars lie on the sides of the parallelogram or their extensions.

| in progress | points |
| :---: | :---: |
| Correct and justified answer | 7 points |
| A correct approach but an incorrect answer due to errors in formulas or arithmetic | 6 points |
| Both facts are proven: a) The quadrilateral with vertices $B D$ and $M N$ is cyclic; b) $\angle M B N=\alpha$, but the solution is not completed to the answer | 5 points |
| One of the facts a) or b) from the previous point is proven | 3 points |
| Correct answer without justification OR specific cases of parallelograms (e.g., a rectangle) are considered | 1 point |
| Several cases of the positions of points $N$ and $N$ on the lines containing sides $C D$ and $A D$ are considered | only one case is evaluated |
| Any statements and/or calculations not leading to a solution | not evaluated |