8. Variant 1.

Given a parallelogram $A B C D$. Let $B P$ and $C Q$ be the perpendiculars dropped from vertices $B$ and $C$ to diagonals $A C$ and $B D$ respectively (point $P$ lies on segment $A C$, and point $Q$ lies on segment $B D$). Find the ratio $\frac{10 B D}{A C}$, if $\frac{A P}{A C}=\frac{4}{9}$ and $\frac{D Q}{D B}=\frac{28}{81}$.
Answer: 6.

Solution: Let $O$ be the point of intersection of the diagonals. Note that points $B, C, Q, P$ lie on the same circle (segment $B C$ is seen from points $P$ and $Q$ at a right angle). Therefore, triangles $B O P$ and $C O Q$ are similar. Let $A C=2 a, B D=2 b$. Then $P O=a-\frac{8 a}{9}=\frac{a}{9}, Q O=$ $b-\frac{56 b}{81}=\frac{25 b}{81}$. From the similarity of triangles $B O P$ and $C O Q$, we have $\frac{P O}{Q O}=\frac{B O}{C O} \cdot \frac{a}{9}: \frac{25 b}{81}=\frac{b}{a} \cdot \frac{81 a}{9 \cdot 25 b}=\frac{b}{a}$, from which $\frac{b}{a}=\frac{3}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_eca953bacaf67be756dbg-7.jpg?height=786&width=957&top_left_y=198&top_left_x=541)