Through the vertices $A, B$ and $C$ of the parallelogram $A B C D$ with sides $A B=3$ and $B C=5$, a circle is drawn, intersecting the line $B D$ at point $E$, and $B E=9$. Find the diagonal $B D$.

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Using the theorem on the sum of the squares of the diagonals of a parallelogram and the theorem on the product of segments of intersecting chords, form a system of equations with respect to the diagonals of the parallelogram.

## Solution

Let the diagonals of the parallelogram intersect at point $O$. Denote $O B=O D=x, A O=O C=y$. According to the theorem on the sum of the squares of the diagonals of a parallelogram,

$$
A C^{2}+B D^{2}=2 A B^{2}+2 B C^{2}
$$

According to the theorem on the product of segments of intersecting chords,

$$
B O \cdot O E=A O \cdot O C
$$

Thus, we obtain the system

$$
\left\{\begin{array}{l}
4 x^{2}+4 y^{2}=2 \cdot 34 \\
x(9-x)=y^{2} x
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=17 \\
9 x=x^{2}+y^{2}
\end{array}\right.
$$

From here, we find that $B D=2 x=\frac{34}{9}$.

## Answer

$\frac{34}{9}$.

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