3. Given a cyclic quadrilateral $A B C D$ with perpendicular diagonals. On the circle circumscribed around it, a point $E$ is marked, diametrically opposite to $D$, and the segments $A B$ and $D E$ do not intersect. Find the ratio of the areas of triangle $B C D$ and quadrilateral $A B E D$.
Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-08.jpg?height=502&width=480&top_left_y=1094&top_left_x=822)

Solution. Let $N$ be the intersection point of $A C$ and $B D$, and $H$ be the foot of the perpendicular dropped from point $E$ to line $A C$. By the problem's condition, $D E$ is the diameter of the circle. Therefore, $\angle D B E=90^{\circ}$, which implies $B E \| A C$. Additionally, $\angle D C E=90^{\circ}$ and

$$
\angle C D E=90^{\circ}-\angle D E C=90^{\circ}-\angle D A C=\angle A D B
$$

from which it follows that $C E=A B$. Thus, $A B E C$ is an isosceles trapezoid, and

$$
C N=A C-A N=A C-C H=A H
$$

It remains to note that

$$
2 \cdot S_{A B E D}=B D \cdot A N+B D \cdot B E=B D \cdot A H=B D \cdot C N=2 \cdot S_{B C D}
$$