3. The diagonal $AC$ of the inscribed quadrilateral $ABCD$ is the diameter of the circumscribed circle $\omega$ around it. A line perpendicular to the segment $BC$ was drawn from point $D$, intersecting the circle $\omega$ again at point $E$. Find the ratio of the areas of triangle $BCD$ and quadrilateral $ABEC$.
Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-12.jpg?height=511&width=537&top_left_y=1058&top_left_x=777)

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

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

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

$$
D N=D E-E N=D E-D H=E H
$$

It remains to note that

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