3. On the sides $\overline{B C}$ and $\overline{C D}$ of the rectangle $A B C D$, points $G$ and $E$ are given such that $|D E|=|B G|$ and $|D E|=\frac{1}{2}|E C|$. Point $F$ is inside the rectangle $A B C D$ such that $C E F G$ is also a rectangle.

If the area of the hexagon $A B G F E D$ (the shaded region in the figure) is equal to the area of the rectangle $C E F G$ (the unshaded part in the figure), how many times longer is $\overline{B C}$ than $\overline{D E}$?

![](https://cdn.mathpix.com/cropped/2024_06_03_02b8c2d6c56c85f46c17g-1.jpg?height=434&width=397&top_left_y=1448&top_left_x=1412)
First method:

Let point $H$ be such that $E D H F$ and $A B G H$ are rectangles. Denote their areas by $P_{1}$ and $P_{2}$, respectively, and the area of rectangle $C E F G$ by $P_{3}$.

We have

$$
P_{1}=x \cdot y, \quad P_{2}=3 x \cdot x, \quad P_{3}=2 x \cdot y
$$

From the condition, we know that $P_{1}+P_{2}=P_{3}$, which leads to the equation: $x \cdot y+3 x \cdot x=2 x \cdot y$.

On the left side, apply the distributive property of multiplication over addition, and then divide both sides of the equation by $x$ (which is permissible since the length $x$ is not 0):

$x \cdot(y+3 x)=2 x \cdot y$

$y+3 x=2 y, \text{i.e.} \quad$ we get $y=3 x$.