## Task B-4.3.

In rectangle $A B C D$, point $E$ divides side $\overline{C D}$ in the ratio $1: 3$ from vertex $D$. If line $A C$ is perpendicular to line $B E$, determine the ratio of the lengths of the sides of this rectangle.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6dc7744662d888032875g-25.jpg?height=749&width=808&top_left_y=1116&top_left_x=544)

We will solve the problem using vectors. If the line $AC$ is perpendicular to the line $BE$, then the dot product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BE}$ is zero, i.e., $\overrightarrow{AC} \cdot \overrightarrow{BE}=0$. Let's represent vectors $\overrightarrow{AC}$ and $\overrightarrow{BE}$ as linear combinations of vectors $\vec{a}=\overrightarrow{AB}$ and $\vec{b}=\overrightarrow{BC}$.

$$
\begin{aligned}
& \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\vec{a}+\vec{b} \\
& \overrightarrow{BE}=\overrightarrow{BC}+\overrightarrow{CE}=\vec{b}-\frac{3}{4} \vec{a}
\end{aligned}
$$

Now the condition $\overrightarrow{AC} \cdot \overrightarrow{BE}=0$ becomes:

$$
(\vec{a}+\vec{b}) \cdot\left(\vec{b}-\frac{3}{4} \vec{a}\right)=0
$$

or

$$
\vec{a} \cdot \vec{b} - \frac{3}{4} (\vec{a})^2 + (\vec{b})^2 - \vec{b} \cdot \frac{3}{4} \vec{a} = 0
$$

Vectors $\vec{a}=\overrightarrow{AB}$ and $\vec{b}=\overrightarrow{BC}$ are perpendicular, so $\vec{a} \cdot \vec{b}=0$. Therefore, and because $(\vec{a})^2=|\vec{a}|^2=|AB|^2=a^2$, $(\vec{b})^2=|\vec{b}|^2=|BC|^2=b^2$, we get

$$
0 - \frac{3}{4} a^2 + b^2 - 0 = 0
$$

or $b^2 = \frac{3}{4} a^2$, from which we obtain $\frac{b}{a} = \frac{\sqrt{3}}{2}$ (or $\frac{a}{b} = \frac{2}{\sqrt{3}}$).