## Similar triangles (other). $\quad[$ [Ratio of areas of triangles with a common base or common height] $]$ Difficult [The area of a figure is equal to the sum of the areas of the figures into which it is divided Cauchy's inequality

The diagonals of a convex quadrilateral $ABCD$ intersect at point $E$. It is known that the area of each of the triangles $ABE$ and $DCE$ is 1, the area of the entire quadrilateral does not exceed 4, and $AD=3$. Find the side $BC$.

#
Prove that $B C \| A D$.

## Solution

According to problem 35162, $B C \| A D$.

Let $B C = x$. From the similarity of triangles $B E C$ and $D E A$, it follows that $S_{B E C} = \frac{B E}{E D} \cdot S_{D C E} = \frac{x}{3}, S_{D E A} = \frac{D E}{B E} \cdot S_{A B E} = \frac{3}{x}$.

By the condition $S_{A B C D} \leq 4$, therefore $\frac{x}{3} + \frac{3}{x} \leq 2$.

On the other hand, according to the Cauchy inequality, this sum is not less than 2, and equality is achieved if and only if each of these numbers is 1. Therefore, $\frac{x}{3} + \frac{3}{x} = 2, x = 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_579637707648b9944b26g-12.jpg?height=277&width=646&top_left_y=-1&top_left_x=709)

## Answer

3.