[ The ratio of the areas of triangles with a common base or common height ] Difficult [ Pythagorean Theorem (direct and inverse). ] Class

Given a parallelogram $A B C D$ with sides $A B=2$ and $B C=3$. Find the area of this parallelogram if it is known that the diagonal $A C$ is perpendicular to the segment $B E$, which connects vertex $B$ to the midpoint $E$ of side $A D$.

#
Find the ratio $\frac{B O}{O E}$.

## Solution

Let $O$ be the intersection point of segments $A C$ and $B E$. From the similarity of triangles $B O C$ and $E O A$, it follows that $B O=2 O E$ and $O C=2 O A$. Let $O E=x, A O=y$. Then $B O=2 x$. From the right triangles $A O B$ and $A O E$, by the Pythagorean theorem, we have

$$
\left\{\begin{array}{l}
4 x^{2}+y^{2}=4 \\
x^{2}+y^{2}=\frac{9}{4}
\end{array}\right.
$$

From this system, we find that $x=\frac{7}{2 \sqrt{3}}, y=\frac{\sqrt{5}}{\sqrt{3}}$. Then

$$
S_{\triangle \mathrm{AOE}}=\frac{1}{2} x y=\frac{\sqrt{35}}{12}, S_{\triangle \mathrm{ABE}}=3 S_{\triangle \mathrm{AOE}}=\frac{\sqrt{35}}{4}
$$

Therefore,

$$
S_{\mathrm{ABCD}}=4 S_{\Delta \mathrm{ABE}}=\sqrt{35}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-25.jpg?height=337&width=744&top_left_y=2490&top_left_x=657)

## Answer

Problem