With the help of additional constructions, obtain similar triangles.

## Solution

The first method. Let $P$ be the intersection point of $MN$ with the line passing through point $C$ parallel to $AB$, and $Q$ be the intersection point of $AD$ with the line passing through point $N$ parallel to $AB$. Denote $MN = x; h_1$ and $h_2$ as the heights of the similar triangles $PCN$ and $QND$ (see the left figure).

Let $b > a$. The ratio of the areas of trapezoids $BMNC$ and $MADN$ is $2:3$, so $3(x + a)h_1 = 2(b + x)h_2$, from which $\frac{h_1}{h_2} = \frac{2}{3} \cdot \frac{b + x}{x + a}$.

From the similarity of triangles $CPN$ and $NQD$, it follows that $\frac{h_1}{h_2} = \frac{x - a}{b - x}$. Therefore, $\frac{2}{3} \cdot \frac{b + x}{x + a} = \frac{x - a}{b - x}$. From this, $x = \sqrt{\frac{3a^2 + 2b^2}{5}}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_60e93adedf73503f106cg-29.jpg?height=786&width=1758&top_left_y=1&top_left_x=161)

The second method. Let $O$ be the intersection point of the extensions of the lateral sides $AB$ and $DC$ (see the right figure), $S$ the area of triangle $BOC$.

$MN = x$ is the desired segment. Then $3(S_{MNO} - S) = 2(S_{AOD} - S_{MNO})$, or $3\left(\frac{x^2}{a^2} - 1\right) = 2\left(\frac{b^2}{a^2} - \frac{x^2}{a^2}\right)$.

From this, $x^2 = \frac{1}{2}(3a^2 + 2b^2)$.

## Answer

$\sqrt{\frac{3a^2 + 2b^2}{5}}$

Submit a comment
[Parallel lines, properties and signs. Transversals]

Points $M$ and $N$ are located on the lateral sides $A B$ and $C D$ of trapezoid $A B C D$, respectively, such that $M N \| A D$. It is known that the area of trapezoid $M B C N$ is to the area of trapezoid $A M N D$ as $2: 3$. Find $M N$, if $B C = a, A D = b$.