Consider two pairs of similar triangles: $D N L$ and $D C B$, $C N K$ and $C D A$.

## Solution

Let segment $M N$ intersect diagonal $A C$ at point $K$, and diagonal $B D$ at point $L$. Denote $M K=$ $K L=L N=x$.

From the similarity of triangles $D N L$ and $D C B$, it follows that ${ }^{x / b}{ }_{b}=D N / D C$.

From the similarity of triangles $C N K$ and $C D A$, it follows that ${ }^{2 x / a}={ }^{C N} /{ }_{C D}$. Therefore, $x\left({ }^{2} / a+1 / b\right)=1$. Hence, $M N$ $=3 x=3 a b / a+2 b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-48.jpg?height=406&width=734&top_left_y=1183&top_left_x=662)

Answer

$3 a b / a+2 b$.
[ Two pairs of similar triangles ]

Points $M$ and $N$ are on the lateral sides $A B$ and $C D$ of trapezoid $A B C D$, line $M N$ is parallel to $A D$, and segment $M N$ is divided into three equal parts by the diagonals of the trapezoid. Find the length of segment $M N$, if $A D=a, B C=b$, and the point of intersection of the diagonals of the trapezoid lies inside quadrilateral $M B C N$.