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[ The area of a figure is equal to the sum of the areas of the figures into which it is divided ]

The diagonals of quadrilateral $A B C D$ intersect at point $O, M$ and $N$ are the midpoints of sides $B C$ and $A D$ respectively. Segment $M N$ divides the area of the quadrilateral in half. Find the ratio $O M: O N$, if $A D=2 B C$.

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$N M$ - the median of triangle $B N C$ (see fig.), therefore $S_{B M N}=S_{C M N}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-29.jpg?height=429&width=757&top_left_y=63&top_left_x=656)

From the condition, it follows that $S_{A B N}=S_{D C N}$. Moreover, triangles $A B N$ and $D C N$ have equal sides $A N$ and $D N$, so their heights $B H$ and $C T$ are also equal. Therefore, $A D \| B C$, which means $A B C D$ is a trapezoid.

Thus, point $O$ lies on segment $M N$, and triangles $A O D$ and $C O B$ are similar with a ratio of $A D / B C = 2$. Since $O N$ and $O M$ are the corresponding medians of these triangles, $O M: O N = 1: 2$.

## Answer

$1: 2$.