10.5. On the sides $B C$ and $A D$ of a convex quadrilateral $A B C D$, their midpoints - points $M$ and $N$ respectively - are marked. Segments $M N$ and $A C$ intersect at point $O$, and $M O=O N$. It is known that the area of triangle $A B C$ is 2019. Find the area of quadrilateral $A B C D$.
Answer: 4038.

Solution. Let $K$ and $L$ be the midpoints of sides $AB$ and $CD$ respectively. Then $KMLN$ is a parallelogram, the diagonals of which intersect at point $O$.

Let $S_{BKM}=S$. Then $S_{OKM}=S$, since triangles $OKM$ and $BKM$ have the same bases and equal heights. Similarly,

![](https://cdn.mathpix.com/cropped/2024_05_06_7a782b7297b78d1fc01ag-2.jpg?height=636&width=897&top_left_y=2132&top_left_x=1088)
$S_{DNL}=S_{ONL}$. But triangles $ONL$ and $OKM$ are equal (for example, $KO=OL$, $ON=OM$ (as diagonals of parallelogram $KMLN$), $\angle KOM=\angle NOL$ (as vertical angles)). Therefore, $S_{DNL}=S_{ONL}=S_{OKM}=S_{BKM}=S$.

Further, from the similarity of triangles $BKM$ and $ABC$ (with a coefficient of $\frac{1}{2}$), it follows that $S_{ABC}=4S_{BKM}=4S$. Similarly, $S_{ADC}=4S_{NDL}=4S$.

Thus, triangles $ABC$ and $ADC$ have equal areas. $S_{ABCD}=2S_{ABC}$.

## Criteria.

2 points. It is proven that the midpoints of the sides form a parallelogram, the diagonals of which intersect at point $O$.

Comment. It is necessary to prove that KLMN is a parallelogram, as well as the equality of all pairs of triangles mentioned in the solution.