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Auxiliary similar triangles Properties and criteria of a parallelogram
[The ratio of the areas of triangles with a common base or common height]

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

The area of trapezoid $A B C D$ is 90. The diagonals intersect at point $O$, segments connecting the midpoint $P$ of base $A D$ with vertices $B$ and $C$ intersect the diagonals of the trapezoid at points $M$ and $N$. Find the area of quadrilateral $O M P N$, if one of the bases of the trapezoid is twice the other.

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Let $O$ be the point of intersection of the diagonals of trapezoid $ABCD$ with bases $BC$ and $AD = 2BC$ (see figure). Triangles $ACD$ and $BAC$ have equal heights. Therefore, $S_{ACD} : S_{BAC} = 2 : 1$, which means $S_{ACD} = \frac{2}{3} S_{ABCD} = 60$. Triangles $BOC$ and $DOA$ are similar with a ratio of $1/2$. Consequently, $S_{AOD} = \frac{2}{3} S_{ACD} = 40$.

$BCDP$ is a parallelogram, and $N$ is its center, so $S_{DNP} = \frac{1}{4} S_{ACD} = 15$. Similarly, $S_{AMP} = 15$. Therefore, $S_{OMPN} = S_{AOD} - S_{AMP} - S_{DNP} = 10$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-05.jpg?height=551&width=755&top_left_y=-1&top_left_x=663)

Consider the case when $BC = 2AD$. Similarly, we get $S_{ACD} = \frac{1}{3} S_{ABCD} = 30. S_{ACP} = \frac{1}{2} S_{ACD} = 15, S_{AOD} = \frac{1}{3} S_{ACD} = 10$.

$S_{DNP} = S_{AMP} = \frac{1}{5} S_{ACP} = 3$. Therefore, $S_{OMPN} = S_{AOD} - S_{AMP} - S_{DNP} = 4$.

## Answer

10 or 4.