Problem 3. In a convex quadrilateral, the lengths of the diagonals are 2 and 4 cm. Find the area of the quadrilateral, knowing that the lengths of the segments connecting the midpoints of opposite sides are equal.
Answer: 4 cm$^2$.

Solution: The segment connecting the midpoints of adjacent sides of a quadrilateral is parallel to its diagonal (as the midline of the corresponding triangle). Therefore, quadrilateral $P Q R T$ is a parallelogram. By the condition, the diagonals of this parallelogram are equal, so it is a rectangle. But then the diagonals $A C$ and $B D$, which are parallel to its sides, are perpendicular to each other.

Draw lines through the vertices of the rectangle parallel to the diagonals. We get the circumscribed rectangle $K L M N$. Note that triangle $A M D$ is equal to $A D O$, and triangle $C D N$ is equal to $C D O$, and so on. Therefore, the area of $K L M N$ is twice the area of $A B C$. Thus, the desired area is $S = 1 / 2 \cdot 2 \cdot 4 = 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cc70e1780740a896c313g-1.jpg?height=365&width=465&top_left_y=1824&top_left_x=1383)

Remark: There are other ways to calculate $S$, for example, as twice the area of $P Q R T$. Also, the formula $S = 1 / 2 \cdot A C \cdot B D \cdot \sin \alpha$, where $\alpha$ is the angle between the diagonals, can be used.

Criteria: 0 points for solving a particular case. 3 points for proving that $S P Q R$ is a rectangle.