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The area of the given convex quadrilateral is $S$. Find the area of the quadrilateral with vertices at the midpoints of the sides of the given one.

#
A quadrilateral with vertices at the midpoints of the sides of a given quadrilateral is a parallelogram.

## Solution

## First method.

Let $d_{1}$ and $d_{2}$ be the diagonals of the given quadrilateral, and $\alpha$ be the angle between them. The quadrilateral with vertices at the midpoints of the sides of the given quadrilateral is a parallelogram with sides $\frac{1}{2} d_{1}$ and $\frac{1}{2} d_{2}$, and the angle $\alpha$ between them. Its area is

$$
\frac{1}{2} d_{1} \cdot \frac{1}{2} d_{2} \sin \alpha = \frac{1}{2} \left( \frac{1}{2} d_{1} d_{2} \sin \alpha \right) = \frac{1}{2} S
$$

Second method.

Let $S$ be the area of the given quadrilateral $ABCD$, and $s$ be the area of the quadrilateral whose vertices are the midpoints $K, L, M$, and $N$ of the sides $AB, BC, CD$, and $AD$ respectively.

Since $KL$ and $MN$ are the midlines of triangles $ABC$ and $ADC$, we have

$$
S_{\Delta \mathrm{KBL}} = \frac{1}{4} S_{\Delta \mathrm{ABC}}, \quad S_{\Delta \mathrm{MDN}} = \frac{1}{4} S_{\Delta \mathrm{ADC}}
$$

Therefore,

$$
S_{\Delta \mathrm{KBL}} + S_{\Delta \mathrm{MDN}} = \frac{1}{4} S_{\Delta \mathrm{ABC}} + \frac{1}{4} S_{\Delta \mathrm{ADC}} = \frac{1}{4} \left( S_{\Delta \mathrm{ABC}} + S_{\Delta \mathrm{ADC}} \right) = \frac{1}{4} S
$$

Similarly,

$$
S_{\Delta \mathrm{KAN}} + S_{\Delta \mathrm{MCL}} = \frac{1}{4} S
$$

Thus,

$$
s = S - S_{\Delta \mathrm{KBL}} - S_{\Delta \mathrm{MDN}} - S_{\Delta \mathrm{KAN}} - S_{\Delta \mathrm{MCL}} = S - \frac{1}{4} S - \frac{1}{4} S = \frac{1}{2} S
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-22.jpg?height=448&width=552&top_left_y=279&top_left_x=753)

## Answer

$\frac{1}{2} S$.

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