Let's transform a given parallelogram into a rhombus with equal area and equal angles.
Consider the task as solved, and let's denote the sides of the parallelogram by $a$ and $b$, and the side of the rhombus with the same area by $x$ (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_259c4c9a58aa61f06d7eg-1.jpg?height=382&width=488&top_left_y=270&top_left_x=813)

Draw the height $m_{1}$ of the parallelogram and the height $m_{2}$ of the rhombus. From the equality of the areas of the two figures,

$$
a m_{1}=x m_{2}
$$

From the similarity of the two right-angled triangles formed by the heights:

$$
b: m_{1}=x: m_{2} .
$$

Multiplying the corresponding sides of this equality by the sides of equation (1), we get

$$
a b=x^{2}
$$

Thus, the side of the rhombus we are looking for can be constructed as the geometric mean of the sides of the parallelogram, and the rhombus can be drawn from the side $x$.

Fejes László (Makó, József A. g. II. o. t.)