## Task $5 / 62$

Given is a trapezoid with the parallel sides $a$ and $c$, the height $h$, and the angle $\alpha$ (figure). Required is the parallel to $a$ and $c$ that divides the area of the trapezoid in half. Solution 1. by calculation, 2. by construction.

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-034.jpg?height=389&width=554&top_left_y=1730&top_left_x=1208)
1. Calculation: Since the area of a trapezoid depends solely on the lengths of the parallel sides and the height, not on the angles, the angle $\alpha$ does not play a role in this problem, and the considerations can be made without loss of generality for a right-angled trapezoid (figure).

The dividing parallel has length $x$, and its distance from $c$ is $y$. Then we have

$$
\frac{x+c}{2} \cdot y=\frac{a+c}{4} \cdot h
$$

By the intercept theorem, $\frac{y}{h}=\frac{x-c}{a-c}$, so $y=\frac{x-c}{a-c} h$. This leads to

$$
x^{2}-c^{2}=\frac{a^{2}-c^{2}}{2} \rightarrow x=\sqrt{\frac{a^{2}+c^{2}}{2}}
$$

It is evident that the length of the parallel is independent of the height $h$. Its distance $y$ from $c$ is then

$$
y=\frac{x-c}{a-c} h \rightarrow y=\frac{\sqrt{\frac{a^{2}-c^{2}}{2}}-c}{a-c} \cdot h
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-035.jpg?height=762&width=419&top_left_y=681&top_left_x=333)

2. Construction: The calculation provides the key to the construction. From

$$
x=\sqrt{\frac{a^{2}+c^{2}}{2}} \quad \text { it follows that } \quad x=\sqrt{\left(\frac{a}{2} \sqrt{2}\right)^{2}+\left(\frac{c}{2} \sqrt{2}\right)^{2}}
$$

This means that $x$ can be obtained as the hypotenuse of a right-angled triangle whose legs are $\frac{a}{2} \sqrt{2}$ and $\frac{c}{2} \sqrt{2}$. These legs are the halves of the diagonals of the squares constructed on the parallel sides of the trapezoid.

Construction description:

1. Construct squares on the parallel sides of the trapezoid, draw a diagonal in each, and bisect these diagonals.
2. Construct a right-angled triangle with the bisected diagonals as legs. The hypotenuse has the length $x$ of the desired parallel.
3. Mark the segment $x$ from $A$ on $a=AB$; the endpoint is $E$. Draw a line through $E$ parallel to $d=DA$; its intersection with $b=BC$ is $F$. The line through $F$ parallel to $a=AB$ and $c=CD$ is the desired parallel.

Determination: All constructions are always feasible and unique.