4. Construct a trapezoid $A B C D$ with a height of $3 \mathrm{~cm}$ and equal sides $B C, C D$ and $D A$, such that: There exists a point $E$ on the base $A B$ for which the segment $D E$ has a length of $5 \mathrm{~cm}$ and divides the trapezoid into two parts with equal areas.
SOLUTION. Analysis: If we denote $|A B|=a,|C D|=c$ and the height of the trapezoid $v$ (Fig. 2),

![](https://cdn.mathpix.com/cropped/2024_04_17_83dc5de7df9f3bd6e502g-3.jpg?height=244&width=434&top_left_y=1286&top_left_x=598)

Fig. 2

we can write the area $S$ of the trapezoid as

$$
S=\frac{1}{2}(a+c) v
$$

The area of triangle $A E D$ is, according to the problem statement, equal to

$$
\frac{|A E| \cdot v}{2}=\frac{1}{2} \cdot S=\frac{1}{2} \cdot \frac{1}{2}(a+c) v,
$$

from which it follows that $|A E|=\frac{1}{2}(a+c)$ (i.e., the segment $A E$ has the same length as the midline of trapezoid $A B C D$). Since point $E$ lies on segment $A B$, we have

$$
|E B|=|A B|-|A E|=a-\frac{1}{2}(a+c)=\frac{1}{2}(a-c),
$$

so $a>c$. If we denote by $B^{\prime}$ the point on segment $A B$ such that $\left|A B^{\prime}\right|=c$, then $\left|B^{\prime} B\right|=a-c$, and since the sought trapezoid $A B C D$ is isosceles, the triangle $B^{\prime} B C$ is also isosceles, so the midpoint $E$ of segment $B^{\prime} B$ is also the foot of the perpendicular from vertex $C$ to the base $A B$ (Fig. 2). Using the Pythagorean theorem, we calculate that

$$
c=\sqrt{|D E|^{2}-v^{2}}=\sqrt{5^{2}-3^{2}} \mathrm{~cm}=4 \mathrm{~cm} .
$$

Description of the construction: