Task B-2.6. The height of an isosceles trapezoid is $h$, and the area of the trapezoid is $h^{2}$. At what angle do the diagonals of the trapezoid intersect?
First solution. Sketch...

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-10.jpg?height=689&width=1450&top_left_y=558&top_left_x=359)

Since $P=\frac{a+c}{2} \cdot h$ and $P=h^{2}$, it follows that $a+c=2 h$.

(1 point)

Extend $\overline{A B}$ beyond point $B$ to point $E$ such that $|B E|=|D C|=c$. Then $B E C D$ is a parallelogram, so $|C E|=|D B|$.

On the other hand, the trapezoid is isosceles, so $|A C|=|D B|$. Therefore, $|C E|=|A C|$. This means that triangle $\triangle A E C$ is isosceles with base $\overline{A E}$.

The height $\overline{C F}$ drawn to the base lies on the perpendicular bisector of the base.

Thus, point $F$ is the midpoint of segment $\overline{A E}$. Therefore,

$$
|A F|=\frac{1}{2}|A E|=\frac{1}{2}(a+c)=h
$$

This means that triangle $\triangle A F C$ is a right isosceles triangle.

It follows that

$$
\angle C A F=\angle F C A=45^{\circ}
$$

The angle between the diagonals is then:

$$
\angle A C E=2 \angle F C A=90^{\circ}
$$