3. The height of an isosceles trapezoid is equal to $h$, and its area is equal to $h^{2}$. Determine the angle between the diagonals of this trapezoid.
Solution. Let $ABCD$ be an isosceles trapezoid ($AD = BC$), $CE$ be the perpendicular dropped from $C$ to the base $AB$, i.e., $CE = h$ (draw a diagram). From the condition of the problem, we have $\frac{a+b}{2} h = h^2$, from which we get $\frac{a+b}{2} = h$, which is the midline of the trapezoid. Then

$$
AE = AB - BE = a - \frac{a-b}{2} = \frac{a+b}{2} = h
$$

so $\measuredangle EAC = \measuredangle ABD = 45^{\circ}$. The angle between the diagonals, as the third angle of $\triangle ABO$, where $O$ is the intersection of the diagonals, is equal to $90^{\circ}$, which means that the diagonals intersect at a right angle.