Problem 4. The height of an isosceles trapezoid is equal to $h$, and its area is $h^{2}$. At what angle do the diagonals of the trapezoid intersect?
Solution. Let $ABCD$ be an isosceles trapezoid with bases $AB$ and $CD$ and height $h$ whose area is $P=h^2$. The point $S$ is the intersection of the diagonals, and $S$ and $S'$ are the feet of the perpendiculars dropped from $S$ to the bases $AB$ and $CD$, respectively. The triangles $SS''D$ and $SS'B$ are similar because they have equal

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angles. Therefore, $\overline{SS''} : \overline{DS''} = \overline{SS'} : \overline{BS'}$. If $a$ and $b$ are the lengths of the bases, and $\overline{SS'} = h_1$ and $\overline{SS''} = h_2$, then $h_1 + h_2 = h$ and $h_2 : \frac{b}{2} = h_1 : \frac{a}{2}$. Thus,

$$
\frac{a}{2} = \frac{b}{2} \frac{h_1}{h_2}
$$

From the condition of the problem, we have $\frac{a+b}{2} h = h^2$, or $\frac{a}{2} + \frac{b}{2} = h$. From equation (1) and the previous formula, we get $\frac{b}{2} \frac{h_1}{h_2} + \frac{b}{2} = h$, or $\frac{b}{2} = h_2$. Similarly, we get $\frac{a}{2} = h_1$. Therefore, $SS''D$ and $SS'B$ are isosceles right triangles. The angle at which the diagonals of the trapezoid intersect is $90^\circ$.