Let $ABCD$ be the isosceles trapeze with height $h$ and area $P = h^2$. Let $S$ be the intersection point of the diagonals and $S'$ and $S''$ are the bases of the altitudes from $S$ to $AB$ and $CD$ correspondently. The triangles $SS''D$ and $SS'B$ are similar because they have equal angles. So we get that $\overline{SS''} : \overline{DS} = \overline{SS'} : \overline{BS'}$. Let $a$ and $b$ be the lengths of $AB$ and $CD$ correspondently and $\overline{SS'} = h_1$, $\overline{SS''} = h_2$. Then we have that $h_1 + h_2 = h$ and $h_2 : \frac{b}{2} = h_1 : \frac{a}{2}$. Hence $\frac{a}{2} = \frac{b}{2} \frac{h_1}{h_2}$. (1)

From the condition in the problem we have $\frac{a+b}{2} h = h^2$, or $\frac{a+b}{2} = h$. From the equality (1) and the last equality we have $\frac{b}{2} \frac{h_1}{h_2} + \frac{b}{2} = h$ i.e. $\frac{b}{2} = h_2$. Similarly $\frac{a}{2} = h_1$. So $SS''D$ and $SS'B$ are isosceles and right-angled triangles. The angle between the diagonals is right angle.