What relationship holds between the sides of an isosceles trapezoid if the quadrilateral determined by the midpoints of its sides is a square?
The quadrilateral $E G F H$ is a square if its diagonals are equal, that is, if $E F = G H$.

![](https://cdn.mathpix.com/cropped/2024_05_02_62396a74e273a9004b79g-1.jpg?height=517&width=572&top_left_y=224&top_left_x=771)

But

$$
E F = \frac{A B + C D}{2}
$$

and

$$
\overline{G H^{2}} = \overline{D I^{2}} = \overline{D B^{2}} - \left(\frac{A B - C D}{2}\right)^{2}
$$

so

$$
\frac{(A B + C D)}{4} = \overline{D B^{2}} - \frac{(A B - C D)^{2}}{4}
$$

from which

$$
\overline{A B^{2}} + \overline{C D^{2}} = 2 \overline{D B^{2}}
$$

Thus, the quadrilateral determined by the midpoints of the sides of an isosceles trapezoid is a square if the sum of the squares of the parallel sides is equal to the sum of the squares of the non-parallel sides.

(Freibauer Ede.)

The problem was also solved by: Groffits G., Krisztián Gy., Lukhaub Gy., Perl Gy., Weisz Á., Weisz J.