A geometry problem by Áron Bán-Szabó

Equation of the line $AB$ can be written as $y=a+ax$ and that of the line $CD$ as $y=b-bx$ .

Coordinates of the midpoints then are

$E=\left(-\dfrac a2,\dfrac a2\right), F=\left(0,\dfrac {a+b}{2}\right), G=\left(\dfrac b2,\dfrac b2\right), H=\left(\dfrac {b-a}{2},0\right)$

For simplicity of calculations it is possible to double all these coordinates. If the double sized object is a square, the original had to be a square as well.

$E'=\left(-a,a\right), F'=\left(0,a+b\right), G'=\left(b,b\right), H'=\left(b-a,0\right)$

The squares of pairwise distances are

$\overline{E'F'}^2=a^2+(a+b-a)^2=a^2+b^2$

$\overline{F'G'}^2=b^2+(b-a-b)^2=a^2+b^2$

$\overline{G'H'}^2=(b-a-b)^2+b^2=a^2+b^2$

$\overline{H'E'}^2=(b-a+a)^2+a^2=a^2+b^2$

So all the sides are the same length.

The pairwise slopes are

$m_{E'F'}=\dfrac ba$

$m_{F'G'}=\dfrac {-a}{b}=-\dfrac ab$

$m_{G'H'}=\dfrac {-b}{-a}=\dfrac ba$

$m_{H'E'}=\dfrac {-a}{b-a+a}=-\dfrac ab$

So there are two pairs of parallel lines at right angles to each other.

The quadrilateral $EFGH$ is therefore a square.