Prove that a parallelepiped with the property that the distance between each two opposite edges is the same is a cube.

Question

Prove that a parallelepiped with the property that the distance between each two opposite edges is the same is a cube.

(managed to prove that the sides are congruent rhombs and don't know how to continue, any help would be appreciated)

Answer 1

$\def\a{{\bf a}}\def\b{{\bf b}}\def\c{{\bf c}}$ Let the parallelepiped be defined by a triple of non-collinear vectors $a,b,c$.

There are two pairs of opposite $c$-edges, the distances between the edges in these pairs being: $$ \frac{|(a+b)\times c|}{|c|}\quad\text{and}\quad \frac{|(a-b)\times c|}{|c|}, $$ respectively, so one concludes: $$ |a\times c+b\times c|=|a\times c-b\times c|. $$ This means $(a\times c)\perp(b\times c)$. Similarly $(a\times b)\perp(c\times b)$ and $(b\times a)\perp(c\times a)$.

Geometrically this means that the three (non-parallel) faces of the parallelepiped are perpendicular to each other. This in turn means that the three (non-parallel) edges of the parallelepiped are perpendicular to each other, so that it is a rectangular parallelepiped (cuboid).

Finally it follows from $a^2+b^2=b^2+c^2=c^2+a^2$ that $|a|=|b|=|c|$.