Let $ABCDEFGH$ be a cube where $ABCD$ is the ground face and $E$ is about $A$, $F$ above $B$, and so on. Consider the space diagonal $AG$, and call $P$ the orthogonal projection of $B$ on $AG$.
The question is to show that $2|AP| = |PG|$. One can easily calculate that the space diagonal has length $\sqrt{3}$, and by the congruence of $ABG$ and $ABP$ we find that $|AP| = \sqrt{3}/3$. Hence $|PG| = 2\sqrt{3}/3$, and we are done. This, however, requires some calculations.
Can we show $2|AP| = |PG|$, without calculating any of the lengths? So, by some similarity or symmetry argument. One observation is that $P$ is in the plane $BDE$, so by symmetry there is a point $P'$ in the plane $CFH$, and in this setup it is sufficient to prove that $|AP| = |PP'|$.