Prove that if in a tetrahedron if two pairs of opposite edges are perpendicular then the third pair is also perpendicular. 
Prove that if in a tetrahedron if two pairs of opposite edges are
  perpendicular then the third pair is also perpendicular.

Method: 
$\vec a+ \vec b  + \vec c + \vec d +\vec e + \vec f = 0 \tag0$ where the vectors represent the sides. 
also, $(\vec a + \vec b). (\vec c+\vec d)= 0 \tag{1}$
$(\vec e+\vec f) .(\vec c+\vec d)= 0 \tag{2}$
From 1 and 2, 
$\vec e + \vec f = k(\vec a+ \vec b)$
Substituting in $0$,
$\vec a + \vec b = -\dfrac{\vec c + \vec d}{k+1}$
Now, $(\vec e+ \vec f). (\vec a+\vec b) = -\dfrac 1{k+1}((\vec a + \vec b). (\vec c+\vec d))= 0$ 
Hence, the third pair too is perpendicular. 
QED. 
Is my method to solve it correct? 

Let the  tetrahedron $T$ be spanned from $0$ by the three vectors $a$, $b$, $c$. The three equations 
$$\eqalign{v+w&=a\cr u\!\qquad+w&=b\cr u+v\qquad &=c\cr}\tag{1}$$
 determine three linearly independent vectors $u$, $v$, $w$. From $(1)$ it then follows that the vertices of $T$ are vertices of the parallelepiped $P$ spanned by $u$, $v$, $w$, and opposite edges of $T$ are diagonals of opposite faces of $P$. Two such edges of $T$ are orthogonal iff the corresponding faces of $P$ are rhombi. If we have two pairs of orthogonal opposite edges of $T$ we can therefore conclude that $|u|=|v|=|w|$, which then implies that the third pair of opposite edges of $T$ is orthogonal as well.

