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## Equal angles in tetrahedron

If the angles between the three pairs of opposite sides of a tetrahedron are equal, prove that these angles are right.

Solution We use the identity $\vec{AB}\cdot \vec{CD}+\vec{AC}\cdot \vec{DB}+\vec{AD}\cdot \vec{BC}=0$. Since the angles between opposite sides is the same for all three pairs of opposite edges, let’s say $\alpha$, we can conclude from the inner product properties that
$\cos \alpha (\pm AB\cdot CD \pm AC\cdot BD \pm AD\cdot BC )=0$.
Now suppose the second factor is 0. Then we can’t have all signs to be + or all to be – and we will get a formula like $AB\cdot CD - AC\cdot BD + AD\cdot BC =0$ or something equivalent with this. So we have equality in Ptolemeu’s inequality, which means that $A,B,C,D$ are in the same plane and on the same circle, which is a contradiction with the fact that $ABCD$ is a tetrahedron.
By the arguments above, the second factor of our product can never be 0, meaning that $\cos \alpha=0$, which means that all the angles between opposite sides are right ( $90^\circ$)