Let the vertex position vectors be $\mathbf a,\mathbf b,\mathbf c,\mathbf d$. Suppose
$$
(\mathbf b-\mathbf a)\cdot(\mathbf d-\mathbf c)=0,
\qquad
(\mathbf c-\mathbf a)\cdot(\mathbf d-\mathbf b)=0.
$$
Subtracting the second scalar product from the first gives
$$
-(\mathbf d-\mathbf a)\cdot(\mathbf c-\mathbf b)=0,
$$
so the remaining pair of <opposite edges of a tetrahedron> is also perpendicular.

Put
$$
S_1=|\mathbf b-\mathbf a|^2+|\mathbf d-\mathbf c|^2,
\quad
S_2=|\mathbf c-\mathbf a|^2+|\mathbf d-\mathbf b|^2,
\quad
S_3=|\mathbf d-\mathbf a|^2+|\mathbf c-\mathbf b|^2.
$$
Direct expansion gives, cyclically,
$$
S_1-S_2=-2(\mathbf d-\mathbf a)\cdot(\mathbf c-\mathbf b),
$$
with the other two differences equal to minus twice the scalar products of the other opposite-edge pairs. All three scalar products vanish, hence
$$
\boxed{S_1=S_2=S_3}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 5C / c / Solution

Two edges of a tetrahedron are opposite when they share no vertex. The six edges split into three opposite pairs.


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