This is a self-answered question, after some playing around. I would be happy to see alternative solutions.
Let $x_1,x_2,x_3,x_4 \in \mathbb{R}^3$ be the vertices of a regular tetrahedron, which lie on the unit sphere, i.e. $|x_i-x_j|$ is constant for $i \neq j$, and $|x_i|=1$.
Claim: $x_4$ is perpendicular to the opposite face, i.e. $x_4 \perp x_1-x_2, x_4 \perp x_1-x_3$.
Question: How to prove this? I would like to find a slick elegant proof.
It is "geometrically obvious" that on a sphere $\mathbb{S}^2 \subseteq \mathbb{R}^3$, the Euclidean (or spherical) distance between two points is determined by the angle between them, so it follows that $x_4$ has equal angles with $x_1,x_2$. Since $|x_1|=|x_2|$, this implies that $x_4$ is perpendicular to the line connecting $x_1$ and $x_2$, i.e. to $x_1-x_2$.
In shorter (modern) phrasing:
Since $|x_i-x_j|^2=2-2\langle x_i,x_j \rangle$ does not depend on $i,j$, $\langle x_1,x_4 \rangle=\langle x_2,x_4 \rangle$, thus $\langle x_1-x_2,x_4 \rangle=0$ as required.