Let $T$ be a regular tetrahedron, and $\ell$ a line from the midpoint of one edge $E_1$, through the center $C$, to the midpoint of another edge $E_2$.
How do I prove that the reflection of each point on the tetrahedron across this line $\ell$ is also on the tetrahedron? I can see this visually (especially when I rotate the model so as to look at $\ell$ head on), but don't know how to prove it.
My start is: $E_1$ and $E_2$ are each normal to $\ell$, and $\ell$ passes through their midpoints, so that any point on either of these edges clearly has its reflection on the tetrahedron. The rest of the tetrahedron is generated by sliding $E_1$ across $\ell$ while rotating it to match $E_2$, so the reflection of each point along the way here is likewise on the tetrahedron. While I believe this last sentence true, it itself needs to be proven.
Update
Based on Blue and Dan Asimov's suggestions, I've developed the following proof, which I request verification of (hence the new proof-verification tag).
A regular tetrahedron is the polyhedron, unique up to similarity, with four vertices, each equidistant from all the others. Therefore, any four mutually equidistant points form a regular tetrahedron.
On a cube, choose an arbitrary vertex. On each of this vertex's three faces, there is exactly one other vertex not on a common edge, thereby defining a set of four vertices. Since any two of these four vertices are on a common face but not a common edge, they are all equidistant from each other ($\sqrt 2$ on a unit cube), and hence circumscribe a regular tetrahedron. The cube's six faces will each have one of the tetrahedron's six edges coincident with one of the face's two diagonals.
Consequently, a line through the center of two opposite faces of the cube will also go through the midpoint of two opposite (i.e. non-adjacent) edges of the tetrahedron. Rotating around this line by $90^\circ$ will map the cube to itself but switch the diagonal on each face to which the tetrahedron's edge is coincident. Rotating around this line by another $90^\circ$ will switch this again, thereby restoring the tetrahedron to its original position. This completes the proof.