Bisecting line segments in a tetrahedron.

Question

Suppose that $OABC$ is a regular tetrahedron with base $ABC$. Suppose further that $T$ is the mid-edge of $AC$, $Q$ is the mid-edge of $OB$, $P$ is the mid-edge of $OA$, and $U$ is the mid-edge of $CB$.

How can one show that $QT$ bisects $PU$?

Answer 1

One nice way to see this is to realize you can determine a regular tetrahedron by taking a set of (four) non-adjacent vertices of a cube. Each of the six edges of this tetrahedron is a diagonal of a face of that cube, so each "mid-edge" point of the tetrahedron is the center of a cube-face. Any of the segments joining opposite "mid-edge" points, then, is perpendicular to the corresponding cube-faces, and passes through the cube's center (which bisects each of those segments).

Answer 2

The midpoint of either $QT$ or $PU$ is simply $\frac{O+A+B+C}4$.