$ABCD$ a tetrahedron. $P$ and $Q$ be points on the sides $AB$ and $CD$ respectively. What is the locus of the midpoint of the line segment $PQ$? Let $ABCD$ a tetrahedron. $P$ and $Q$ be points on the sides $AB$ and $CD$ respectively. What is the locus of the midpoint of the line segment $PQ$?
The locus is the set of points $X$ satisfying the condition that $X$ is the midpoint of the line segment $PQ$. How do I go about finding and describing this set?

Denote points of tetrahedron with $A(0, 0, 0)$, $B(a, 0, 0)$, $C(\frac{a}2,\frac{a\sqrt 3}{2}, 0)$, $D(\frac{a}2,\frac{a\sqrt3}{6},\frac{a\sqrt6}3)$
Arbitrary points $P\in AB$ and $Q\in CD$ have the following coordinates:
$$P(pa,0,0), Q(\frac{a}2, \frac{a\sqrt3}{6}(3-2q),\frac{aq\sqrt6}3)$$
...with:
$$p,q\in[0,1]$$
Coordinates of the midpoint are:
$$M(\frac{a}4(2p+1), \frac{a\sqrt3}{12}(3-2q),\frac{aq\sqrt6}6)$$
or:
$$x=\frac{a}4(2p+1)\tag{1}$$
$$y=\frac{a\sqrt3}{12}(3-2q),\space z=\frac{aq\sqrt6}6\tag{2}$$
Obviously, you can pick the value for $x$ independently of $y,z$. 
The range for $x$ is:
$$x_{p=0}=\frac{a}4,\space x_{p=1}=\frac{3a}4,$$
For any given $x$ the locus of points is given with (2) which is actually a straight line going from:
$$y_{q=0}=a\frac{\sqrt{3}}4, \space z_{q=0}=0$$
...to:
$$y_{q=1}=a\frac{\sqrt{3}}{12}, \space z_{q=1}=a\frac{\sqrt 6}6$$
So the locus of midpoints is actually a rectangle with the following coordinates:
$$(\frac{a}4, a\frac{\sqrt{3}}4, 0),\space(\frac{a}4, a\frac{\sqrt{3}}{12}, a\frac{\sqrt 6}6)$$
$$(\frac{3a}4, a\frac{\sqrt{3}}4, 0),\space(\frac{3a}4, a\frac{\sqrt{3}}{12}, a\frac{\sqrt 6}6)$$
