Tetrahedron vector problem

Question

Can you give me hints on how to solve this problem. Prove that three line segments which connect middle of sides of tetrahedron which don't lie on the same plane go through the same point. How to prove intersection with vectors? I don't even know how to attack this problem.

Answer 1

The geometry of the problem is not affected by a rotation or by a scaling. so without loss of generality we can tabulate the four point as: $$ \left( {\begin{array}{*{20}c} A \\ B \\ C \\ D \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} 0 & 0 & 0 \\ 2 & 0 & 0 \\ {2a} & {2b} & 0 \\ {2c} & {2d} & {2e} \\ \end{array} } \right) $$ The midpoint of the segment $AB$ will be $$ M_{A\,B} = \left( {A + B} \right)/2 $$ and analoguously for the remaining 5 combinations.
The three segments that connect the middle of two non-complanary sides will be those with end points $$ \left( {M_{A\,X} ,M_{Y\,Z} } \right) $$ (one end on one of the 3 segments from $A$ to $X$, the other end on the segment connecting the other 2 points) i.e. those lying on the three lines having parametric equations $$ M_{A\,X} + \lambda _X \left( {M_{Y\,Z} - M_{A\,X} } \right) $$ and we shall demonstrate that such 3 lines are concurrent:
can you take on from here?