i. Show that the lines joining the vertices of a tetrahedron to the centroids of opposite faces are concurrent.
ii. Show that the joins of the midpoints of the opposite edges of a tetrahedron intersect and bisect each other.

Explanation

i. `G_(1)`, the centroid of `DeltaBCD`, is `(vecb+vecc+vecd)/(3) and A` is `veca`.
The position vector of point G which divides `AG_(1)` in the ratio 3 : 1 is
`" "(3*(vecb+vecc+vecd)/(3)+1*veca)/(3+1)=(veca+vecb+vecc+vecd)/(4)`
The symmetry of the result shows that this point will also lie on `BG_(2)`, `CG_(3) and DG_(4)` (where `G_(2), G_(3), G_(4)` are centroids of faces ACD, ABD and ABC, respectively). Hence, these four lines are concurrent at point `(veca+vecb+vecc+vecd)/(4)` , which is called the centroid of the tetrahedron.
ii. The midpoint of `DA` is `(veca+vecd)/(2)` and the of `BC` is `(vecb+vecc)/(2)` and the midpoint of these midpoint is `(veca+vecb+vecc+vecd)/(4)` and symmetry of the result proves the fact.