OABC is a tetrahedron with OA = a, OB = b, OC = c. P and Q are mid points of OA and BC respectively. Find in terms of a, b, c the position vector of the mid-point of PQ relative to O as the origin, and hence deduce that the lines joining the mid-points of opposite edges of a tetrahedron are concurrent and bisect each other.
First, $OA = a$, since P is at the mid-point of OA, $$OP = \frac{a}{2}$$
Using section formula:
$$OQ=\frac{b+c}{2}$$
Mid-point of PQ has a position vector of: $$OM=\frac{\frac{a+b+c}{2}}{2}=\frac{a+b+c}{4}$$
I don't know how to prove that they are concurrent and that they bisect each other.