The points A, B, C, D have position vectors a, b, c and d respectively. Prove that the lines joining the midpoints of opposite edges of the tetrahedron ABCD bisect each other and give the position vector of the point of intersection.
I have started by working out the position vectors of E and F:
b + $\vec BA$ = a
so $\vec BA$ = a – b and position vector of E is $\frac{(a – b)}{2}$
and that for F is $\frac{d – c}{2}$.
But I can proceed no further.
Best Answer
Let $M_1$ be the midpoint of $[EF]$ and $M_2$ the midpoint of $[GH]$. Let us prove that $M_1=M_2$.
The following sum $$\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}$$
can be written in two ways:
$$1) \ \ (\vec{OA}+\vec{OB})+(\vec{OC}+\vec{OD})=2\vec{OE}+2\vec{OF}=4\vec{OM_1}$$
$$2) \ \ (\vec{OA}+\vec{OC})+(\vec{OB}+\vec{OD})=2\vec{OH}+2\vec{OG}=4\vec{OM_2}$$
Idintifying these results, one gets $\vec{OM_1}=\vec{OM_2}$, therefore $M_1=M_2$.
Consequently, the position vector is:
$$\vec{OM_1}=\vec{OM_2}=\dfrac14(\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD})$$
Remark: This issue, like many others, is based on a barycentric reasoning: if $A,B,C,D$ have weight $1$, midpoints $E$ and $F$ for example have weights $2$, therefore one can concentrate the total weight, $4$: on $M_1$, the midpoint of midpoints.