Prove that the three lines joining the midpoints of opposite…

Prove that the three lines joining the midpoints of opposite edges of a tetrahedron meet at one point, and express that point through the vertices.

Explanation of Method

The segments joining midpoints of opposite edges of a tetrahedron meet at one point.

Step 1: Midpoints of opposite edges

Pair	Midpoints	Midpoint of the connector
$AB,\ CD$	$\tfrac{\vec a+\vec b}2,\ \tfrac{\vec c+\vec d}2$	$\tfrac{\vec a+\vec b+\vec c+\vec d}4$
$AC,\ BD$	$\tfrac{\vec a+\vec c}2,\ \tfrac{\vec b+\vec d}2$	$\tfrac{\vec a+\vec b+\vec c+\vec d}4$
$AD,\ BC$	$\tfrac{\vec a+\vec d}2,\ \tfrac{\vec b+\vec c}2$	$\tfrac{\vec a+\vec b+\vec c+\vec d}4$

Step 2: Conclusion

All three connecting segments have the same midpoint $G$, so they pass through $G$ and bisect each other there. $\blacksquare$

\[G=\frac{\vec a+\vec b+\vec c+\vec d}{4}=\left(\tfrac{x_1+x_2+x_3+x_4}4,\ \tfrac{y_1+y_2+y_3+y_4}4,\ \tfrac{z_1+z_2+z_3+z_4}4\right)\]

$G$ is the centroid of the tetrahedron. It also lies on each segment from a vertex to the centroid of the opposite face, dividing it in the ratio $3:1$.

$A$ $B$ $C$ $D$ $G$

$M_1M_2\ (AB,CD)$
$M_3M_4\ (AC,BD)$
$M_5M_6\ (AD,BC)$

The three segments joining midpoints of opposite edges meet at the centroid $G=\tfrac14(\vec a+\vec b+\vec c+\vec d)$ and bisect each other there.

Result

\[\boxed{\,G=\frac{\vec a+\vec b+\vec c+\vec d}{4}=\left(\tfrac{x_1+x_2+x_3+x_4}4,\ \tfrac{y_1+y_2+y_3+y_4}4,\ \tfrac{z_1+z_2+z_3+z_4}4\right)\,}\]