Taking a rather more sophisticated approach we may get another proof without vectors - using a generalisation found in 2013.
First redefine the term
bimedian for
![](https://latex.codecogs.com/png.latex?\bg_white n)
points as the segment joining the midpoint of one segment to the barycenter of the remaining
![](https://latex.codecogs.com/png.latex?\bg_white n-2)
points. Then the sum of the squares of the
![](https://latex.codecogs.com/png.latex?\bg_white n(n-1)/2)
bimedians is equal to
![](https://latex.codecogs.com/png.latex?\bg_white n/(4n-8))
times the sum of the squares of all segments joining the
![](https://latex.codecogs.com/png.latex?\bg_white n)
points.
In the case
![](https://latex.codecogs.com/png.latex?\bg_white n=4)
, the six bimedians coincide two by two. This explains why, in this case, we now have one half of the sum of the squares of its edges, instead of one fourth.
Fonda, A.,
On a Geometrical Formula Involving Medians and Bimedians, Mathematics Magazine, Vol. 86, No. 5 (December 2013), pp. 351-357