The Medians of a Quadrilateral:

Let PQRS be a quadrilateral. Let's define the medians of the quadrilateral as the diagonals (segments PR and QS) and the segments joining midpoints of opposite sides. Show that the 4 medians of PQRS are concurrent iff PQRS is a parallelogram.

Solution:

A word on notation. I'll denote the position vector for a point P as P. Thus the vector from P to Q is -P + Q = Q- P.

If the 4 medians are concurrent, then change coordinates if necessary so that the point of intersection is 0. Since the diagonals intersect at 0, P = aR and Q = bS for negative numbers a and b. The other two medians, the "midpoint connectors" also intersect at 0, so (P+Q)/2 = c(R+S)/2 and (P+S)/2 = d(Q+R)/2 for negative numbers c and d. Then

P + Q = c(R + S)
aR + bS = cR + cS
(a - c)R = (c - b)S

But R and S cannot be parallel, since then P, Q, R and S would all be collinear. Therefore a - c = 0 = c - b, and a = b = c.

P + S= d(Q + R)
aR + S = bdS + dR
(a - d)R = (bd - 1)S

Once again, R and S cannot be parallel, so a = d and bd = 1. Thus a = b = c = d = -1, which means P = -R and Q = -S. Subtracting, Q - P = R - S, so edges PQ and SR are parallel and congruent, and PQRS is a parallelogram.

Conversely, if PQRS is a parallelogram, then Q - P = R - S, and P + R = Q + S. We can see that the diagonals bisect each other, since the midpoints of the diagonals are (P + R)/2 and (Q + S)/2 = (P + R)/2. The midpoints of the other two medians are [(P + Q)/2 + (R + S)/2]/2 = (P + R)/2 and [(P + S)/2 + (Q + R)/2]/2 = (P + R)/2. Thus the 4 medians are concurrent (and they all bisect each other!) Q.E.D.

If you're still interested, a third equivalent condition for the quadrilateral is that the center of gravity of the interior of the quadrilateral coincides with the center of gravity of its vertices; in other words the face mean is the same as the vertex mean.