A simple geometric proof

Prove that the three medians of a triangle intersect at a single point. Please show work. Thanks in advance!

The proof using co-ordinate geometry is as under:

Let the vertices of the triangle be A (x1, y1), B (x2, y2) and C (x3, y3).
Mid-point, D, of BC is [(x2+x3)/2, (y2+y3)/2]
A point G on AD dividing it in the ratio 2 : 1 from A has x-coordinate given by
[2 * (x2+x3)/2 + x1]/(2 + 1)
= (1/3) (x1+x2+x3).
Similarly, y-coordinate is given by
(1/3) (y1+y2+y3).
If E and F are the midpoints of CA and AB, it can be shown that the points dividing BE and CF from B and C respectively in the ratio 2 : 1 has the same x- and y-coordinates as G
=> G is the point of concurrence of the medians of the triangle ABC.

The proof using vectors is as under:

Let the position vectors of A, B and C be a, b and c respectively.
The midpoint, D of BC has position vector (b+c)/2.
The point G, dividing AD in the ratio 2:1 has position vector given by
[2*(b+c)/2 + a]/3 = (a+b+c)/3.
Similarly, if E and F are the midpoints of CA and AB, then the points dividing BE and CF in the ratio 2 : 1 from B and C respectively will have position-vectors given by (a+b+c)/3 which is the same as of G
=> G is the point of concurrence of the medians of the triangle ABC.