Draw to the vertices of a triangle ABC parallels and equals to sides of another triangle DEF. The resulting triangle GHI has the same centroid with the triangle ABC.
Consider the sides of triangle DEF as free vectors: DE, EF, FD. Their sum is zero. Taking an arbitrary origin O, the centroid or barycenter of the triangle ABC is the point J, such that OJ = (1/3)(OA+OB+OC). Analogously the centroid of GHI would be point J', such that OJ' = (1/3)(OG+OH+OI) = (1/3)([OA+AG] + [OB+BH] + [OC+CI]) = OJ, since AG+BH+CI=0. Look at the file Centroids.html for a nice application.