We are given a triangle ABC with medial triangle DEF (D on BC, E on AC, F on AB). It is easy to prove, assuming that the medians of a triangle concurr at the centroid G, that there is a homothety with ratio -2 centered at G that maps DEF to ABC. The proof is as follows: parallel lines are clearly homothetic; given XY and X'Y' parallel, there is an external center of homothety at the intersection of XX' and YY' and an internal center at the intersection of XY' and X'Y. Thus the intersection of AD and CF (G) is an internal center of homothety which maps FD to AC with ratio -2 (since FD is half of AC). But AD and BE also intersect at G so, G maps DE to AB. Similarly G maps FE to BC.
There is an excercise in my notes which says to prove that the medians of a triangle concurr using homothety. The proof given is as follows: the midline of a triangle is parallel to the corresponding side and half the length, so there exists a homothety mapping ABC to DEF. The center of homothety is the point of concurrency of the medians.
The second statement makes sense, since if A maps to D then A,D, and the center of homothety are collinear (by definition). But the first statement doesn't seem obvious to me. Why does corresponding sides parallel imply there exists a homothety mapping ABC to DEF?