Proof

Let M_a be the midpoint of side BC. Drop perpendiculars BD, M_aE, and CF onto the given line. Obviosly

Let also AL be perpendicular to MN. Triangles ALG and M_aEG are similar and GA = 2·M_aG. Therefore, LA = 2M_aE, or

Triangles BDM and ALM are similar, as are triangles CFN and ALN, from where we get

Let's now tackle the converse. Assume point P is such that

holds for any line through P that intersects AB in M' and AC in N'. We have to show that P = G.. To this end assume that P is different from G and that the line is different from GP. Let MN passes through G parallel to M'N'. Then, with the reference to the diagram above,

This is because, in the diagram, BM/MA < BM'/M'A and CN/NA < CN'/N'A. For other locations of P or the straight line, the inequalities may have to be simultaneously reversed. (3) implies

which contradicts the proven part.