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A Characteristic Property of Centroid
In
Here, BM, MA, CN, NA are considered as signed segments. In a certain sense the identity even holds when the line in question is parallel to, say, AB. In this case, M is a point at infinity and 
Copyright © 1996-2008 Alexander Bogomolny
Proof
Let Ma be the midpoint of side BC. Drop perpendiculars BD, MaE, and CF onto the given line. Obviosly
Let also AL be perpendicular to MN. Triangles ALG and MaEG are similar and
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
This is because, in the diagram,
which contradicts the proven part.
Copyright © 1996-2008 Alexander Bogomolny |
ABC, a line is drawn through centroid G. Assume the line intersects AB in M and AC in N. Then
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