# A Characteristic Property of Centroid

In ΔABC, a line is drawn through centroid G. Assume the line intersects AB in M and AC in N. Then

(1) | BM/MA + CN/NA = 1. |

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

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## Proof

Let M_{a} be the midpoint of side BC. Drop perpendiculars BD, M_{a}E, and CF onto the given line. Obviosly

(2) | M_{a}E = (BD + CF)/2. |

Let also AL be perpendicular to MN. Triangles ALG and M_{a}EG are similar and _{a}G._{a}E, or

(2') | LA = BD + CF. |

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

BM/MA + CN/NA | = BD/LA + CF/LA |

= (BD + CF)/LA | |

= LA/LA (from 2') | |

= 1. |

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

(1') | BM'/M'A + CN'/N'A = 1 |

holds for any line through P that intersects AB in M' and AC in N'. We have to show that

(3) | BM/MA + CN/NA | < BM'/M'A + CN'/N'A. |

This is because, in the diagram,

BM/MA + CN/NA < 1,

which contradicts the proven part.

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Copyright © 1996-2018 Alexander Bogomolny