# The Medians

Below I offer a proof to the well known

### Theorem

Three medians of a triangle meet at a point - *centroid* of the triangle.

The proof is due to David Ramsey, a graduate student at the University of Chicago; it is based on the following

### Lemma

The medians of a triangle serve as the medians of its medial triangle.

### Proof of Lemma

Let MM_{a}, MM_{b}, MM_{c} be the three medians of ΔABC such that M_{a}M_{b}M_{b} is its medial triangle. I'll prove that MM_{a} is also a median of ΔM_{a}M_{b}M_{c}.

The midlines in a triangle are parallel to the corresponding sides, in particular, M_{b}M_{c}||BC and M_{a}M_{c}||AC, implying that the quadrilateral CM_{b}M_{c}M_{a} is parallelogram. In a parallelogram the diagonals are divided into halves by their point of intersection, D in the above diagram. Therefore, M_{c}D is a median of ΔM_{a}M_{b}M_{c}.

In addition, we may observe that the medial lines cut a triangle into four smaller ones, all equal, such that the area of each is 1/4 of the area of the reference triangle.

### Proof of Theorem

Assume to the contrary that the medians in a triangle are not concurrent:

Their three points of intersection form a triangle, say, ΔDEF which is located entirely within (i.e., in the interior) of ΔABC. This implies that the area ΔDEF is less than that of ΔABC. Observe that, by our assumption, the area of ΔDEF could not be zero, for that would imply that it's degenerate. However, if the three points of intersection of the medians are collinear some two intersect on the third, meaning that the three are concurrent.

By Lemma, the medians of the medial triangle ΔM_{a}M_{b}M_{c} are those of ΔABC. Thus we can iterate: the medians of ΔM_{a}M_{b}M_{c} form (the same) ΔDEF which lies entirely within ΔM_{a}M_{b}M_{c} and so its area is less than that of ΔM_{a}M_{b}M_{c}. But the latter is just 1/4 of the area of ΔABC. Repeating the steps leads to

Area(ΔDEF) < 4^{-n}·Area(ΔABC),

for any positive integer n, implying that Area(ΔDEF) = 0. But a nondegenerate triangle has positive area; and we are done.

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

65261782 |