Proving the Existence of the Centroid

Date: 11/16/1999 at 13:02:17
From: Stephen Orlando
Subject: Geometry: Proving Centroid Exists

The median in a triangle is a segment that connects a vertex to the 
midpoint of the opposite side. If you draw a median from each vertex 
in a triangle they always intersect at one point, where they are 
concurrent, called the centroid. This happens to be the center of 
gravity. There are other properties about this point such as the fact 
that the centroid is always 2/3 the distance from the vertex to the 
opposite side. I would like to know how to prove that the third median 
will always intersect the existing intersection of the other two 
medians. Most books seem to take this for granted and frankly I'm not 
sure how to prove it myself. 

Date: 11/16/1999 at 14:31:12
From: Doctor Peterson
Subject: Re: Geometry: Proving Centroid Exists

Hi, Stephen.

There are several ways to prove this; I'll give you the proof I recall 
from high school, which I found very memorable.

First, notice that we can prove the concurrence of the medians (the 
existence of the centroid) by proving that any two medians meet at a 
point 2/3 from the vertex to the side. Can you see why? If AD and BE 
meet 2/3 of the way along AD, and AD and CF meet at a point 2/3 of the 
way along AD, then all three lines meet at that point.

So let's prove that part:

                  B
                  +
                 / \\
                /  \  \
               /   \    \
              /     \     \
             /      \       \
          F +        \        + D
           / \       \    /    \\
          /  \        +M        \ \
         /    \   /    \        \   \
        /     +N       \         \    \
       /  /    \        \         \     \
      +--------+--------+---------+-------+
     A         X        E         Y        C

I want to show that M, where AD and BE intersect, divides AD in the 
ratio 2:1. I just draw two lines parallel to BE through the midpoints 
of the other two sides: FX and DY. There's a theorem that parallel 
lines divide all transversals in the same proportion (you can use 
similar triangles to prove this easily), so since AF = FB, we know 
that AX = XE, and since BD = DC, we know that EY = YC. This makes AX, 
XE, EY, and YC all the same length, since E is the midpoint of AC. 
Now, looking at AD as a transversal of the same parallel lines, we can 
see that AN = NM = MD, so I've divided AD into three equal parts and M 
lies 2/3 of the way from A to D.

This can also be seen as a special case of Ceva's Theorem; you can 
find both that and several explanations of the centroid by searching 
the Dr. Math archives.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/