Medians of a Triangle

Date: 02/16/99 at 22:06:24
From: Jin Park
Subject: Medians of a Triangle

I learned in school that the 3 medians of a triangle divide themselves 
up into a ratio of 1:2. I clearly understand what this means but could 
you give me a proof of it (that the medians of triangle divide 
themselves up into a ratio of 1:2)?  I guess I'm not stuck on anything 
right now because I do not know where to start!

Thanks,
Jin Park

Date: 02/17/99 at 02:54:03
From: Doctor Floor
Subject: Re: Medians of a Triangle

Let us consider a triangle ABC. The midpoints of the sides form a 
triangle A'B'C'. AA', BB' and CC' meet in a point G (known as the 
centroid):

  

First we notice that triangle A'B'C' is similar to ABC; the sides of 
A'B'C' have half the length of the the sidelengths of ABC. We can see 
that, because for instance B'C' is an image of BC from multiplication 
over A with factor 1/2. So indeed B'C' must be half the length of BC.

Also we notice that AA' passes through the midpoint of B'C', since in 
the multiplication mentioned above, A' moves to the midpoint of B'C'. 
Similar results are found for BB' and CC'. But that means that the 
medians of ABC are also medians of A'B'C'.

We now can conclude that the distance AG in ABC corresponds to A'G in 
A'B'C'. From the observation that A'B'C' has lengths half of those in 
ABC, we can conclude that AG:A'G = 2:1. And that is what we wanted to 
prove.

I hope this helps!

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/