Medians of a TriangleDate: 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/ |
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