Proving the Existence of the CentroidDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/