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Triangle's Medians Make Smaller Triangles with Equal Area


Date: 04/15/99 at 00:33:18
From: Courtney Peurasaari
Subject: Area of triangles

I am studying to be a teacher at MSU and I have to take a geometry 
course, which is quite difficult for me. In class recently my teacher 
posed the problem: given a triangle in which you construct the three 
medians, resulting in six triangles, why is the area of each of these 
six triangles the same? Now, I know that if you draw only one median, 
it spilts the triangle in half. I was playing around with this on 
Geometers Sketchpad, by doing things such as getting the triangle to 
form a parallelogram, but that just got complicated. I think there 
must be a simpler answer. So, if you can help me in explaining that no 
matter what kind of triangle it is, when you find the medians six new 
triangles form and the area to these six triangles is the same, it 
would be very helpful.

Thank You, 
Courtney Peurasaari


Date: 04/15/99 at 11:49:05
From: Doctor Rob
Subject: Re: Area of triangles

Thanks for writing to Ask Dr. Math!

I don't know what the simplest way to see this is, but here is how I
figured this out.

Let the triangle be ABC, and let the midpoints of the sides be P (on 
BC), Q (on AC), and R (on AB). Let the intersection of the medians be 
O.

First, the areas of AOR and BOR are the same because OR is the median
from O to AB. The same is true of the areas of AOQ and COQ, and BOP 
and COP.

Next, you know that the areas of APB and APC are equal because AP is 
the median from A to BC. That means that the areas of AOB and AOC are 
equal. Dividing by 2, that means that the areas of AOR and AOQ are 
equal.

Do the same with BQA and BQC, concluding that BOR and BOP are equal.

Do the same with CRA and CRB, concluding that COP and COQ are equal.

Putting this all together, the six small triangles all have the same
area.

Another approach would be to show that the perpendicular distance from 
Q to AP equals half the distance from C to AP, which equals half the
distance from B to AP, which equals the distance from R to AP, so that
AOQ and AOR have the same base AO and equal altitudes, so equal areas.
Then do the same for BOP and BOR, and for COP and COQ.

Another approach would be to use the fact that OP = AP/3, OQ = BQ/3,
and OR = CR/3, provided you are aware of that.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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