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Trisecting an Angle and the Opposite Side in a Triangle

Date: 09/03/2008 at 22:33:18
From: Sean
Subject: Geometry/ Trisector proof help

Prove that it is impossible to have a triangle in which the trisectors
of an angle also trisect the opposite side.

I am unsure how to prove this. It seems that if i trisect the angles 
of an equilateral triangle so that each of the trisected angles is 
20, it would indeed divide the opposite side into 3 equal pieces.

I have completed geometry, and I have tried several things. I got 
started trying to use exterior angle theorem, but got confused. I 
think that may be a good way to do it, but I dead ended after 
extending two of the sides to create isosceles triangles. I think it 
may be all of the criss-crossing lines that is getting me confused.

Date: 09/04/2008 at 12:00:21
From: Doctor Peterson
Subject: Re: Geometry/ Trisector proof help

Hi, Sean.

I don't think I've ever tried proving this, but it's a very nice 
little theorem!

It may seem as if trisecting an angle in an equilateral triangle would
work, but it is not true.

I think I'd approach it by contradiction.  Suppose that you have a 
triangle ABC with trisectors BD and BE, D and E being on AC, and 
further suppose that D and E trisect AC, so that AD = DE = EC:

         / | \
        / | | \
       /  | |  \
      /  |   |  \
     /   |   |   \
    /   |     |   \
   /    |     |    \

(The picture is not meant to suggest that ABC is isosceles; it's just 
easiest to "draw" this way.)

Now focus on triangle ABE.  Here BD bisects angle ABE, while D is the 
midpoint of AE; so BD is both an angle bisector and a median.  What 
does that imply?  Repeat with triangle DBC, and look for a 

There may be many other ways to approach this, so if you see any 
ideas of your own while you try this out, go ahead and pursue them!

If you need more help, please write back and show me how far you got.

- Doctor Peterson, The Math Forum 

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Associated Topics:
High School Triangles and Other Polygons

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