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Geometric Proof of the Steiner-Lehmus Theorem

Date: 05/13/2002 at 01:42:14
From: Jim
Subject: Plane Geometry

Prove:  If two bisectors of two angles of a triangle are equal then 
the triangle is an isosceles triangle.

What I am looking for is a "geometric proof" where algebra isn't 
involved.  I suppose if one were to use trigonometry and analytic 
geometry the proof would be even easier.  

I've seen a geometric proof that was quite long and involved but I've 
misplaced it.  But it seems as if such a simple problem should have a 
simple (geometric) solution!


Date: 05/13/2002 at 16:16:52
From: Doctor Floor
Subject: Re: Plane Geometry

This is called the Steiner-Lehmus theorem.  Here is a geometric proof:

We consider given that ABC is a triangle, AD and BE bisect angles A 
and B such that AD = BE.

Start by constructing F such that AF = AE and DF = AB. Let G be the 
intersection of AD and BE.

We find ADF and EBA are congruent, so 

  <FAD = <AEB and <ADF = 1/2 <B

In triangles AGE and BGD we find

    <AEB + 1/2 <A = <ADB + 1/2 <B

    <FAD + 1/2 <A = <ADB + <ADF

             <FAB = <BDF

We also find that these final angles are 

   <AEB + 1/2 <A = <AGB 

                 = 180 -(<A + <B)/2 > pi/2

From this we can conclude the congruence of BAF and FDB despite only
having SSA, because the unknown angles must be acute. So DB = AF = AE
and thus 1/2 <A = 1/2 <B, and we see that ABC is isosceles.

If you have more questions, just write back.

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


Date: 05/14/2002 at 02:08:55
From: Jim
Subject: Plane Geometry

Thank you very much!  That was exactly what I was looking for.  Very 
short and not too complicated.  Bravo.


Date: 10/14/2003 at 00:34:22
From: Daniel
Subject: Geometric Proof of the Steiner-Lehmus Theorem

I was happy to find a simple geometric proof of the Steiner-Lehmus
theorem on your site, but as I was reading through the proof there 
was one step which I just could not understand how it was done.

In triangles AGE and BGD we find

    <AEB + 1/2 <A = <ADB + 1/2 <B

    <FAD + 1/2 <A = <ADB + <ADF

             <FAB = <BDF    <----- This is the step

I do not understand how this follows from the previous steps.  It 
would evidently follow from <FAD - 1/2 <A = <ADB - <ADF 
but not from <FAD + 1/2<a = <ADB + <ADF

Any help would be appreciated,

Daniel


Date: 10/14/2003 at 12:04:09
From: Doctor Peterson
Subject: Re: Geometric Proof of the Steiner-Lehmus Theorem

Hi, Daniel.

The proof should state which of two possible points is to be chosen 
as F; when I tried going through the theorem just now, I made the 
same mistake you made, and drew F below AD rather than above AD (at 
F' below):

                              C
                             / \
                            /   \
                           /     \
                          /       \
                         /         \
                        /           \
                       /             \
                      /               \
     F-----------------                \
      \             E  -----------------D
      \            /   \             / / \
       \          /       \       /   /   \
       \         /            G      /     \
        \       /          /     \  /       \
         \     /        /          /\        \
         \    /      /            /    \      \
          \  /   /               /         \   \
          \ / /                  /            \ \
           A--------------------/----------------B
             \                 /
                \             /
                  \          /
                     \      /
                        \  /
                          F'

With F on the same side of AD as C, as shown, the angles add up 
so that

  <FAD + 1/2 <A = <FAB
  <ADB + <ADF   = <BDF

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/

Associated Topics:
High School Triangles and Other Polygons

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