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Triangle Proof with Contradiction

Date: 02/21/2004 at 17:47:04
From: Matt
Subject: Proof with triangle ABC

Let D, E lie internally on side BC of triangle ABC and consider the
following conditions:

1) angle BAD = angle DAE = angle EAC
2) |BD| = |DE| = |EC|

Prove that, whatever the shape of triangle ABC, 1) and 2) cannot both
be true, that is, if either is true, then the other is false.

I am struggling with this question and therefore would really
appreciate some help.



Date: 02/21/2004 at 21:20:39
From: Doctor Rob
Subject: Re: Proof with triangle ABC

Thanks for writing to Ask Dr. Math, Matt!

First, assume that both conditions (1) and (2) hold.

Then AD bisects <BAE by (1).  Then |AB|/|BD| = |AE|/|DE|, by the angle
bisector theorem.  That implies that |AB| = |AE| by (2).  Then
triangles ABD and AED are congruent by (2) and Side Side Side.  That
implies that AD is perpendicular to BC.

Use the same argument on AE, the bisector of <DAC to show that AE is
also perpendicular to BC.

Putting these together, the triangle ADE then has angles whose
measures add to more than the measure of two right angles.

This contradiction shows that (1) and (2) cannot both be true.

Feel free to reply if I can help further with this question.

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



Date: 02/28/2004 at 16:22:52
From: Matt
Subject: Thank you (Proof with triangle ABC)

Hi Doctor Rob,

Thanks for you reply.  I found it very helpful.

Matt
Associated Topics:
High School Triangles and Other Polygons

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