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
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.