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/
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