From: Ken.Pledger@vuw.ac.nz (Ken Pledger) Subject: Re: External Bisectors Date: Thu, 01 Apr 1999 10:37:32 +1200 Newsgroups: sci.math Keywords: Steiner-Lehmus Theorem In article <0r1xhi5rul9m@forum.swarthmore.edu>, alpha@ome.ga (Alpha Omega) wrote: > In a triangle ABC, we have that b_2 = b_3 > (external bisectors of angles B,C). > > Is the triangle isosceles? > > Alpha Yes. This is a minor variant of the Steiner-Lehmus Theorem concerning internal bisectors, which is notoriously difficult to prove. The neatest proof I've seen is by Gilbert and MacDonnell, in the M.A.A. "Selected Papers on Geometry," (1979) pp.127-8; reprinted from the American Mathematical Monthly 70 (1963), 79-80. You may like to try adapting this or another proof to your case. Ken Pledger. ============================================================================== From: Andres Koropecki Subject: Re: tricky proof Date: Sat, 24 Apr 1999 15:32:44 GMT Newsgroups: sci.math In article <3721BD4A.22C221BB@accesstoledo.com>, > Six years later I'm still stumped. At this point I'm not even sure it's > a true proposition, though I seem to remember convincing myself it was > at some point. I tried solving it in a Euclidean way and gave up > (flipping through the Elements didn't help) and every once in a while I > try to solve it analytically, but run into nasty irreducible equations, > no proof. > > Anyway, it's pretty simple sounding: > > Given that two of a triangle's angle bisectors are equal in length; > Prove that the triangle is isosceles. This is known as Lehmus-Steiner theorem, and the proof is not hard, not long!!! The hard, maybe, is to conceive it. You can find the proof, and some historical stuff about it in a geometry book from H. S. M. Coxeter. I don't know the name in english, but it's something like "Return to geometry" or "Back into geometry". The idea of the proof is the following: (1) Prove that if two chords on a circle subtend different acute angles from a point on the circle, the smallest angle is subtended by the smallest chord. (2) Prove that if a triangle has two different angles, the smallest angle has a (interior) bisector smaller than the (interior) bisector of the bigest angle. (of course, use (1)). (3) Prove the theorme by proving its contrapositive (if the two angles are different then the bisectors are different). Here's nothing to do, actually. That's it. Regards, Andres Koropecki --- akoro@math.unl.edu.ar -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ============================================================================== From: Clive Tooth Subject: Re: tricky proof Date: Sat, 24 Apr 1999 17:00:45 +0100 Newsgroups: sci.math Andres Koropecki wrote: > In article <3721BD4A.22C221BB@accesstoledo.com>, > > Six years later I'm still stumped. At this point I'm not even sure it's > > a true proposition, though I seem to remember convincing myself it was > > at some point. I tried solving it in a Euclidean way and gave up > > (flipping through the Elements didn't help) and every once in a while I > > try to solve it analytically, but run into nasty irreducible equations, > > no proof. > > > > Anyway, it's pretty simple sounding: > > > > Given that two of a triangle's angle bisectors are equal in length; > > Prove that the triangle is isosceles. > > This is known as Lehmus-Steiner theorem, and the proof is not hard, not > long!!! The hard, maybe, is to conceive it. You can find the proof, and some > historical stuff about it in a geometry book from H. S. M. Coxeter. I don't > know the name in english, but it's something like "Return to geometry" or > "Back into geometry". See Coxeter's "Introduction to Geometry", Second Edition, page 420, for a different proof from that given below. He mentions that J A McBride has asserted that more than 60 proofs have been given. > The idea of the proof is the following: (1) Prove that > if two chords on a circle subtend different acute angles from a point on the > circle, the smallest angle is subtended by the smallest chord. (2) Prove that > if a triangle has two different angles, the smallest angle has a (interior) > bisector smaller than the (interior) bisector of the bigest angle. (of > course, use (1)). (3) Prove the theorme by proving its contrapositive (if the > two angles are different then the bisectors are different). Here's nothing to > do, actually. That's it. Regards, -- Clive Tooth http://www.pisquaredoversix.force9.co.uk/ End of document ============================================================================== From: lk3a@esmeralda.mech.Virginia.EDU (Count Dracula) Subject: Re: tricky proof Date: 25 Apr 1999 16:20:07 GMT Newsgroups: sci.math Matt Fritts writes: > Given that two of a triangle's angle bisectors are equal in length; > Prove that the triangle is isosceles. Let ABC be the given triangle and denote by iA the length of the bisector of the angle at A, by iB the length of the bisector of the angle at B. Let p = (a + b + c)/2 = semi-perimeter of the triangle, a, b, c, being the side lengths opposite A, B, and C, respectively. Formulas are known for iA and iB : iA = 2/(b + c) Sqrt[p b c (p - a)] iB = 2/(a + c) Sqrt[p a c (p - b)] See for example Berger Volume I for these. In most texts, they are discussed where Heron's formula is mentioned. Write the given condition as iA^2 = iB^2. This condition simplifies to (a - b) ( b a^2 + (b^2 + 3 b c + c^2) a + c^2 (b + c) ) = 0 The quadratic factor in a on the right has no positive real root for a. Therefore a = b. -- _________________________________________________________________________ Levent Kitis lk3a@esmeralda.mech.virginia.edu University of Virginia Department of Mechanical and Aerospace Engineering _________________________________________________________________________ ============================================================================== Coxeter's "Introduction to Geometry" (Wiley; 2d Ed., 1980) has this as the last exercise in section 1.1: "Prove the Steiner-Lehmus theorem: Any triangle having two equal internal angle bisectors (each measured from a vertex to the opposite side) is isosceles. (Hint [*] If a triangle has two different angles, the smaller angle has the longer internal bisector.)" [*] The footnote is a citation to Court, "College Geometry", Barnes and Noble, 1952, p.72, and the pointer, "For Lehmus's proof of 1848, see Coxeter and Geitzer [Geometry Revisited, Random House, 1967], p. 15"