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From: bs@faron.UUCP (Robert D. Silverman)
Newsgroups: net.puzzle
Subject: Re: Geometry problem
Message-ID: <382@faron.UUCP>
Date: Thu, 14-Nov-85 13:59:53 EST
Article-I.D.: faron.382
Posted: Thu Nov 14 13:59:53 1985
Date-Received: Fri, 15-Nov-85 20:11:29 EST
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Organization: The MITRE Coporation, Bedford, MA
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> > 
> > take everything over to one side and factor and we get:
> >        (b - a) * (b + a - d) = 0
> > 
> > sooooo, either  b - a = 0, or b+a-d=0...
> > 
> > the second is impossible, bcuz  a>d and b>0 (this is "intuitively obvious")
> 			    ^^^^^^^^^^
> 	This is not necessarily true (a is less than d if a < c), but
> 	a+b-d=0 is still impossible as follows:
> 	a+b-d = AB + BC - AE
> 	      = AB + BE - AE + EC
> 	Now, as ABE form a triangle, AB + BE > AE.
> 	 .
> 	. .  AB + BE - AE + AC > 0
> 
> -- 
> Yogesh Gupta                           Advanced Systems Group,
> {sdcrdcf, sdcsvax}!bmcg!asgb!gupta     Burroughs Corp., Boulder, CO.
> --------------------------------------------------------------------
> 	All opinions contained in this message are my own and do not
> 	reflect those of my employer or the plant on my desk.

It is my opinion that the proof is still incomplete. His original assertion
that the angle bisector divides the side into the same ratio as the adjoining
sides can hardly be called an axiom and is far from obvious. I, in fact
hadn't remembered it as a theorem until the proof reminded me. To qualify
as a rigorous proof either a reference should be made for this 
assertion or it should be proved separately as a lemma.
 
I know of two other proofs. One is purely trigonometric and the other uses
a theorem of Fermat (again far from obvious) that the 3 pairs of opposite
sides of a hexagon which is inscribed in an ellipse , if extended, intersect
at three points which are co-linear.  The relation to an ellipse becomes 
obvious if you study the problem for a little bit. That's all I'm going to say.
 
Bob Silverman   (they call me Mr. 9)