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From: bill@ur-cvsvax.UUCP (Bill Vaughn)
Newsgroups: net.puzzle
Subject: Re: Re: Geometry problem
Message-ID: <242@ur-cvsvax.UUCP>
Date: Thu, 14-Nov-85 01:00:21 EST
Article-I.D.: ur-cvsva.242
Posted: Thu Nov 14 01:00:21 1985
Date-Received: Fri, 15-Nov-85 20:30:20 EST
References: <2966@brl-tgr.ARPA> <264@Navajo.ARPA>
Distribution: net
Organization: Center for Visual Science, U. of Rochester
Lines: 45

> >Given: Triangle ABC with angle bisectors AE and BD such that
> >AE=BD.
> >Prove: AC=BC.
> >                           C
> >                          /\
> >                         /  \
> >                        /    \
> >                      D/.    .\E
> >                      /   ..   \
> >	                /   .  .   \
> >                    /  .      .  \
> >                   / .          . \
> >                  /.              .\
> >                A/ ________________ \B
> 
> 
> >This problem is harder that it looks.
> 
> ok....here we go!!
> 
> let AC=a,BC=B,AB=c,AD=BE=d
> 
> now we all know that an angle bisector divides the side it hits into a ratio
> that is equal to the ratio of the sides....
> 
> so we have:
>        AD/DC = AB/AC  ,   BE/DE = BA/BC
> 
> or, in terms of the lengths:
>        d/(a-d) = c/b  ,  e/(b-d) = c/a
> 
(etc.)

Sorry, but no cigar.  The conditions of the theorem are AE = BD
NOT AD=BE.

One essentially has to PROVE that AD = BE because from this it follows
easily that ABC is isosceles.

(HINT: Proving the contrapositive looks like it should be easier. Thats
 the angle ( :-) )  I'm taking.)

Bill Vaughn
U of Rochester
ur-cvsvax!bill@rochester.arpa