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From: robert@brl-tgr.ARPA (Robert Shnidman )
Newsgroups: net.puzzle
Subject: Re: Geometry problem
Message-ID: <3119@brl-tgr.ARPA>
Date: Tue, 12-Nov-85 08:28:50 EST
Article-I.D.: brl-tgr.3119
Posted: Tue Nov 12 08:28:50 1985
Date-Received: Thu, 14-Nov-85 08:09:53 EST
References: <2966@brl-tgr.ARPA> <264@Navajo.ARPA>
Distribution: net
Organization: Ballistic Research Lab
Lines: 56

> >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
> 
> solve both eqns for c and we get:
>        c = (bd)/(a-d) ,  c = (ad)/(b-d)
> set these equal, eliminate the d in each numerator, cross-multiply (feels so
> good!!), and we get:
>        b^2 - bd = a^2 - ad
> 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")
> 
> so, this means that  a - b = 0  ==>  a = b  ==>  AC = BC      qed
> 
> aaaaahhhhhh!!!!!
> 
> jeff hasn't used geometry in like 4 years.....
> 
> it's nov 9, and it's 12:30 am.......do you know where your children are??

AD=BE was NOT given!  Solve the given problem.