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From: yeff@Navajo.ARPA
Newsgroups: net.puzzle
Subject: Re: Geometry problem
Message-ID: <264@Navajo.ARPA>
Date: Sat, 9-Nov-85 03:29:52 EST
Article-I.D.: Navajo.264
Posted: Sat Nov  9 03:29:52 1985
Date-Received: Wed, 13-Nov-85 03:22:47 EST
References: <2966@brl-tgr.ARPA>
Reply-To: yeff@Navajo.UUCP (Jeff Soesbe)
Distribution: net
Organization: Stanford University
Lines: 54

>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??