Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site brl-tgr.ARPA Path: utzoo!watmath!clyde!cbosgd!ihnp4!qantel!hplabs!hao!seismo!brl-tgr!robert 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.