Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site Navajo.ARPA Path: utzoo!linus!philabs!prls!amdimage!amdcad!amd!pesnta!greipa!decwrl!Glacier!Navajo!yeff 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??