From: David Shield Newsgroups: sci.math Subject: Re: GEOMETRY PROBLEM Date: 30 Nov 1995 03:07:51 GMT rkleiner@moe.coe.uga.edu (Ronen Kleiner M) wrote: >Consider any triangle ABC. Let P be any point in the interior of ABC. >Let D,E, and F be the points of intersection of the lines extended from >AP, BP, and CP, respectively, to the opposite sides. > >Prove: (AF)(BD)EC) = (FB)(DC)(EA) > >Thanks, >Ron > This brings back memories - I remember learning it at school 40 years ago (am I really THAT old?) under the name of Ceva's Theorem. Re-state it slightly: AF BD CE -- . -- . -- = 1 FB DC EA The proof is based on the idea that the areas of two triangles with the same perpendicular height are in the same ratio as the lengths of their bases. Thus AF / FB = area PAF / area PFB = area CAF / area CFB, so that AF / FB = area PCA / area PCB. Now do the corresponding thing for each of the other sides; and the product of the ratios of areas reduces to 1. Wow, my teacher would be proud of me, remembering it after all that time! David Shield. ============================================================================== From: billr@linfield.edu\ (Bill Raddatz) Newsgroups: sci.math Subject: Re: GEOMETRY PROBLEM Date: Fri, 01 Dec 1995 11:01:52 +0400 In article <49j767\$21r@sheoak.bendigo.latrobe.edu.au>, David Shield wrote: > rkleiner@moe.coe.uga.edu (Ronen Kleiner M) wrote: > >Consider any triangle ABC. Let P be any point in the interior of ABC. > >Let D,E, and F be the points of intersection of the lines extended from > >AP, BP, and CP, respectively, to the opposite sides. > > > >Prove: (AF)(BD)EC) = (FB)(DC)(EA) > > > >Thanks, > >Ron > > --A nice proof by David Shield deleted because this (*&^ Newsreader won't let my reply be shorter than the included stuff. -- Don't forget that the converse is also true, making for a nice way to show concurrence of lines related to a triangle. You can also prove Ceva's theorem using Menelaus' Theorem: Given D on BC, E on CA, and F on AB (possibly extended sides) of triangle ABC, D, E, and F are collinear iff > AF BD CE > -- . -- . -- = -1. > FB DC EA In this case you need to use directed distance, so that PQ = -QP. Another sort of related result, which I happen to have here on my desk don't ask me why) is a Theorem by Paul A. Clement, found in The American Mathematical Monthly, v. 65, num.8, October, 1958. Given D on BC, E on CA, and F on AB (possibly extended sides) of triangle ABC the perpendiculars to the sides at D, E, and F are concurrent iff (AF)^2 + (BD)^2 + (CE)^2 = (FB)^2 + (DC)^2 + (EA)^2. Bill Raddatz