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RE: Triangle Proof II
Posted:
Jun 30, 2004 4:59 AM


> Original Message > From: ownergeometrypuzzles@mathforum.org [mailto://ownergeometry > puzzles@mathforum.org] On Behalf Of Eamon > Sent: Wednesday, June 30, 2004 7:08 AM > To: geometrypuzzles@mathforum.org > Subject: Triangle Proof II > > ABC is an ARBITRARY triangle. > D lies on BC and E lies on AC. > Prove that if AD = BE then the triangle is isosceles. > > > > > y C >  /\ >  / \ >  / \ D >  E/ \ >  / \ >  / \ > ____/__________________\B____ > A x
The result is of course false. More conditions should be added.
Example: Take a nonisosceles triangle ABC. From A draw a circle of arbitrary radius cutting BC at D (plus another point). With the same radius draw a circle with B as centre, cutting AC at E (plus another point). That does it. Note, there is an abundance of such radii. All you need is to pick a length greater than the largest of the altidudes of the triangle.
Regards. Michael Lambrou



