Geometry Proof Involving Circle and Triangle
Date: 09/26/2005 at 09:52:33 From: Drew Subject: Geometry Proofs Triangle ABC cuts a circle at points E, E', D, D', F amd F'. Prove that if AD, BF and CE are concurrent, than AD', BF' and CE' are also concurrent. I have researched this and all I come up with is the nine point circle proof. I am not able to work this proof out, can you please help?
Date: 09/28/2005 at 07:55:33 From: Doctor Floor Subject: Re: Geometry Proofs Hi, Drew, Thanks for your question. We can use two important geometry facts here: 1. Ceva's theorem, see Ceva's Theorem http://mathforum.org/library/drmath/view/55095.html 2. The power of a point with respect to a circle: Intersecting Circles http://mathforum.org/library/drmath/view/55125.html Going back to your problem, with Ceva's Theorem we have: BD CE AF -- * -- * -- = 1 .................................... DC EA FB While the power of B with respect to the circle yields that: BD*BD' = BF*BF' And because FB = -BF and F'B= - BF' this gives us BD*BD' = FB*F'B or BD F'B -- = --- ........................................... FB BD' In the same way we derive that CE D'C AF E'A -- = --- and -- = --- ............................. DC CE' EA AF' Now with the help of  and  we can rewrite  into D'C E'A F'B --- * --- * --- = 1 BD' CE' AF' or equivalently BD' CE' AF' --- * --- * --- = 1 D'C E'A F'B With (the converse of) Ceva's Theorem this shows that AD', BE' and CF' are concurrent. See also: Eric W. Weisstein. "Cyclocevian Triangle." http://mathworld.wolfram.com/CyclocevianTriangle.html If you replace "circle" in your question with "conic section" then the statement is still correct. The proof is more difficult and requires some knowledge of "projective geometry". If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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