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 ....................................[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
  -- =  --- ...........................................[2]
  FB    BD'

In the same way we derive that

  CE   D'C       AF   E'A
  -- = ---  and  -- = --- .............................[3]
  DC   CE'       EA   AF'

Now with the help of [2] and [3] we can rewrite [1] 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/