Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By richard1234

Thread: Proof

  1. #1
    Sep 2011


    in triangle ABC The point P lies on the side BC so that AB+BP=AC+CP ,and the point Q lies on the side AC so that:BA+AQ=BC+CQ and the point R lies on the side AB so that:CB+BR=CA+AR .How to prove that the lines AP,BQ,CR are meet
    Last edited by Mhmh96; Jul 12th 2012 at 03:19 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Jun 2012

    Re: Proof

    Suppose AR = a, RB = b, BP = c, PC = d, CQ = e, QA = f. From the conditions of the problem, we know that

    (1) $\displaystyle a+b+c = d+e+f$

    (2) $\displaystyle b+c+d = e+f+a$

    (3) $\displaystyle c+d+e = f+a+b$

    Subtracting (2) from (1), we obtain $\displaystyle a-d = d-a \Rightarrow a = d$. Subtracting (3) from (2), $\displaystyle b-e = e-b \Rightarrow b = e$. Plugging this back into (1), we obtain $\displaystyle c = f$.

    Therefore $\displaystyle \frac{a}{b}*\frac{c}{d}*\frac{e}{f} = \frac{a}{b}*\frac{c}{a}*\frac{b}{c} = 1$. The segments AP, BQ, CR must intersect at a point, due to Ceva's theorem.
    Attached Thumbnails Attached Thumbnails Proof-ceva-s.png  
    Thanks from Mhmh96
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 19th 2010, 10:50 AM
  2. Replies: 0
    Last Post: Jun 29th 2010, 08:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Feb 27th 2010, 10:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: Jun 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Apr 14th 2008, 04:07 PM

Search Tags

/mathhelpforum @mathhelpforum