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Geometric nature of X

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  • Jean-Louis Ayme
    Dear Mathlinkers,   1. ABC a triangle 2. I the incenter of ABC 3. A’B’C’ the I-cevian triangle of ABC 4. LMN the medial triangle of ABC 5. A’’ the
    Message 1 of 3 , Apr 25, 2012
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      Dear Mathlinkers,
       
      1. ABC a triangle
      2. I the incenter of ABC
      3. A’B’C’ the I-cevian triangle of ABC
      4. LMN the medial triangle of ABC
      5. A’’ the point of intersection MI and B’C’, and circularly B’’, C’’
       
      Prouve :                A’’B’’C’’ and ABC are perspective at X
       
       
      Sincerely
      Jean-Louis

      [Non-text portions of this message have been removed]
    • Francisco Javier
      X is the symmedian point (barycentric square of I). If you change I by any other point P then you ll get that X is the the barycentric square of P. Best
      Message 2 of 3 , Apr 25, 2012
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        X is the symmedian point (barycentric square of I).
        If you change I by any other point P then you'll get that X is the the barycentric square of P.

        Best regards,

        Francisco Javier.

        --- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
        >
        > Dear Mathlinkers,
        >  
        > 1. ABC a triangle
        > 2. I the incenter of ABC
        > 3. A’B’C’ the I-cevian triangle of ABC
        > 4. LMN the medial triangle of ABC
        > 5. A’’ the point of intersection MI and B’C’, and circularly B’’, C’’
        >  
        > Prouve :                A’’B’’C’’ and ABC are perspective at X
        >  
        >  
        > Sincerely
        > Jean-Louis
        >
        > [Non-text portions of this message have been removed]
        >
      • Chandan Banerjee
        If you change LMN by cevian triangle of a point P, then X will be the isogonal conjugate of P wrt ABC. Proof:- Suppose, AA1 is the external angle-bisector of
        Message 3 of 3 , Apr 25, 2012
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          If you change LMN by cevian triangle of a point P, then X will be the isogonal conjugate of P wrt ABC.
          Proof:-
          Suppose, AA1 is the external angle-bisector of \angle BAC where A1 lies on BC. IM intersects AA1 at A2. Then note that, (l_a,IA1;B'C',BC)=-1. So (A2I;B"M)=-1. So (AA1,AI;AB",AM)=-1. So AB" is the isogonal conjugate of AM and similar for others. So done.
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