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Triangle inscribed in Steiner Circumellipse

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  • garciacapitan
    Dear friends, Let P an interior point to triangle ABC and DEF its cevian triangle. Draw parallels to BC through E and F, intersecting cevian AP at E_A and F_A.
    Message 1 of 1 , Mar 5, 2007
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      Dear friends,

      Let P an interior point to triangle ABC and DEF its cevian triangle.
      Draw parallels to BC through E and F, intersecting cevian AP at E_A
      and F_A. We call M_{BA} the intersection point of DF and BF_A, and MC_
      {A} the intersection point of DE and CE_{A},

      1. The points M_{BA}, M_{CA} and P are collinear. Call r_a the line
      which these points lies on. Call r_b, r_c the corresponding lines to
      other cevians.

      2. If we construct in a similar way the points M_{AB}, M_{CB}, M_
      {AC}, M_{BC}, the six points M_{??} lie on the same conic G_{M}.

      3. The lines r_{a}, r_{b}, r_{c} cut the sides of ABC at six points N_
      {??} lying on the same conic G_{N}.

      4. Let XYZ the triangle bounded by the lines joining points N_{??}
      nearest to the vertices of ABC. Then XYZ is inscribed in the Steiner
      circumellipse of ABC and XYZ is perspective with ABC, with perspector
      P.

      You can look at

      http://garciacapitan.auna.com/problemas/jb/20070306/

      for a figure.

      I look forward for references and/or related results.

      Thank you,

      ---
      Francisco Javier García Capitán
      http://garciacapitan.auna.com
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