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Re: [EMHL] Midcevian Triangle

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  • Antreas P. Hatzipolakis
    Dear Paul ... The perspectivity can be proven by applying the Fruitful Theorem (nested cevian triangles theorem). Denote T = The reference triangle ABC mT =
    Message 1 of 6 , Jan 21, 2004
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      Dear Paul

      [APH]:
      >>Definition:
      >>
      >>Let ABC be a triangle, P a point, A'B'C' the cevian triangle
      >>of P and A*,B*,C* the midpoints of the cevians AA', BB', CC' of P.
      >>
      >>I call the triangle A*B*C* midcevian triangle of P.
      >>
      >>Locus:
      >>
      >>Which is the locus of P such that the medial triangle of
      >>ABC and the medial triangle of the midcevian triangle of P
      >>are perspective? (The entire plane?)

      [PY]:
      >Yes, this is the point that divides PG in the ratio 3:1.
      >
      >In fact, the midcevian triangle of P is perspective with the medial
      >triangle at the inferior of the isotomic conjugate of P.

      The perspectivity can be proven by applying the Fruitful Theorem
      (nested cevian triangles theorem).

      Denote

      T = The reference triangle ABC

      mT = The Medial Triangle of ABC

      V = The midcevian triangle of P

      mV = The medial triangle of the midcevian triangle V of P


      We have:

      mT is inscribed in T, and T, mT are perspective

      V is inscribed in mT

      T, V are perspective

      ==> V, mT are perspective.

      Now,

      mV is inscribed in V, and V, mV are perspective.

      V is inscribed in mT

      V, mT are perspective [see above]

      ==> mT, mV are perspective

      OED


      Good Night from the rainy Athens

      Antreas
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