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13161Concurrencies on a trianle.

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  • Kostas Vittas
    Jun 1, 2006
      Dear all my friends.

      I would like to present an interesting configuration as follows :
      ( I apologize for my English )

      A triangle ABC is given and let AD, BE, CF be, it's altitudes.
      We draw three circles (K1), (K2), (K3), with side segments BC, AC, AB
      respectively, as diameters.

      We denote as D', D'', the intersection points of line AD from
      circle (K1). Similarly we denote the pair of points E', E'' and F',
      F'', on lines BE, CF, respectively. ( The points D', E', F', inwardly
      to the triangle ABC ).

      We denote A', as the intersection point of the lines E'F'',
      E''F'. Similarly B' as the one of the lines D'F'', D''F' and C' as
      the one of the lines D'E'', D''E'.

      Results :

      1. The points A', B', C', lie on BC, AC, AB, respectively.
      2. The lines AA', BB', CC', are concurrent at a point, so be it M.
      3. The lines A'D',B'E',C'F', are concurrent at a point, so be it L.
      4. The lines A'D'',B'E'',C'F'',are concurrent at a point,so be it N.
      5. The points L, M, N, lie on a line pass through orthocenter H, of
      the triangle ABC.
      6. The points L, N, are the circumcenters of the triangles D'E'F',
      D''E''F'', respectively.
      7. If D1, E1, F1, are the intersection points of circles (K1), (K2),
      (K3) respectively, from the circumcircle of the triangle
      D''E''F'', then the lines AD1, BE1, CF1, are concurrent.
      8. If D2, E2, C2, are the intersection points of circles (K1), (K2),
      (K3) respectively, from the circumcircle of the triangle D'E'F',
      then the lines D1D2, E1E2, F1F2, are concurrent at orthocenter H
      of ABC.

      I will try to post here next time, some proofs I have in mind for
      (1), (2),...(6). I still not have proofs for (7), (8) and I am
      waiting, with pleasure, for any idea. ( My drawing is erected on a
      triangle ABC such that BC = 9.5, AC = 8.5, AB = 6.5 ).

      Best regards.
      Kostas Vittas.
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