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13436Re: Concurrencies on a triangle.

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  • Kostas Vittas
    Jun 27, 2006
      --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
      > ...
      >7f. As a helping proposition, we use the result that the points B',
      >A, S, are collinear ( we shall prove the proposition for this
      >result, below ). Hence, we consider the segments ...
      > ...

      Dear all my friends.

      A friend of mine Kostas Helatiotis, said me that we don't need
      any helping proposition for the proof of (7), in message 13274.

      7f. From cyclic non-convex hexagon D'D''D1CEB, by Pascal's
      theorem, we have that the points B', A, S, are collinear. Hence, we
      consider the segments ...

      I apologize for my blindness and please, cancel all about the
      proof of the helping proposition (7.1), as unnecessary prattlings.

      Best regards.
      Kostas Vittas.
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