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21054Re: Simson lines

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  • Cosmin
    Jun 7, 2012
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      An easier proof would be to just recall that the Simson line bisects the segment determined by the orthocenter of the triangle and the point, thus if we let X, Y be the midpoints of HP and HQ (where H is the orthocenter of ABC), the result follows at once by Desargues' theorem applied to triangles PXM, QYN, where M, N are the projections of P, Q on AB.

      Best,
      Cosmin

      --- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:
      >
      > Dear colleagues!
      > Is next fact known?
      > Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.
      >
      > Sincerely Alexey
      >
      > [Non-text portions of this message have been removed]
      >
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