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21054Re: Simson lines

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  • Cosmin
    Jun 7, 2012
      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.


      --- 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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