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Re: Prependicular lines

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  • Kostas Vittas
    ... BC. ... Dear dam_xoan90 and Francois. A simple elementary proof of this problem, has already been post in Mathlinks Forum, one year ago. Please see at:
    Message 1 of 4 , Sep 1, 2007
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      --- In Hyacinthos@yahoogroups.com, "dam_xoan90" <dam_xoan90@...> wrote:
      >
      > Let ABC be an acute triangle and M be the middle point of the side
      BC.
      > Let H the orthocenter of this triangle and let E the intersection of
      > HM with the angle bisector of angle A. Denote by F the orthogonal
      > projection of E on AC, prove that HF is perpendicular to AE
      >


      Dear "dam_xoan90" and Francois.

      A simple elementary proof of this problem, has already been post in
      Mathlinks Forum, one year ago.

      Please see at: http://www.mathlinks.ro/Forum/viewtopic.php?t=109749

      Best regards.
      Kostas Vittas.
    • michgarl
      ... isosceles in A. ... circumcenter. ... side AC and ... uniform ... as choo-c ... correspondence ... and B, so ... AB and ... (B , C ) ... Is it known that
      Message 2 of 4 , Sep 2, 2007
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        --- In Hyacinthos@yahoogroups.com, "Francois Rideau"
        <francois.rideau@...> wrote:
        >
        > Dear friend
        >
        > As I am in my dotagen I see choo-choo everywhere!
        >
        > First, your problem only makes sense when triangle ABC is
        isosceles in A.
        >
        > So in the sequel, I suppose that AB <> AC.
        >
        > Line HM is on the point A '' symmetric of A wrt the ABC-
        circumcenter.
        >
        > Now I give you a solution based on choo-choo theory.
        >
        > Given any point P on line HM , we project him orthogonally at Q on
        side AC and
        > on R on side AB.
        >
        > When P moves on line HM (as a choo-choo that is to say with a
        uniform
        > motion!), points Q moves on line AC and R moves on line AB, (also
        as choo-c
        > hoos!). Then the correspondence g: Q --> R is an affine
        correspondence
        > between lines AC and AB, that is to say preserves affine ratio.
        >
        > Look now at special pairs of homologous points in this affine
        > correspondence.
        >
        >
        >
        > 1° When P is on A'', its projections on line AC and AB are just C
        and B, so
        > the pair (C,B) is a pair of homologous points.
        >
        > 2) When P is on H, its projections are B' on line AC and C' on line
        AB and
        > remind lines BB' and CC' are altitudes of triangle ABC. So the pair
        (B', C')
        > is another pair of homologous points in this affine correspondence.
        >
        >
        >
        Is it known that the envelop of line (QR) when P moves on line (HM)
        is a parabol
        If P = k H + (1-k) M
        Q = (1+k) S_c A +(b²+S_a - k S_c) C
        and R = (1+k)S_b A + (S²+S_a - k S_b) B
        it is easy to calculate equation of line (QR)
        How can we find equation of the envelop?

        Michel
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