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Re: [EMHL] Triangle Proof

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  • Gilles Boutte
    Dear all, ... Let G be on AC with BE//DG, thus BD/CD = EG/EC by theorem of Thales. F is on the altitude DE of ADG, hence AF is perpendicular to BE or to DG iff
    Message 1 of 2 , Jun 30, 2004
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      Dear all,

      >Given an arbitrary triangle ABC,
      >
      >let D be the foot of the perpendicular from A to BC
      >let E be the foot of the perpendicular from D to AC
      >let F be a point on the line DE
      >
      >Prove that AF is perpendicular to BE iff BD/DC = FE/FD.
      >
      >Regards,
      >Barukh.
      >
      Let G be on AC with BE//DG, thus BD/CD = EG/EC by theorem of Thales.

      F is on the altitude DE of ADG, hence AF is perpendicular to BE or to DG
      iff F is the orthocenter of ADG iff FG is perpendicular to AD, i.e. FG//DC.

      By theorem of Thales FG//DC iff FE/FD = GE/GC.

      Friendly,

      Gilles Boutte
    • Vladimir Dubrovsky
      ... Triangles EAD and EDC are similar. Consider the similarity that takes the first into the second. It is the composition of 90 degrees rotation about E and
      Message 2 of 2 , Feb 2 1:19 AM
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        Barukh.wrote:

        > I would like to know the synthetic proof of the following statement:
        >
        > Given an arbitrary triangle ABC,
        >
        > let D be the foot of the perpendicular from A to BC
        > let E be the foot of the perpendicular from D to AC
        > let F be a point on the line DE
        >
        > Prove that AF is perpendicular to BE iff BD/DC = FE/FD.

        Triangles EAD and EDC are similar. Consider the similarity that takes the
        first into the second. It is the composition of 90 degrees rotation about E
        and dilation with ratio
        EC/ED = ED/EA (= tan CAD).
        So it takes segment AF into the segment DH perpendicular to AF with H on EC
        and EH/HC=EF/FD.

        It remains to notice that EH/HC=BD/DC iff DH is parallel to BE, ie AF is
        perpendicular to BE..

        Regards,

        Vladimir Dubrovsky
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