Loading ...
Sorry, an error occurred while loading the content.

A very simple geometry question..

Expand Messages
  • malisundaresan
    Prove that the line joining the -circumcenter of a triangle to a vertex- is perpendicular to the line -joining the foot of altitudes of adjoining sides.- Some
    Message 1 of 3 , Sep 30, 2007
    • 0 Attachment
      Prove that the line joining the -circumcenter of a triangle to a
      vertex- is perpendicular to the line -joining the foot of altitudes of
      adjoining sides.-
      Some sort of a HINT please....
    • michgarl
      dear colleague There is a very simple response the orthic triangle is homothetic to the tangential one The line joining the -circumcenter of a triangle to a
      Message 2 of 3 , Oct 1, 2007
      • 0 Attachment
        dear colleague
        There is a very simple response
        the orthic triangle is homothetic to the tangential one
        The line joining the -circumcenter of a triangle to a
        vertex- is perpendicular to the tangent line in that vetex to the
        circumcircle, so to the parallel side of the orthic triangle

        There are others proves using barycentrics coordinates

        Best regards

        Michel Garitte



        --- In Hyacinthos@yahoogroups.com, "malisundaresan"
        <malisundaresan@...> wrote:
        >
        > Prove that the line joining the -circumcenter of a triangle to a
        > vertex- is perpendicular to the line -joining the foot of altitudes
        of
        > adjoining sides.-
        > Some sort of a HINT please....
        >
      • vprasad_nalluri
        ... altitudes of ... Let E, F be the feet of the altitudes through B, C upon AC, AB respectively, Let X be the midpoint of AH (H orthocenter) X is the center
        Message 3 of 3 , Oct 1, 2007
        • 0 Attachment
          --- In Hyacinthos@yahoogroups.com, "malisundaresan"
          <malisundaresan@...> wrote:
          >
          > Prove that the line joining the -circumcenter of a triangle to a
          > vertex- is perpendicular to the line -joining the foot of
          altitudes of
          > adjoining sides.-
          > Some sort of a HINT please....
          >
          Let E, F be the feet of the altitudes through
          B, C upon AC, AB respectively,
          Let X be the midpoint of AH (H orthocenter)
          X is the center of the circle through A, E, H, F.
          XE = XF
          The nine-point circle (N) passes through E and F
          NE = NF (each = R/2)
          Follows NX is the perpendicular bisector of EF.
          In the triangle AOH :
          X, N being the midpoints of AH, OH we have
          NX is parallel to OA.
          Together with the fact that NX is perpendicular to EF,
          it now follows that OA is perpendicular to EF

          Vijayaprasad
        Your message has been successfully submitted and would be delivered to recipients shortly.