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Re: Two equivalent theorems (an old and a new(?))

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  • ben_goss_ro
    ... HP. ... CC* (resp.) ... I m sure the fact that the Simson line of P is parallel to HP helps, but it s 4 AM here (in Romania), so I m going to think it
    Message 1 of 3 , Apr 1 5:05 PM
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      --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"
      <xpolakis@o...> wrote:
      > Let ABC be a triangle, H its orthocenter and P a point.
      >
      > OLD THEOREM:
      > The reflections of the line HP in the sidelines of ABC concur
      > at P' on the circumcircle of ABC.
      >
      > NEW(?) THEOREM:
      > Let A', B', C' be the orthogonal projections of A,B,C on the line
      HP.
      > The reflections of AA', BB', CC' in the angle bisectors AA*, BB*,
      CC* (resp.)
      > of ABC concur at P' on the circumcircle of ABC.


      I'm sure the fact that the Simson line of P' is parallel to HP helps,
      but it's 4 AM here (in Romania), so I'm going to think it through
      tomorrow morning :).
    • ben_goss_ro
      ... helps, ... In fact, the following holds (easy to prove, just a quick angle chase): If ABC is a triangle and M is a point on the circumcircle of ABC then
      Message 2 of 3 , Apr 1 10:52 PM
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        --- In Hyacinthos@yahoogroups.com, "ben_goss_ro" <ben_goss_ro@y...>
        wrote:
        > --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"
        > <xpolakis@o...> wrote:
        > > Let ABC be a triangle, H its orthocenter and P a point.
        > >
        > > OLD THEOREM:
        > > The reflections of the line HP in the sidelines of ABC concur
        > > at P' on the circumcircle of ABC.
        > >
        > > NEW(?) THEOREM:
        > > Let A', B', C' be the orthogonal projections of A,B,C on the line
        > HP.
        > > The reflections of AA', BB', CC' in the angle bisectors AA*, BB*,
        > CC* (resp.)
        > > of ABC concur at P' on the circumcircle of ABC.
        >
        >
        > I'm sure the fact that the Simson line of P' is parallel to HP
        helps,
        > but it's 4 AM here (in Romania), so I'm going to think it through
        > tomorrow morning :).


        In fact, the following holds (easy to prove, just a quick angle
        chase):

        If ABC is a triangle and M is a point on the circumcircle of ABC then
        the perpendicular drawn through A to the Simson line of M and AM are
        isogonal with respect to angle BAC.

        The fact that the Simson line and HP are parallel is again easy to
        show by using another angle computation.

        In our case take M=P' and from the sentence above we get that AA',
        BB', CC', the perpendiculars from A, B, C to the simson line of P'
        (and thus to HP, which is parallel to the sSimson line of P') are
        exactly the isogonal cevians of AP', AB', AC' respectively.
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