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Re: [EMHL] Re: Three collinear centers

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  • Antreas Hatzipolakis
    Dear Angel Beautiful..... !!!! Thanks Antreas http://amontes.webs.ull.es/geogebra/Hyacinthos21421.html ... [Non-text portions of this message have been
    Message 1 of 37 , Jan 17, 2013
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      Dear Angel

      Beautiful..... !!!!

      Thanks

      Antreas



      http://amontes.webs.ull.es/geogebra/Hyacinthos21421.html


      >
      > http://amontes.webs.ull.es/geogebra/Hyacinthos21421Pedal.html
      >


      >
      >
      >
      > http://amontes.webs.ull.es/geogebra/Hyacinthos21428.html
      >


      [Non-text portions of this message have been removed]
    • Antreas
      It is Another Seven Circles Theorem ! The Seven Circles Theorem states: Let a,b ,c, a ,b,c be a closed chain of six circles, all toouching a base circle
      Message 37 of 37 , Mar 21, 2013
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        It is Another "Seven Circles Theorem" !

        The "Seven Circles Theorem" states:

        Let a,b',c, a',b,c' be a closed chain of six circles,
        all toouching a base circle w[omega], and suppose that
        their points of contact with w are six distinct points
        A,B',C,A',B,C' respectively. Then, subject of a certain
        extra condition to be discussed below, AA',BB',CC' are concurrent.
        (C J A Evelyn, B G Money - Coutts, J A Tyrrell:
        The Seven Circles Theorem and other theorems. London 1974, p. 31)

        See also
        http://en.wikipedia.org/wiki/Seven_circles_theorem
        http://mathworld.wolfram.com/SevenCirclesTheorem.html

        APH


        --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
        >
        > Dear Antreas
        >
        > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
        > >
        > > Dear Alex
        > >
        > > Angel has tested the loci problems and it seems that
        > > for all P's the lines are concurrent for the cevian case,
        > > and for the pedal case as well except for the points P on the
        > > bisectors and on the circumcircle.
        > >
        > > Hmmmmmmm...... Maybe it is true in general ie for any
        > > circle intersecting the sidelines of the triangle!!
        > >
        >
        >
        > A check with GeoGebra:
        >
        >
        >
        > Let ABC be a triangle and P a point.
        >
        > Let (Q) be the CEVIAN circle of P and (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the points Tbc, Tba and Tca, Tcb.
        >
        > For all P (except for the points P on the bisectors) the lines TabTac, TbcTba, TcaTcb are concurrent.
        >
        > http://amontes.webs.ull.es/geogebra/Hyacinthos21421.html
        >
        > ----------------------
        >
        > Let (Q) be the PEDAL circle of P and (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the points Tbc, Tba and Tca, Tcb.
        >
        > For all P (except for the points P on the
        > bisectors and on the circumcircle) the lines TabTac, TbcTba, TcaTcb are concurrent.
        >
        > http://amontes.webs.ull.es/geogebra/Hyacinthos21421Pedal.html
        >
        > ------------------------------------------------
        >
        > Let (Q) be the circle Intersecting the sidelines of the triangle in D, E and F. (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the
        > points Tbc, Tba and Tca, Tcb.
        >
        > The lines TabTac, TbcTba, TcaTcb are concurrent.
        >
        > http://amontes.webs.ull.es/geogebra/Hyacinthos21428.html
        >
        > Best regards,
        > Angel M.
        >
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