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Re: Two isogonal points

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  • garciacapitan
    Dear Quim, in the general case, if P={u,v,w} (barycentric coordinates), then X={a^2*b^2*u*w^2, b^2*c^2*u^2*v, a^2*c^2*v^2*w} Y={a^2*c^2*u*v^2, a^2*b^2*v*w^2,
    Message 1 of 13 , Aug 4, 2006
      Dear Quim,

      in the general case, if P={u,v,w} (barycentric coordinates), then

      X={a^2*b^2*u*w^2, b^2*c^2*u^2*v, a^2*c^2*v^2*w}

      Y={a^2*c^2*u*v^2, a^2*b^2*v*w^2, b^2*c^2*u^2*w}

      From this, your statement follows.



      --- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer" <qcastell@...>
      wrote:
      >
      > Dear friends,
      >
      > I have noticed this fact:
      > P, Q being two isogonal points, the cevians of P and the cevians of Q
      > intersect at P, Q and six more points. The lines (other than the
      > cevians) joining each vertex with these points are concurrent at two
      > isogonal points X, Y. When P, Q are G, K the new points X, Y are the
      > Brocard points.
      >
      > Greetings from Bsrcelona,
      >
      > Quim Castellsaguer
      >
      ´
    • Kostas Vittas
      Dear Quim. Let PaPbPc, QaQbQc be the cevian triangles of isogonal points P, Q, wrt ABC respectively, in the general case. We denote the six more intersection
      Message 2 of 13 , Aug 5, 2006
        Dear Quim.

        Let PaPbPc, QaQbQc be the cevian triangles of isogonal points P,
        Q, wrt ABC respectively, in the general case. We denote the six more
        intersection points of their cevians as follows:

        - As D, the intersection point of BQb, CPc.
        - As D', the intersection point of BPb, CQc.
        - As E, the intersection point of CQc, APa.
        - As E', the intersection point of CPc, AQa.
        - As F, the intersection point of AQa, BPb.
        - As F', the intersection point of APa, BQb.

        Results mentioned in your message:

        1 - The lines AD, BE, CF, are concurrent at a point so be it, X.
        2 - The lines AD', BE', CF', are concurrent at a point sp be it, Y.
        3 - The points X, Y, are isogonal wrt ABC.

        Another interesting ( I think ) result is that the lines DD',
        EE', FF', XY', are concurrent at point so be it, O.

        When the points X, Y, are coincided with the Brocard points
        ( when P, Q, are G, K, respectively ), then the point O coincides
        with the midpoint of their segment.

        Some notes for the proof about the concurrency point O, have been
        posted in message #13310 ( in more general configuration of two
        arbitrary isogonal lines through each vertex of a given triangle ).

        Best regards.
        Kostas Vittas.



        --- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
        <qcastell@...> wrote:
        >
        > Dear friends,
        >
        > I have noticed this fact:
        > P, Q being two isogonal points, the cevians of P and the cevians of
        Q
        > intersect at P, Q and six more points. The lines (other than the
        > cevians) joining each vertex with these points are concurrent at
        two
        > isogonal points X, Y. When P, Q are G, K the new points X, Y are
        the
        > Brocard points.
        >
        > Greetings from Bsrcelona,
        >
        > Quim Castellsaguer
        >
      • Kostas Vittas
        Dear Quim. It is clear that the points D, D , are isogonal with respect to the triangle ABC. Similarly about the pairs of points E, E and F, F . So, we can
        Message 3 of 13 , Aug 5, 2006
          Dear Quim.

          It is clear that the points D, D', are isogonal with respect to
          the triangle ABC. Similarly about the pairs of points E, E' and F,
          F'. So, we can easy to prove the concurrencies at points X, Y and
          that these points are also isogonal wrt ABC.

          But how about the proof that the points X, Y, coincided with
          the Brocard points, when P, Q, are G, K respectively and that the
          point O, is the midpoint of their segment?

          Best regards.
          Kostas Vittas.



          --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
          wrote:
          >
          > Dear Quim.
          >
          > Let PaPbPc, QaQbQc be the cevian triangles of isogonal points
          P,
          > Q, wrt ABC respectively, in the general case. We denote the six
          more
          > intersection points of their cevians as follows:
          >
          > - As D, the intersection point of BQb, CPc.
          > - As D', the intersection point of BPb, CQc.
          > - As E, the intersection point of CQc, APa.
          > - As E', the intersection point of CPc, AQa.
          > - As F, the intersection point of AQa, BPb.
          > - As F', the intersection point of APa, BQb.
          >
          > Results mentioned in your message:
          >
          > 1 - The lines AD, BE, CF, are concurrent at a point so be it, X.
          > 2 - The lines AD', BE', CF', are concurrent at a point sp be it, Y.
          > 3 - The points X, Y, are isogonal wrt ABC.
          >
          > Another interesting ( I think ) result is that the lines DD',
          > EE', FF', XY', are concurrent at point so be it, O.
          >
          > When the points X, Y, are coincided with the Brocard points
          > ( when P, Q, are G, K, respectively ), then the point O coincides
          > with the midpoint of their segment.
          >
          > Some notes for the proof about the concurrency point O, have
          been
          > posted in message #13310 ( in more general configuration of two
          > arbitrary isogonal lines through each vertex of a given triangle ).
          >
          > Best regards.
          > Kostas Vittas.
          >
          >
          >
          > --- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
          > <qcastell@> wrote:
          > >
          > > Dear friends,
          > >
          > > I have noticed this fact:
          > > P, Q being two isogonal points, the cevians of P and the cevians
          of
          > Q
          > > intersect at P, Q and six more points. The lines (other than the
          > > cevians) joining each vertex with these points are concurrent at
          > two
          > > isogonal points X, Y. When P, Q are G, K the new points X, Y are
          > the
          > > Brocard points.
          > >
          > > Greetings from Bsrcelona,
          > >
          > > Quim Castellsaguer
          > >
          >
        • Steve Sigur
          Quim, This is particularly interesting when the isogonal points are O and H, in which case the 6 intersections form two triangles antisimilar to ABC. Happy
          Message 4 of 13 , Aug 6, 2006
            Quim,

            This is particularly interesting when the isogonal points are O and
            H, in which case the 6 intersections form two triangles antisimilar
            to ABC.

            Happy geometrizing,

            Steve

            On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:

            >
            > I have noticed this fact:
            > P, Q being two isogonal points, the cevians of P and the cevians of Q
            > intersect at P, Q and six more points. The lines (other than the
            > cevians) joining each vertex with these points are concurrent at two
            > isogonal points X, Y. When P, Q are G, K the new points X, Y are the
            > Brocard points.

            Steve

            Notation:
            http://paideiaschool.org/TeacherPages/Steve_Sigur/resources/VoodooPad%
            20web/notation.html


            Triangle web page:
            http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

            Other math:
            http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





            [Non-text portions of this message have been removed]
          • Jeff Brooks
            Dear Steve and Quim and all, I am happy to see this old friend . Please see also: http://groups.yahoo.com/group/Hyacinthos/message/8045 Sincerely, Jeff PS
            Message 5 of 13 , Aug 6, 2006
              Dear Steve and Quim and all,

              I am happy to see this 'old friend'.
              Please see also:

              http://groups.yahoo.com/group/Hyacinthos/message/8045

              Sincerely, Jeff
              PS It's nice to see something in a new light.


              On Sun Aug 6, 2006 12:58 pm, Steve Sigur wrote:

              Quim,

              This is particularly interesting when the isogonal points are O and
              H, in which case the 6 intersections form two triangles antisimilar
              to ABC.

              Happy geometrizing,

              Steve

              On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:

              >
              > I have noticed this fact:
              > P, Q being two isogonal points, the cevians of P and the cevians of
              > Q intersect at P, Q and six more points. The lines (other than the
              > cevians) joining each vertex with these points are concurrent at two
              > isogonal points X, Y. When P, Q are G, K the new points X, Y are the
              > Brocard points.
            • Kostas Vittas
              ... Q ... two ... the ... Dear Quim and all my friends. I would like to present a proof, of the result that the isogonals point X, Y, become Brocard s points
              Message 6 of 13 , Aug 7, 2006
                --- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
                <qcastell@...> wrote:
                >
                > Dear friends,
                >
                > I have noticed this fact:
                > P, Q being two isogonal points, the cevians of P and the cevians of
                Q
                > intersect at P, Q and six more points. The lines (other than the
                > cevians) joining each vertex with these points are concurrent at
                two
                > isogonal points X, Y. When P, Q are G, K the new points X, Y are
                the
                > Brocard points.
                >
                > Greetings from Bsrcelona,
                >
                > Quim Castellsaguer
                >


                Dear Quim and all my friends.

                I would like to present a proof, of the result that the
                isogonals point X, Y, become Brocard's points wrt ABC, when the P, Q,
                are G, K, respectively, based on the below proposition, as follows:

                1 – PROPOSITION. A triangle is given and let Ka, Ma be an
                arbitrary point and the midpoint of BC, respectively. If K is a point
                on AKa, we draw the segment lines BK, CK and we denote as D, E, their
                intersection points from the median AMa, respectively. If P is the
                intersection point of BE, CD, prove that:

                a) – The segment KP is parallel to BC and that it is bisected from
                the median AMa.
                b) – If K is the Lemmoine point of ABC, then the point P, lies on the
                midperpendicular of BC.

                2 – PROOF. ( in my drawing AB = 10.0, AC = 15.0, BC = 13.00 ).
                Let P be, the intersection point of CD, from the line through K and
                parallel to BC. So, it is easy to prove that the points B, E, P, are
                collinear ( from trapezium BKPC ) and that the median AMa, passes
                through the midpoint F, of the segment KP.

                We denote as K'a, the intersection point of BC from the segment
                line AP and then we have that this point is isotomic of Ka, with
                respect to BC. That is, the locus of P, when K is moving on segment
                line AKa, is the line AK'a where K'a, is the reflexion of Ka wrt Ma.

                As K, in our configuration, is the Lemoine point of ABC, we will
                prove that the line through K and parallel to BC, intersects the
                median AMa and the midperpenticular of BC, at the points F, P
                respectively, such that KF = FP.

                We denote as Ba, Ca, the intersection points of sidelines AB, AC
                respectively, from the segment line KP. It is well known that the
                segment BaCa, is a cord ( one of the three ), of the first Lemoine's
                circle (L) of ABC, centered at midpoint L of KO, where O is the
                circumcenter of ABC.

                The point F ( as intersection point of AMa, BaCa ), is the
                midpoint of BaCa, because of BaCa//BC and Ma, is the midpoint of BC.
                So, from circle (L) we have that the point F is the orthogonal
                projection of L, on BaCa. Hence, the point F is also the midpoint of
                KP, because of LF//OP and L, is the midpoint of KO.

                That is, we have that KF = FP. If K'a, is the intersection point
                of BC, from the segment line AP, we have that KaMa = MaK'a and the
                proof is completed.

                3 – We use now the configuration which is mentioned in message #
                13879 and we will prove that the isogonal points X, Y ( when P, Q,
                are G, K ), are the Brocard's points of ABC.

                ( They have been considered the isogonal points G, K, instead of
                P,Q and similarly about the traces of their cevians on the sidelines
                of ABC. Ma, instead of Pa, Ka, instead of Qa, etc. Also I have made a
                new drawing: AB = 10.0, AC = 13.0, BC = 15.0 ).

                3a – We denote as D the intersection point of BKb, CMc ( Mc is
                the midpoint of AC ). So, if D' is the intersection point of BMb ( Mb
                is the midpoint of AC ), CKc, we conclude that the point D, D', are
                isogonal conjugates wrt ABC. Similarly about the pairs of points E,
                E' and F, F' ( where E is the intersection point of CKc, AMa, etc ).

                3b – We denote as Y the intersection point of the segment lines
                AD', CF' ( F' is the intersection point of AMa, BKb ). So, by
                the ''isogonic'' theorem ( see messages #13680, #13685 ), we have
                that the lines AF ( where F is the intersection point of AKa, BMb ),
                BY, CD, are concurrent. Hence, the line BY passes through the point
                E' ( the intersection point of CMc, AKa ).

                3c – We conclude now, that the lines AD ( isogonal of AD' ), BE
                ( isogonal of BE' ), CF ( isogonal of CF' ), are concurrent at a
                point so be it X, isogonal conjugate of Y, wrt ABC.

                3d – We consider the median BMb and the points F, D', as it's
                intersection points, from the symmedians AKa, CKc, respectively. So,
                if B' is the intersection point of AD', CF, this point, by the above
                already has been proved proposition, lies on the midperpendicular of
                AC. Hence, we have that <ACX = <CAB', (1).

                <BAX = <CAB',(2).( because of AX, AB' are isogonal wrt angle <A ).

                (1),(2), => <BAX = <ACX, (3).

                By the same way we can prove that <BAX = <CBX, (4).

                From (3),(4), we conclude that the point X, is the first
                Brocard's point wrt ABC. Hence the point Y ( as isogonal of X ), is
                the second Brocard's point wrt ABC and the proof is completed.

                NOTES.

                a) – Based on the lemma 1, which has been stated and proved in
                message #13310, we can prove that the lines DD', EE', FF', XY, are
                concurrent at a point so be it S. I still not have a proof of the
                result that the point S, is the midpoint of the segment XY, in this
                configuration and I am eager for your advices.

                I have noticed another interesting result in this configuration.
                Through K ( Lemoine point ), we draw three lines parallel to BC, AC,
                AB and we denote as A', B', C', their intersection points, with the
                corresponded midperpendiculars, respectively. The triangle A'B'C', is
                antisimilar of ABC and it has as centroid, the centroid G of ABC.

                It is clear that the triangles ABC, A'B'C', are perspective at a
                point so be it T, isotomic of K with respect to ABC.

                Best regards.
                Kostas Vittas.
              • Kostas Vittas
                Dear Quim and all my friends. I have noticed another interesting result, out of our configuration and I don t know if it is already well known. If T, is the
                Message 7 of 13 , Aug 7, 2006
                  Dear Quim and all my friends.

                  I have noticed another interesting result, out of our
                  configuration and I don't know if it is already well known.

                  If T, is the isotomic point of the Lemoine point K, with respect
                  to a given triangle ABC, then the line connecting the point T, with
                  the orthocenter H of ABC, is parallel to the line KO, where O is the
                  circumcenter of ABC.

                  Best regards.
                  Kostas Vittas.



                  --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
                  wrote:
                  >
                  > --- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
                  > > <qcastell@> wrote:
                  > >
                  > > Dear friends,
                  > >
                  > > I have noticed this fact:
                  > > P, Q being two isogonal points, the cevians of P and the cevians
                  > > of Q intersect at P, Q and six more points. The lines (other than
                  > > the cevians) joining each vertex with these points are
                  > > concurrent at two isogonal points X, Y. When P, Q are G, K the
                  > > new points X, Y are the Brocard points.
                  > >
                  > > Greetings from Bsrcelona,
                  > >
                  > > Quim Castellsaguer
                  > >
                  >
                  >
                  > Dear Quim and all my friends.
                  >
                  > ...
                  >
                  > I have noticed another interesting result in this
                  >configuration. Through K ( Lemoine point ), we draw three lines
                  >parallel to BC,AC,AB and we denote as A', B', C', their intersection
                  >points, with the corresponded midperpendiculars, respectively. The
                  >triangle A'B'C', >is antisimilar of ABC and it has as centroid, the
                  >centroid G of ABC.
                  >
                  > ...
                • Jeff Brooks
                  Dear friends, With these thoughts in mind, can somebody please help me with a problem I am having: Given a known point P, what point or points Q allow
                  Message 8 of 13 , Aug 8, 2006
                    Dear friends,

                    With these thoughts in mind, can somebody please help me with a
                    problem I am having:

                    Given a known point P, what point or points Q allow triangles A1B1C1
                    and A2B2C2 to be similar with one another?

                    A1 := BP /\ CQ, A2 := BQ /\ CP,
                    B1 := CP /\ AQ, B2 := CQ /\ AP,
                    C1 := AP /\ BQ, C2 := AQ /\ BP;

                    I have tried numerous times to get the coordinates of Q in terms of
                    those given by P, but my computer program's kernel keeps shutting
                    down and I am not getting anywhere fast with hand calculations!

                    Your help is greatly appreciated.
                    Sincerely, Jeff


                    >
                    >
                    > On Aug 6, 2006, at 4:10 PM, Jeff Brooks wrote:
                    >
                    > > I am happy to see this 'old friend'.
                    > > Please see also:
                    > >
                    > > http://groups.yahoo.com/group/Hyacinthos/message/8045
                    > >
                    > > Sincerely, Jeff
                    >
                    > An old friend for me too.
                    >
                    > Steve
                    >
                  • Kostas Vittas
                    ... Dear Quim, Steve, Jeff, Francois and all. Let PaPbPc, QaQbQc be the cevian triangles of two arbitrary points P, Q, in the plane of a given triangle ABC.
                    Message 9 of 13 , Aug 10, 2006
                      On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:
                      >
                      > I have noticed this fact:
                      > P, Q being two isogonal points, the cevians of P and the cevians of
                      > Q intersect at P, Q and six more points. The lines (other than the
                      > cevians) joining each vertex with these points are concurrent at
                      > two isogonal points X, Y. When P, Q are G, K the new points X, Y
                      > are the Brocard points.
                      >

                      Dear Quim, Steve, Jeff, Francois and all.

                      Let PaPbPc, QaQbQc be the cevian triangles of two arbitrary
                      points P, Q, in the plane of a given triangle ABC. We denote the
                      intersection points of the lines connecting the vertices of ABC ( the
                      other than cevians of P, Q ), with the six , furthermore P, Q,
                      intersection points of their cevians, as follows:

                      - As D, the intersection point of BQb, CPc.
                      - As D', the intersection point of BPb, CQc.
                      - As E, the intersection point of CQc, APa.
                      - As E', the intersection point of CPc, AQa.
                      - As F, the intersection point of AQa, BPb.
                      - As F', the intersection point of APa, BQb.

                      RESULTS:
                      a)- The lines AD,BE,CF, are concurrent at a point so be it, X.
                      b)- The lines AD',BE',CF', are concurrent at a point so be it, Y.
                      c)- The lines DD',EE',FF',XY, are concurrent at a point so be it S.

                      PROOF. Let S be the intersection point of DD', FF'. We consider
                      the triads of points D, Q, F' and F, P, D', on the segment lines BQb,
                      BPb, respectively. So, by Pappus theorem, we have that the points E',
                      S, E, are collinear. Hence the lines DD', EE', FF', are concurrent at
                      point S.

                      Because of the concurrency at point S, by Desarques theorem, we
                      conclude that the triangles E'DF, ED'F', are perspective and so, the
                      intersection point S', of the segment lines DF, D'F', lies on the
                      sideline AC of ABC.

                      From the collinearity of A, C, S', we conclude that the triangles
                      E'DF, YF'D', where Y is the intersection point of AD', CF', are
                      perspective and hence, the line E'Y passes through the vertex B of
                      ABC ( also intersection point of DF', FD' ).

                      From the collinearity of A, C, S', we also conclude that XDF,
                      EF'D', where X is the intersection point of AD, CF, are perspective
                      and hence, the line EX passes through the vertex B of ABC.

                      From the collinearity of A, C, S', we also conclude that XDF,
                      YD'F', are perspective and hence, the line XY passes through the
                      point S ( intersection point of DD', FF' ) and the proof is completed.

                      It is easy to prove that if the points P, Q, are isotomic
                      ( instead isogonal ) with respect to ABC, then the two new
                      concurrency point X, Y, are also isotomic wrt ABC.

                      So, we can play with the special cases of the given pairs of
                      points P, Q, in two categories of configuration ( isogonal and
                      isotomic ).

                      I present a quick result in the isotomic configuration. When the
                      point Q coincides with the orthocenter of ABC, then the line
                      connecting the concurrency point S, with the centroid G of ABC,
                      bisects the segment HP ( where now, P is the isotomic of H wrt ABC ).

                      Best regards.
                      Kostas Vittas.
                    • Kostas Vittas
                      ... Dear all my friends. As, in message #13935, it already has been proved, we don t need any Lemma for the proof. Best regards. Kostas Vittas.
                      Message 10 of 13 , Aug 10, 2006
                        On Aug 7, 2006, at 4:42 PM, Kostas Vittas wrote:
                        >
                        > NOTES.
                        >
                        > a) – Based on the lemma 1, which has been stated and proved in
                        > message #13310, we can prove that the lines DD', EE', FF', XY, are
                        > concurrent at a point so be it S.
                        >

                        Dear all my friends.

                        As, in message #13935, it already has been proved, we don't need
                        any Lemma for the proof.

                        Best regards.
                        Kostas Vittas.
                      • Eric Danneels
                        Dear Steve you wrote: This is particularly interesting when the isogonal points are O and H, in which case the 6 intersections form two triangles antisimilar
                        Message 11 of 13 , Aug 13, 2006
                          Dear Steve

                          you wrote:

                          This is particularly interesting when the isogonal points are O and
                          H, in which case the 6 intersections form two triangles antisimilar
                          to ABC.

                          *************

                          These antisimilar triangles (let's call them A1B1C1 and A2B2C2) have
                          some interesting properties:
                          1) Their circumcircles intersect at the orthocenter H of ABC and at
                          a second point S;

                          Take A' on A1A2 such that A'A1 = k.A2A1 (with k any number) and
                          define B' and C' similarly

                          2) The triangles A'B'C' are all antisimilar to ABC and their
                          circumcircles form a pencil through H and S
                          3) The triangles A'B'C' and ABC orthological. H is the fixed
                          orthology center. The other orthology center lies on the
                          circumcircle of ABC.
                          4) S is the center of similarity of the triangles A'B'C': a rotation
                          around S combined with a homothety with center S generates all
                          triangles A'B'C'

                          If we take A', B' and C' at the midpoints of A1A2, B1B2 and C1C2

                          5) The circumcircle of A'B'C' goes through the circumcenter O of ABC
                          so its circumcenter is X(5) of ABC.
                          6) The corresponding orthology center on the circumcircle of ABC is
                          X(74)
                          7) The medial triangle of ABC is perspective with A'B'C' at X(1147)

                          The barycentric coordinates of S are quite complex.
                          If my computations are right the first barycentric should be

                          (b^2+c^2-a^2)(a^12 - (b^2+c^2)a^10 + (b^4-b^2c^2+c^4)a^8
                          - 3(b^2-c^2)^2(b^2+c^2)a^6
                          + 2(b^2-c^2)(b^6-c^6)a^4 + b^2c^2(b^2-c^2)^4)


                          Kind regards

                          Eric
                        • Wilson Stothers
                          Dear All, I have been away for a bit, and am just catching up with this thread. As Kostas indicates, the isogonal case is just a special case of a more
                          Message 12 of 13 , Aug 17, 2006
                            Dear All,

                            I have been away for a bit, and am just catching up with this
                            thread. As Kostas indicates, the "isogonal" case is just a
                            special case of a more general result. This apears in my paper

                            http://forumgeom.fau.edu/FG2006volume6/FG200613.pdf

                            See especially Lemma 11 and Theorems 12, 13, 14. I have
                            summarized some of the relevant bits below Kostas's result

                            --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
                            wrote:
                            >
                            > On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:
                            > >
                            > > I have noticed this fact:
                            > > P, Q being two isogonal points, the cevians of P and the cevians
                            of
                            > > Q intersect at P, Q and six more points. The lines (other than
                            the
                            > > cevians) joining each vertex with these points are concurrent at
                            > > two isogonal points X, Y. When P, Q are G, K the new points X, Y
                            > > are the Brocard points.
                            > >
                            >
                            > Dear Quim, Steve, Jeff, Francois and all.
                            >
                            > Let PaPbPc, QaQbQc be the cevian triangles of two arbitrary
                            > points P, Q, in the plane of a given triangle ABC. We denote the
                            > intersection points of the lines connecting the vertices of ABC (
                            the
                            > other than cevians of P, Q ), with the six , furthermore P, Q,
                            > intersection points of their cevians, as follows:
                            >
                            > - As D, the intersection point of BQb, CPc.
                            > - As D', the intersection point of BPb, CQc.
                            > - As E, the intersection point of CQc, APa.
                            > - As E', the intersection point of CPc, AQa.
                            > - As F, the intersection point of AQa, BPb.
                            > - As F', the intersection point of APa, BQb.
                            >
                            > RESULTS:
                            > a)- The lines AD,BE,CF, are concurrent at a point so be it, X.
                            > b)- The lines AD',BE',CF', are concurrent at a point so be it, Y.
                            > c)- The lines DD',EE',FF',XY, are concurrent at a point so be it S.
                            >

                            Note that much of result c) appears in the glossary for ETC -
                            the point S is the crosspoint of P and Q.

                            We also observe that the triangles ABC, DEF, D'E'F' are all
                            triply perspective - the other perspectors are P and Q in each
                            case. We have a desmic structure containing A,B,C,D,E,F,D',E',F'
                            with perspectors X,Y,S. Indeed, any such desmic structure with
                            triply perspective triangles arises this way from a pair P,Q.

                            An alternative proof uses barycentrics :
                            Suppose that P = p1:p2:p3, and Q = q1:q2:q3.
                            Then X, Y are given by
                            x1:x2:x3 = 1/p2q3:1/p3q1:1/p1q2 and
                            y1:y2:y3 = 1/q2p3:1/q3p1:1/q1p2

                            In fact, we also have that the barycentric products P*Q and X*Y
                            are equal. If we write W for the common value, then we can say
                            that P,Q are W-isoconjugates, as are X,Y. W = p1q1:p2q2:p3q3.
                            This is obvious from the coordinates.

                            Further S = x1+y1:x2+y2:x3+y3 has the properties in c), and
                            patently lies on XY.

                            The point R = x1-x2:x2-y2:x3-y3 is also on XY - it is the
                            "harmon" of the desmic structure. It can be described as
                            the intersection of the tripolars of P and Q or as the
                            perspector of C(R), the circumconic through P and Q.

                            Alternatively, given R and W, we can recover P and Q as the
                            intersections of C(R) and the tripolar of R*, the W-isoconjugate
                            of R. Of course, P and Q may be complex. We can also recover S
                            as the R*-ceva conjugate of R.

                            X and Y are the intersections (other than A,B,C and the points
                            fixed by W-isoconjugation) of the pivotal isocubics pK(W,R)
                            and pK(W,S).

                            The points D,E,F,D',E',F' are intersections (other than A,B,C)
                            of pK(W,S) and nK0(W,R).

                            That's probably enough for now ....

                            Regards,

                            Wilson
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