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  • Antreas
    Let ABC be a triangle and A B C , A B C the cevian, pedal triangles of P, resp. Denote: Ab, Ac = the reflections of A in BB , CC Bc, Ba = the reflections of
    Message 1 of 22 , Feb 25, 2013
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      Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
      triangles of P, resp.

      Denote:

      Ab, Ac = the reflections of A' in BB', CC'

      Bc, Ba = the reflections of B' in CC', AA'

      Ca, Cb = the reflections of C' in AA', BB'

      Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)

      La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.

      For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
      (antipode of Feuerbach point)

      For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
      on the pedal circle of H (=NPC).

      Which is the point of concurrence?

      In general:
      Which is the locus of P such that the lines La,Lb,Lc
      are concurrent?

      Antreas
    • Angel
      Dear Antreas, If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle. If P=H the lines La,Lb,Lc concur in X(1986)=
      Message 2 of 22 , Feb 26, 2013
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        Dear Antreas,

        If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle.

        If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry Wolk,Hyacinthos 7876,9/13/03)

        X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)


        In general:
        The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma) of degree nine (SA, SB, SC usual Conway notation):


        (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-

        (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+

        (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-

        (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+

        (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
        2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
        4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
        2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
        (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+

        (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
        (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
        SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
        (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
        (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
        (SA-4SB)(SA+SB)SC)y^5)+

        (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
        (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
        SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*

        (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
        x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
        SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
        x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
        SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+

        ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+

        4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-

        (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
        SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+

        (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
        SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-

        4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-

        (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-

        (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
        2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
        (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+

        (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6 =0


        (There must be a better simplification of this equation!)


        This curve (Gamma) contains points isodynamic (X13, X14) but the corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.

        The triangle center X(74) -isogonal conjugate of Euler infinity point- is on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
        passing through A'', B'', C'' resp. (concurrent at X74 = common circumcenter of the triangles).


        Best regards
        Angel Montesdeoca

        --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
        >
        > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
        > triangles of P, resp.
        >
        > Denote:
        >
        > Ab, Ac = the reflections of A' in BB', CC'
        >
        > Bc, Ba = the reflections of B' in CC', AA'
        >
        > Ca, Cb = the reflections of C' in AA', BB'
        >
        > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)
        >
        > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
        >
        > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
        > (antipode of Feuerbach point)
        >
        > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
        > on the pedal circle of H (=NPC).
        >
        > Which is the point of concurrence?
        >
        > In general:
        > Which is the locus of P such that the lines La,Lb,Lc
        > are concurrent?
        >
        > Antreas
        >
      • Angel
        More information on the algebraic curve of degree nine (Gamma): - Passes through the vertices of the triangle ABC. - The vertices are multiple points of order
        Message 3 of 22 , Feb 26, 2013
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          More information on the algebraic curve of degree nine (Gamma):

          - Passes through the vertices of the triangle ABC.

          - The vertices are multiple points of order 3.

          - The real tangents at the vertices of ABC intersect at the point X(74).


          Angel M.

          --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
          >
          >
          > Dear Antreas,
          >
          > If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle.
          >
          > If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry Wolk,Hyacinthos 7876,9/13/03)
          >
          > X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)
          >
          >
          > In general:
          > The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma) of degree nine (SA, SB, SC usual Conway notation):
          >
          >
          > (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-
          >
          > (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+
          >
          > (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-
          >
          > (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+
          >
          > (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
          > 2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
          > 4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
          > 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
          > (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+
          >
          > (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
          > (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
          > SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
          > (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
          > (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
          > (SA-4SB)(SA+SB)SC)y^5)+
          >
          > (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
          > (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
          > SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*
          >
          > (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
          > x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
          > SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
          > x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
          > SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+
          >
          > ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+
          >
          > 4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-
          >
          > (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
          > SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+
          >
          > (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
          > SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-
          >
          > 4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-
          >
          > (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-
          >
          > (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
          > 2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
          > (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+
          >
          > (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6 =0
          >
          >
          > (There must be a better simplification of this equation!)
          >
          >
          > This curve (Gamma) contains points isodynamic (X13, X14) but the corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.
          >
          > The triangle center X(74) -isogonal conjugate of Euler infinity point- is on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
          > passing through A'', B'', C'' resp. (concurrent at X74 = common circumcenter of the triangles).
          >
          >
          > Best regards
          > Angel Montesdeoca
          >
          > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
          > >
          > > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
          > > triangles of P, resp.
          > >
          > > Denote:
          > >
          > > Ab, Ac = the reflections of A' in BB', CC'
          > >
          > > Bc, Ba = the reflections of B' in CC', AA'
          > >
          > > Ca, Cb = the reflections of C' in AA', BB'
          > >
          > > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)
          > >
          > > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
          > >
          > > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
          > > (antipode of Feuerbach point)
          > >
          > > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
          > > on the pedal circle of H (=NPC).
          > >
          > > Which is the point of concurrence?
          > >
          > > In general:
          > > Which is the locus of P such that the lines La,Lb,Lc
          > > are concurrent?
          > >
          > > Antreas
          > >
          >
        • Antreas Hatzipolakis
          Dear Angel Thank you. I came to this configuration trying to find three homocentric (concentric) circles but not by construction (ie not by taking a point as
          Message 4 of 22 , Feb 26, 2013
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            Dear Angel

            Thank you.

            I came to this configuration trying to find three homocentric (concentric)
            circles
            but not by construction (ie not by taking a point as center).

            Antreas



            On Tue, Feb 26, 2013 at 2:57 PM, Angel <amontes1949@...> wrote:

            > **
            >
            >
            >
            > Dear Antreas,
            >
            > If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of
            > Feuerbach point on the incircle.
            >
            > If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT
            > (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry
            > Wolk,Hyacinthos 7876,9/13/03)
            >
            > X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)
            >
            > In general:
            > The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma)
            > of degree nine (SA, SB, SC usual Conway notation):
            >
            > (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-
            >
            > (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+
            >
            > (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-
            >
            > (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+
            >
            > (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
            > 2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
            > 4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
            > 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
            > (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+
            >
            > (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
            > (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
            > SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
            > (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
            > (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
            > (SA-4SB)(SA+SB)SC)y^5)+
            >
            > (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
            > (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
            > SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*
            >
            > (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
            > x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
            > SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
            > x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
            > SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+
            >
            > ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+
            >
            > 4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-
            >
            > (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
            > SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+
            >
            > (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
            > SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-
            >
            > 4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-
            >
            > (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-
            >
            > (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
            > 2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
            > (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+
            >
            > (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6
            > =0
            >
            > (There must be a better simplification of this equation!)
            >
            > This curve (Gamma) contains points isodynamic (X13, X14) but the
            > corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.
            >
            > The triangle center X(74) -isogonal conjugate of Euler infinity point- is
            > on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
            > passing through A'', B'', C'' resp. (concurrent at X74 = common
            > circumcenter of the triangles).
            >
            > Best regards
            > Angel Montesdeoca
            >
            >
            > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
            > >
            > > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
            > > triangles of P, resp.
            > >
            > > Denote:
            > >
            > > Ab, Ac = the reflections of A' in BB', CC'
            > >
            > > Bc, Ba = the reflections of B' in CC', AA'
            > >
            > > Ca, Cb = the reflections of C' in AA', BB'
            > >
            > > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp.
            > (concurrent at P = common circumcenter of the triangles)
            > >
            > > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
            > >
            > > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
            > > (antipode of Feuerbach point)
            > >
            > > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
            > > on the pedal circle of H (=NPC).
            > >
            > > Which is the point of concurrence?
            > >
            > > In general:
            > > Which is the locus of P such that the lines La,Lb,Lc
            > > are concurrent?
            > >
            > > Antreas
            > >
            >
            >
            >



            --
            http://anopolis72000.blogspot.com/


            [Non-text portions of this message have been removed]
          • Antreas Hatzipolakis
            [APH] ... More for P = H: The NPCs of A AbAc, B BcBa, C CaCb are concurrent on the NPC of A B C (On the Poncelet point of H with respect A B C ie the point
            Message 5 of 22 , Feb 26, 2013
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              [APH]


              >
              > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
              > triangles of P, resp.
              >
              > Denote:
              >
              > Ab, Ac = the reflections of A' in BB', CC'
              >
              > Bc, Ba = the reflections of B' in CC', AA'
              >
              > Ca, Cb = the reflections of C' in AA', BB'
              >
              > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent
              > at P = common circumcenter of the triangles)
              >
              > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
              >
              > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
              > (antipode of Feuerbach point)
              >
              > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
              > on the pedal circle of H (=NPC).
              >
              >
              >
              More for P = H:

              The NPCs of A'AbAc, B'BcBa, C'CaCb are concurrent on
              the NPC of A'B'C' (On the Poncelet point of H with respect
              A'B'C' ie the point of concurrence of the NPCs of
              A'B'C', HB'C', HC'A', HA'B'. So we have seven concurrent NPCs)

              The parallels through A,B,C to the (concurrent at H)
              Euler lines Ea, Eb, Ec of A'AbAc, B'BcBa, C'CaCb, resp. concur on the
              circumcircle of ABC on the antipode of the Euler line
              reflection point. And the perpendiculars, on the Euler line reflection
              point.


              APH


              [Non-text portions of this message have been removed]
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