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Re: [EMHL] Feuerbach

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  • Alexey Zaslavsky
    Dear Randy and Barry! Yes this is well known. The synthetic proof is in our with Akopjan book Geometry of conics . Sincerely
    Message 1 of 23 , May 1 10:48 PM
      Dear Randy and Barry!
      Yes this is well known. The synthetic proof is in our with Akopjan book "Geometry of conics".

      Sincerely Alexey

      One generalization:

      Let P be any point on the rectangular hyperbola ABCHQ.
      Then the cevian circle of P goes through the center of the hyperbola (on the nine-point circle).

      Randy

      >________________________________
      > From: Barry Wolk <wolkbarry@...>
      >To: "Hyacinthos@yahoogroups.com" <Hyacinthos@yahoogroups.com>
      >Sent: Monday, April 30, 2012 5:08 PM
      >Subject: [EMHL] Feuerbach
      >
      >
      >
      >I have a strange result. Let P be any point on the conic through ABCHI.
      >Then the cevian circle of P goes through the Feuerbach point.
      >
      >Generalizations?
      >--
      >Barry Wolk
      >
      >
      >
      >

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





      [Non-text portions of this message have been removed]
    • Antreas Hatzipolakis
      Let ABC be a triangle and A B C the cevian triangle of I. Ab := the Orthogonal projection of A on BB Ac := the Orthogonal projection of A on CC Similarly
      Message 2 of 23 , Aug 10, 2016

        Let ABC be a triangle and A'B'C' the cevian triangle of I.

        Ab := the Orthogonal projection of A on BB'
        Ac := the Orthogonal projection of A on CC'

        Similarly Bc,Ba; Ca,Cb

        The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
        the Feuerbach point.

        APH
        https://groups.yahoo.com/neo/ groups/hyacinthos/ conversations/topics/10485

        Now, denote: L1, L2, L3 = the Euler lines of AAbAc, BBcBa, CCaCb, resp.

        The parallels to L1, L2, L3 through:

        1. A,B,C, resp.

        2. A', B', C' ,resp. (where A'B'C' = pedal triangle of I)

        3. Ia, Ib, Ic, resp. (where IaIbIc = excentral triangle)

        4. A",B",C", resp. (where A", B", C' = the orth. proj. of Ia,Ib,Ic on BC,CA,AB,resp.)

        are concurrent.

        [ 1 = reflection of I in F,  2 = reflection of F in I, 3 = reflection of 1 in 4]

        APH



      • Antreas Hatzipolakis
        ... Furthermore: Denote: A b, A c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.) L 1 = the Euler line of AA bA c.
        Message 3 of 23 , Aug 10, 2016

          Let ABC be a triangle and A'B'C' the cevian triangle of I.

          Ab := the Orthogonal projection of A on BB'
          Ac := the Orthogonal projection of A on CC'

          Similarly Bc,Ba; Ca,Cb

          The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
          the Feuerbach point.

          APH
          https://groups.yahoo.com/neo/g roups/hyacinthos/conversations /topics/10485

          Now, denote: L1, L2, L3 = the Euler lines of AAbAc, BBcBa, CCaCb, resp.

          The parallels to L1, L2, L3 through:

          1. A,B,C, resp.

          2. A', B', C' ,resp. (where A'B'C' = pedal triangle of I)

          3. Ia, Ib, Ic, resp. (where IaIbIc = excentral triangle)

          4. A",B",C", resp. (where A", B", C' = the orth. proj. of Ia,Ib,Ic on BC,CA,AB,resp.)

          are concurrent.

          [ 1 = reflection of I in F,  2 = reflection of F in I, 3 = reflection of 1 in 4]

          APH


          Furthermore:

          Denote:

          A'b, A'c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.)

          L'1 = the Euler line of AA'bA'c. Similarly L'2, L'3

          L'1, L'2, L'3 are concurrent at a point F'
          (they pass through the exFeuerbach points Fa, Fb, Fc, resp.)

          The parallels to L'1, L'2, L'3 through:

          1. A,B,C, resp.

          2. A', B', C' ,resp. (where A'B'C' = pedal triangle of I)

          3. Ia, Ib, Ic, resp. (where IaIbIc = excentral triangle)

          4. A",B",C", resp. (where A", B", C' = the orth. proj. of Ia,Ib,Ic on BC,CA,AB,resp.)

          are concurrent.

          [ 1 = reflection of I in 2,  4 = reflection of F' in 3]

          APH






          --
        • Antreas Hatzipolakis
          Thanks to Angel for his responses !! Now, let s continue..... Let ABC be a triangle. Denote: A b, A c = the orthogonal projections of A on the external
          Message 4 of 23 , Aug 12, 2016

            Thanks to Angel for his responses !!

            Now, let's continue.....



            Let ABC be a triangle.

            Denote:

            A'b, A'c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.)

            L'1 = the Euler line of AA'bA'c. Similarly L'2, L'3

            L'1, L'2, L'3 are concurrent at a point F'
            (they pass through the exFeuerbach points Fa, Fb, Fc, resp.)

            The parallels to L'1, L'2, L'3 through:

            [...]

            5. A1, B1, C1 (where A1, B1, C1 are the orthogonal projections of the excenters Ia, Ib, Ic, resp.  on the sides B'C', C'A', A'B' of the pedal triangle of I)

            6. A2, B2, C2 (where A2, B2, C2 are the orthogonal projections of the vertices A', B', C', resp. of the pedal triangle of I on the sides IbIc, IcIa, IaIb of the excentral triangle)

            are concurrent (I think !).

            APH

          • Antreas Hatzipolakis
            [APH]: Let ABC be a triangle. Denote: A b, A c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.) L 1 = the Euler line
            Message 5 of 23 , Aug 13, 2016
              [APH]:

              Let ABC be a triangle.

              Denote:

              A'b, A'c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.)

              L'1 = the Euler line of AA'bA'c. Similarly L'2, L'3

              L'1, L'2, L'3 are concurrent at a point F'
              (they pass through the exFeuerbach points Fa, Fb, Fc, resp.)

              The parallels to L'1, L'2, L'3 through:

              [...]

              5. A1, B1, C1 (where A1, B1, C1 are the orthogonal projections of the excenters Ia, Ib, Ic, resp.  on the sides B'C', C'A', A'B' of the pedal triangle of I)

              6. A2, B2, C2 (where A2, B2, C2 are the orthogonal projections of the vertices A', B', C', resp. of the pedal triangle of I on the sides IbIc, IcIa, IaIb of the excentral triangle)

              are concurrent (I think !).

              [Angel Montesdeoca]:

              5. The parallels to L'1, L'2, L'3 through A1, B1, C1 (where A1, B1, C1 are the orthogonal projections of the excenters Ia, Ib, Ic, resp.   on the sides B'C', C'A', A'B' of the pedal triangle of I) are concurrent in
              W5 =  (r+2R) X(1) - r X(21) = (r+4R) X(7) - (r+2R) X(79).

              W5 = (a (a^4 (b-c)^2-a^5 (b+c)+(b^2-c^2)^2 (b^2-b c+c^2)-a (b-c)^2 (b^3+4 b^2 c+4 b c^2+c^3)+a^3 (2 b^3+3 b^2 c+3 b c^2+2 c^3)+a^2 (-2 b^4+3 b^3 c+6 b^2 c^2+3 b c^3-2 c^4)):...:...)

              with (6-9-13)-search number  (1.52086596235072, 1.62091170244161, 1.81655670528601).

              W5 is   the midpoint of X(i) and X(j) for these {i,j}:  {3647,3874}.

              W5 lies on lines X(i)X(j) for {i,j}: {1,21}, {7,79}, {20,5441}, {27,1844}, {30,553}, {46,7675}, {57,3651}, {65,4304}, {72,5325}, {142,442}, {226,6841}, {354,946}, {377,5883}, {938,2475}, {1012,5884}, {1100,3284}, {1387,2771}, {1697,8000}, {1699,9960}, {1729,2280}, {2646,5427}, {3085,5686}, {3336,7411}, {3555,5837}, {3584,4015}, {3648,9965}, {3833,4197}, {4313,5903}, {5044,6675}, {5249,6701}, {5273,5904}, {5570,6744}, {5719,10021}, {5735,7671}, {5836,8261}.


              6. The parallels to L'1, L'2, L'3 through A2, B2, C2 (where A2, B2, C2 are the orthogonal projections of the vertices A', B', C', resp. of the pedal triangle of I on the sides IbIc, IcIa, IaIb of the excentral triangle) are concurrent in
              W6 =  (r+3R)X(21) - (r+4R)X(142) = (2r+R) X(35)   - (2r+5R) X(79).

              W6 = (2 a^7-a^6 (b+c)-(b-c)^4 (b+c)^3+5 a b c (b^2-c^2)^2+a^3 (b+c)^2 (2 b^2-3 b c+2 c^2)-a^5 (4 b^2+6 b c+4 c^2)+a^4 (b^3-4 b^2 c-4 b c^2+c^3)+a^2 (b-c)^2 (b^3+6 b^2 c+6 b c^2+c^3):...:...)

              with (6-9-13)-search number  (-1.13150673357715, -1.02446396734287, 4.87214264402659).

              W6 is the midpoint of X(i) and X(j) for these {i,j}:  {79,1770}.

              W6 lies on lines X(i)X(j) for {i,j}: {4, 1768}, {9, 3648}, {21, 142}, {30, 553}, {35, 79}, {63, 2475}, {191, 3474}, {442, 1155}, {516, 3649}, {1708, 7701}, {1836, 5248}, {3874, 7354}, {3911, 6841}, {5122, 10021}, {5325, 6175}, {5441, 5557}

               
                 Angel Montesdeoca




              --
            • Antreas Hatzipolakis
              Let ABC be a triangle and A B C the cevian triangle of I. Denote: Ab, Ac = the reflections of A in BB , CC , resp. La = the Euler line of A AbAc. Similarly
              Message 6 of 23 , Aug 17, 2016

                Let ABC be a triangle and A'B'C' the cevian triangle of I.

                Denote:

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

                La = the Euler line of A'AbAc. Similarly Lb, Lc.

                A1, A2, A3 = the orthogonal projections of A on La, Lb, Lc, resp.

                L1 = the Euler line of A1A2A3. Similarly L2, L3 = the Euler lines of B1B2B3 and C1C2C3 resp.

                The L1, L2, L3 concur at Feuerbach point of ABC.

                Note:
                Let A'b, A'c be the orthogonal projections of A on BB', CC' and L'a the Euler line of AA'bA'c.
                Similarly L'b, L'c.
                The L'a, L'b, L'c concur at Feuerbach point (*)
                Now, the triangles AA'bA'c and A1A2A3 share the same Euler line (the one is the reflection of the other in its Euler line). So L'a = L1 and similarly L'b = L2, L'c = L3.
                Therefore L1, L2, L3 concur at Feuerbach point.

                APH
              • Antreas Hatzipolakis
                Let ABC be a triangle. Denote: A = the intersection of BC and the I-bisector of the triangle BIC B = the intersection of CA and the I-bisector of the
                Message 7 of 23 , Aug 26, 2016
                  Let ABC be a triangle.

                  Denote:

                  A' = the intersection of BC and the I-bisector of the triangle BIC
                  B' = the intersection of CA and the I-bisector of the triangle CIA
                  C' = the intersection of AB and the I-bisector of the triangle AIB.

                  I think the circumcircle of A'B'C' passes through the Feuerbach point.
                  If so we have four concurrent circles
                  (the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')

                  APH


                • Antreas Hatzipolakis
                  [APH]: Let ABC be a triangle. Denote: A = the intersection of BC and the I-bisector of the triangle BIC B = the intersection of CA and the I-bisector of the
                  Message 8 of 23 , Aug 27, 2016
                    [APH]:

                    Let ABC be a triangle.

                    Denote:

                    A' = the intersection of BC and the I-bisector of the triangle BIC
                    B' = the intersection of CA and the I-bisector of the triangle CIA
                    C' = the intersection of AB and the I-bisector of the triangle AIB.

                    I think the circumcircle of A'B'C' passes through the Feuerbach point.
                    If so we have four concurrent circles
                    (the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')

                    [Peter Moses]:

                    Hi Antreas,
                     
                    >I think the circumcircle of A'B'C' passes through the Feuerbach point.
                    It does !
                     
                    Best regards,
                    Peter.




                  • Antreas Hatzipolakis
                    [APH]: Let ABC be a triangle. Denote: A = the intersection of BC and the I-bisector of the triangle BIC B = the intersection of CA and the I-bisector of the
                    Message 9 of 23 , Aug 27, 2016

                      [APH]:

                      Let ABC be a triangle.

                      Denote:

                      A' = the intersection of BC and the I-bisector of the triangle BIC
                      B' = the intersection of CA and the I-bisector of the triangle CIA
                      C' = the intersection of AB and the I-bisector of the triangle AIB.

                      I think the circumcircle of A'B'C' passes through the Feuerbach point.
                      If so we have four concurrent circles
                      (the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')

                      [Peter Moses]:

                      Hi Antreas,
                       
                      >I think the circumcircle of A'B'C' passes through the Feuerbach point.
                      It does !
                       
                      Best regards,
                      Peter.

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

                      Thanks, Peter!

                      Which point is the center of the circle?

                      I guess it is complicated and not listed in ETC.

                      APH





                      --
                    • Antreas Hatzipolakis
                      [APH]: Let ABC be a triangle. Denote: A = the intersection of BC and the I-bisector of the triangle BIC B = the intersection of CA and the I-bisector of the
                      Message 10 of 23 , Aug 27, 2016



                        [APH]:

                        Let ABC be a triangle.

                        Denote:

                        A' = the intersection of BC and the I-bisector of the triangle BIC
                        B' = the intersection of CA and the I-bisector of the triangle CIA
                        C' = the intersection of AB and the I-bisector of the triangle AIB.

                        I think the circumcircle of A'B'C' passes through the Feuerbach point.
                        If so we have four concurrent circles
                        (the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')

                        [Peter Moses]:

                        Hi Antreas,
                         
                        >I think the circumcircle of A'B'C' passes through the Feuerbach point.
                        It does !
                         
                        Best regards,
                        Peter.

                        [APH]:

                        Thanks, Peter!

                        Which point is the center of the circle?

                        I guess it is complicated and not listed in ETC.

                        [Peter Moses]:

                        Hi Antreas,
                         
                        The triangle A’B’C’ is the cevian of X(174).
                         
                        >I guess it is complicated and not listed in ETC.
                        Yes, it appears that way.  things quicky became very long and I decided not to bother wrestling with 20 pages of square roots!
                         
                        Best regards,
                        Peter.
                         






                        --



                        --
                      • Antreas Hatzipolakis
                        [APH]: Let ABC be a triangle. Denote: A = the intersection of BC and the I-bisector of the triangle BIC B = the intersection of CA and the I-bisector of the
                        Message 11 of 23 , Aug 27, 2016




                          [APH]:

                          Let ABC be a triangle.

                          Denote:

                          A' = the intersection of BC and the I-bisector of the triangle BIC
                          B' = the intersection of CA and the I-bisector of the triangle CIA
                          C' = the intersection of AB and the I-bisector of the triangle AIB.

                          I think the circumcircle of A'B'C' passes through the Feuerbach point.
                          If so we have four concurrent circles
                          (the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')

                          [Peter Moses]:

                          The triangle A’B’C’ is the cevian of X(174).

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

                          Whose, other than I, X(174),  points the cevian circle passes through the Feuerbach point?


                          APH
                           









                          --
                        • Antreas Hatzipolakis
                          [APH]: Let ABC be a triangle. Denote: A = the intersection of BC and the I-bisector of the triangle BIC B = the intersection of CA and the I-bisector of the
                          Message 12 of 23 , Aug 31, 2016
                             

                            [APH]:

                            Let ABC be a triangle.

                            Denote:

                            A' = the intersection of BC and the I-bisector of the triangle BIC
                            B' = the intersection of CA and the I-bisector of the triangle CIA
                            C' = the intersection of AB and the I-bisector of the triangle AIB.

                            I think the circumcircle of A'B'C' passes through the Feuerbach point.
                            If so we have four concurrent circles
                            (the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')

                            [Peter Moses]:

                            The triangle A'B'C' is the cevian of X(174).

                            [APH]:

                            Whose, other than I, X(174),  points the cevian circle passes through the Feuerbach point?


                            [Angel Montesdeoca]:


                            The cevian circle of P passes through the Feuerbach point
                            If P lies on the Feuerbach Hyperbola or on the quartic (Q): a(b-c)y^2z^2+b(c-a)z^2x^2+c(a- b)x^2y^2 =0.

                            ((Q) passes through the vertices (double) of ABC, the vertices of the anticomplementary triangle, X(2), X(7), X(174), X(189), X(366), X(1029), X(5451), X(8046), X(8051))

                            Triangle center X(i) such that the cevian circle passes through the Feuerbach point, i:
                            {1, 2, 4, 7, 8, 9, 21, 79, 80, 84, 90, 104, 174, 177, 189, 256, 294, 314, 366, 885, 941, 943, 981, 983, 987, 989, 1000, 1029, 1039, 1041, 1061, 1063, 1156, 1172, 1251, 1320, 1389, 1392, 1476, 1896, 1937, 2298, 2320, 2335, 2344, 2346, 2481, 2648, 2997, 3062, 3065, 3254, 3255, 3296, 3307, 3308, 3427, 3467, 3495, 3551, 3577, 3680, 4180, 4866, 4876, 4900, 5377, 5424, 5451, 5551, 5553, 5555, 5556, 5557, 5558, 5559, 5560, 5561, 5665, 6595, 6596, 6597, 6598, 6599, 6601,
                            7003, 7049, 7091, 7126, 7133, 7149, 7155, 7160, 7161, 7162, 7261, 7284, 7285, 7317, 7319, 7320, 7595, 7707, 8046, 8051, 8372, 8759, 8809, 9365, 9372, 9442}


                            Angel Montesdeoca




                            --
                          • Antreas Hatzipolakis
                            Let ABC be a triangle. Denote: Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp. Aaa, Aab, Aac = the orthogonal projections of A on INa,INb,INc, resp. L1 =
                            Message 13 of 23 , Sep 28, 2016

                              Let ABC be a triangle.

                              Denote:

                              Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

                              Aaa, Aab, Aac = the orthogonal projections of A on INa,INb,INc, resp.

                              L1 = the Eueler line of AaaAabAac. Similarly L2, L3.

                              L1, L2, L3 concur at the Feuerbach point of ABC.

                              If Ab, Ac are the orthogonal projections of A on IB, IC, then Euler line of AaaAAbAAac = Euler line of AAbAc.
                              Similarly Bc,Ba and Ca, Cb
                              We know that the Euler lines of AAbAc, BBcBa, CCaCb concur at the Feuerbach point.

                              APH
                            • Antreas Hatzipolakis
                              Let ABC be a triangle, P a point and A B C the pedal triangle of I. Denote: Oa, Ob, Oc = the circumcenters of IB C , IC A , IA B , resp. Ooa, Oob, Ooc = the
                              Message 14 of 23 , Oct 22, 2016
                                Let ABC be a triangle, P a point and A'B'C' the pedal triangle of I.

                                Denote:

                                Oa, Ob, Oc = the circumcenters of IB'C', IC'A', IA'B', resp.

                                Ooa, Oob, Ooc = the circumcenters of IObOc, IOcOa, IOaOb, resp.

                                The reflections of A'Ooa, B'Oob, C'Ooc in BC,CA, AB (or equivalently in IA', IB', IC'), resp. are concurrent at the antipode in the incircle of the Feuerbach point.

                                APH
                              • Antreas Hatzipolakis
                                Another construction of the Feuerbach point: Let ABC be a triangle and A B C the pedal triangle of I. Denote: T = the excentral triangle of A B C T1, T2, T3
                                Message 15 of 23 , Apr 21
                                  Another construction of the Feuerbach point:

                                  Let ABC be a triangle and A'B'C' the pedal triangle of I.

                                  Denote:

                                  T = the excentral triangle of A'B'C'

                                  T1, T2, T3 = the pedal triangles of A, B, C wrt triangle T, resp.

                                  The circumcircles of T1, T2, T3 concur at Feuerbach point of ABC.

                                  APH
                                • Antreas Hatzipolakis
                                  Another construction of the Feuerbach point: Let ABC be a triangle and A B C the pedal triangle of I. Denote: T = the excentral triangle of A B C T1, T2, T3
                                  Message 16 of 23 , Apr 22


                                    Another construction of the Feuerbach point:

                                    Let ABC be a triangle and A'B'C' the pedal triangle of I.

                                    Denote:

                                    T = the excentral triangle of A'B'C'

                                    T1, T2, T3 = the pedal triangles of A, B, C wrt triangle T, resp.

                                    The circumcircles of T1, T2, T3 concur at Feuerbach point of ABC.


                                    Variation:

                                    Let IaIbIc be the excentral triangle of ABC and A', B', C' the orthogonal projections of Ia, Ib, Ic on BC, CA, AB, resp.

                                    T = the excentral triangle of A'B'C'

                                    Let Ta, Tb, Tc be the pedal triangles of A, B, C wrt triangle T.

                                    The circumcircles of Ta, Tb, Tc concur at the Feuerbach point of ABC. ABC.


                                    APH



                                  • Antreas Hatzipolakis
                                    [APH]: Another construction of the Feuerbach point: Let ABC be a triangle and A B C the pedal triangle of I. Denote: T = the excentral triangle of A B C T1,
                                    Message 17 of 23 , Apr 22
                                      [APH]:

                                      Another construction of the Feuerbach point:

                                      Let ABC be a triangle and A'B'C' the pedal triangle of I.

                                      Denote:

                                      T = the excentral triangle of A'B'C'

                                      T1, T2, T3 = the pedal triangles of A, B, C wrt triangle T, resp.

                                      The circumcircles of T1, T2, T3 concur at Feuerbach point of ABC.


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

                                       

                                      Dear Mr Antreas,

                                      I see a general problem

                                      Let P is a point on Feuerbach Hyperbola of a triangle ABC and A'B'C' is the cevian triangle of P wrt ABC.

                                      T = the excentral triangle of A'B'C'

                                      Let Ta, Tb, Tc be the pedal triangles of A, B, C wrt triangle T.

                                      The circumcircles of Ta, Tb, Tc concur at the Feuerbach point of ABC.

                                      And more general

                                      Let ABC be a triangle and A'B'C' is the cevian triangle of any P wrt ABC.
                                       
                                      H is the orthocenter of ABC.

                                      T = the excentral triangle of A'B'C'.

                                      Let Ta, Tb, Tc be the pedal triangles of A, B, C wrt triangle T.

                                      The circumcircles of Ta, Tb, Tc concur at the center of the Hyperbola through A,B,C,H,P.

                                      Best regards,
                                      Tran Quang Hung.


                                    • Antreas Hatzipolakis
                                      Let ABC be a triangle and A B C the pedal triangles of I. Denote: A*, B*, C* = the reflections of A , B , C in AI, BI, CI resp. Ba = B*C Intersection BI Ca
                                      Message 18 of 23 , May 14 4:10 AM

                                        Let ABC be a triangle and A'B'C' the pedal triangles of I.

                                        Denote:

                                        A*, B*, C* = the reflections of A', B', C' in AI, BI, CI resp.

                                        Ba = B*C' Intersection BI
                                        Ca = C*B' Intersection CI

                                        Similarly Cb, Ab, Ac, Bc 

                                        The Euler lines of ABaCa, BCbAb, CAcBc concur at Feuerbach point.

                                        Note:
                                        It is equivalent to Hyacinthos #10485
                                        (for Ba, Ca are the orthogonal projections of A on BI, CI, resp.)

                                        APH
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