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Another Fruitful Theoerem

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  • xpolakis@otenet.gr
    THEOREM: Let P, Q be two points on the plane of ABC, and A B C , A B C their cevian triangles. Denote: B C / B C := A* C A / C A := B* A B / A B :=
    Message 1 of 2 , Aug 31, 2000
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      THEOREM:

      Let P, Q be two points on the plane of ABC, and A'B'C', A"B"C" their cevian
      triangles. Denote:
      B'C' /\ B"C" := A*
      C'A' /\ C"A" := B*
      A'B' /\ A"B" := C*

      Then:
      A'A*, B'B* , C'C* concur.
      A"A*, B"B* , C"C* concur.
      (or in other words: The triangle A*B*C* is in perspective with both
      triangles A'B'C', A"B"C")

      PROOF:

      I will prove that A"A*, B"B* , C"C* concur.
      (The proof of the concurrence of A'A*, B'B* , C'C* is similar)

      Denote B*B" /\ C*C" := J

      I shall prove that J belongs in the line A*A" as well
      (that is, A*,J,A" are collinear points).

      Triangle (A*A"C"); Transversal: B*JB":

      B*A" B"C" JA*
      ---- x ----- x ----- = 1 (1)
      B*C" B"A* JA"

      Triangle (C*C"A"); Trans. B*JB":

      B*C" B"A" JC*
      ---- x ----- x ----- = 1 (2)
      B*A" B"C* JC"

      Triangle (A*A"B"); Trans. C*JC":

      C*B" C"A* JA"
      ---- x ----- x ----- = 1 (3)
      C*A" C"B" JA*


      (1) X (2) X (3) ==>

      A*C" A"B" JC* C*A" JB" C"B* C*B" A*C" B*A"
      ( ---- x ---- x ---- ) x ( ---- x ---- x ---- ) x ( ---- x ----- x ---- ) = -1
      A*B" A"C* JC" C*B" JB* C"A" C*A" A*B" B*C"

      (4)
      We have:

      Triangle (B*B"A"); Trans. C*JC":

      C*A" JB" C"B*
      ---- x ----- x ----- = 1 (5)
      C*B" JB* C"A"


      Triangle (A"B"C); Trans. C*A'B':

      C*B" B'C A'A"
      ---- x --- x ---- = 1 (6a)
      C*A" B'B" A'C


      Triangle (B"C"A); Trans. A*B'C':

      A*C" B'B" C'A
      ---- x ---- x ----- = 1 (6b)
      A*B" B'A C'C"


      Triangle (C"A"B); Trans. B*A'C':

      B*A" C'C" A'B
      ---- x ----- x ----- = 1 (6c)
      B*C" C'B A'A"


      (6a) X (6b) X (6c) ==>

      C'A A'B B'C C*B" A*C" B*A"
      ( ---- x ---- x ---- ) x ( ---- x ----- x ----- ) = 1 (6)
      C'B A'C B'A C*A" A*B" B*C"

      Now:

      Triangle (ABC); Cevians: AA', BB', CC':

      C'A A'B B'C
      --- x ---- x ---- = -1 (7)
      C'B A'C B'A


      (6) and (7) ==>

      C*B" A*C" B*A"
      --- x ---- x ---- = -1 (8)
      C*A" A*B" B*C"

      (4) and (5) and (8) ==>

      A*C" A"B" JC*
      ---- x ---- x ---- = 1 ==> A*,J, A" are collinear.
      A*B" A"C* JC"

      (converse of Menemaus theorem for the triangle: C*C"B").

      O(/per E)/dei Dei=cai


      Antreas
    • Angel
      ... Additional properties: Let C(P,Q) be the conic through the traces of P and Q: C(P,Q) is the conic through the points A ,B ,C ,A ,B ,C . Let A,A*,A** be
      Message 2 of 2 , Jul 18, 2011
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        --- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
        >
        > THEOREM:
        >
        > Let P, Q be two points on the plane of ABC, and A'B'C', A"B"C" their cevian
        > triangles. Denote:
        > B'C' /\ B"C" := A*
        > C'A' /\ C"A" := B*
        > A'B' /\ A"B" := C*
        >
        > Then:
        > A'A*, B'B* , C'C* concur.
        > A"A*, B"B* , C"C* concur.
        > (or in other words: The triangle A*B*C* is in perspective with both
        > triangles A'B'C', A"B"C")


        > Antreas


        Additional properties:


        Let C(P,Q) be the conic through the traces of P and Q: C(P,Q) is the conic through the points A',B',C',A'',B'',C''.

        Let A,A*,A** be the diagonal points of quadrilateral B'C'B''C''.

        Let B,B*,B** be the diagonal points of quadrilateral C'A'C"A".

        Let C,C*,C** be the diagonal points of quadrilateral A'B'A"B".

        Let X' be the perspector of A'B'C' and A*B*C*, and X'' the perspector de A''B''C'' and A*B*C*.

        Let DEF be the triangle with sides A*A**, B*B**, C*C**.


        Then:


        1) Points A**, B**, and C** are on the line PQ.

        2) Te side B*C* is the line AA**; the side C*A* is the line BB**; and the side A*B* is the line CC**.

        3) The triangles ABC and A*B*C* are perspective; let Y* be the perspector.

        4) The triangles ABC and DEF are perspective; let Z be the perspector.


        5) The triangles A*B*C* and DEF are perspective; let Z* be the perspector perspector.


        6) The line PQ is the polar of Y* with respect to the conic C(P,Q).


        7) The line PQ is the perspective axis (or homology axis, or perspectrix) of the triangles A*B*C* and DEF.


        8) Points X' and X'' are on the conic C(P,Q).

        9) The line X'X'' is the perspectrix of ABC and A*B*C*.

        10) The line X'X'' is the polar of Z* with respect to the conic C(P,Q).

        11) The perspectrix of ABC and DEF is the polar of Z with respect to the conic C(P,Q).



        Particular cases:


        === If C(G,H)=the nine-point circle, then:

        Y*=X(523);

        X'=X(115) = CENTER OF KIEPERT HYPERBOLA; lie on the nine-point circle

        X''=X(125) = CENTER OF JERABEK HYPERBOLA; lie on the nine-point circle

        Z=( 1/(a^4-a^2(b^2+c^2)-b^2c^2): .... : ... )

        Z*=( (b^2-c^2)^2(a^4-a^2(b^2+c^2)-b^2c^2): ... : ...)



        === If C(X(1),X(1029))=The incentral circle, then:


        Y*=( a(b-c)(a^3+a^2(b+c)-a(b^2+b*c+c^2)-(b+c)^3): ...:...);

        X'=( a(b-c)^2(a^3+a^2(b+c)-a(b^2+b*c+c^2)-(b+c)^3)^2:...:...); lie on the incentral circle



        Z= (a/(a^8-a^7(b+c)-a^6(7b^2+11b*c+7c^2) -
        a^5(b^3+13b^2c+13b*c^2+c^3) +
        a^4(11b^4+11b^3c+3b^2c^2+11b*c^3+11c^4)+
        a^3(5b^5+19b^4c+23b^3c^2+23b^2c^3+19b*c^4+5c^5) +
        a^2(-5b^6+b^5c+15b^4c^2+19b^3c^3+15b^2c^4+b*c^5-5c^6)-
        a(b-c)^2(b+c)^3(3b^2+2b*c+3c^2) - b(b-c)^2c(b+c)^4 ):...:...);


        === If C(X(8),X(189))=the Mandart circle, then:

        Y*=X(522);

        X'=X(11) = FEUERBACH POINT; lie on the Mandart circle

        X''=((b-c)^2(-a+b+c)^2(a^3+a^2(b+c)-(b-c)^2(b+c)-a(b+c)^2): ... :...); lie on the Mandart circle


        Angel
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