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Re: [EMHL] Another Fruitful Theoerem

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  • xpolakis@otenet.gr
    ... I confused with the equalities! will look at it later. Sorry! APH
    Message 1 of 2 , Sep 1, 2000
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      I 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")
      >
      >PROOF:

      I confused with the equalities! will look at it later. Sorry!

      APH
    • xpolakis@otenet.gr
      ... distinct ... Here is the proof (I deleted what was wrong or not needed in my previous posting): Triangle (A B C); Transversal C*A B : C*B B C A A
      Message 2 of 2 , Sep 2, 2000
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        I wrote:

        >THEOREM:
        >
        >Let P, Q be two points on the plane of ABC, and A'B'C', A"B"C" their cevian
        distinct
        >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:

        Here is the proof (I deleted what was wrong or not needed in my previous
        posting):

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

        C*B" B'C A'A"
        ---- x --- x ---- = 1
        C*A" B'B" A'C


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

        A*C" B'B" C'A
        ---- x ---- x ----- = 1
        A*B" B'A C'C"


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

        B*A" C'C" A'B
        ---- x ----- x ----- = 1
        B*C" C'B A'A"


        ==>

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

        Now:

        Triangle (ABC); Concurrent Cevians: AA', BB', CC':
        (A'B'C' is cevian triangle of some point P)

        C'A A'B B'C
        --- x ---- x ---- = -1
        C'B A'C B'A

        ==>

        C*B" A*C" B*A"
        --- x ---- x ---- = -1
        C*A" A*B" B*C"

        From this, we conclude that in the triangle (A"B"C") the cevians
        A*A", B*B", C*C" are concurrent (A*B*C* is in perspective with A"B"C")
        [Converse of Ceva Theorem]
        Similarly for the other triad (A'A*, B'B*, C'C*)

        Note that in the proof above, we used the "cevianity" of (A'B'C') only.
        This means that the theorem is actually a porism (corollary) of
        a general one. Namely:

        Let A'B'C', A"B"C" be two triangles inscribed in a reference triangle ABC.
        Denote:
        B'C' /\ B"C" := A*
        C'A' /\ C"A" := B*
        A'B' /\ A"B" := C*

        If A'B'C' is in perspective with ABC, then A"B"C" in perspective with A*B*C*
        (ie A"A*, B"B* , C"C* concur)

        Antreas
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