## Re: [EMHL] Another Fruitful Theoerem

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• ... 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
• ... 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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