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