- 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 - --- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
>

Additional properties:

> 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

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