Is it a new discovery on conic section? I discovered a problem in plane geometry (there are some nice special cases) as follows:

Let $ABC$ be a triangle and $\Omega$ be arbitrary circumconic. Let two points $A_b, A_c \in BC$, $B_c, B_a \in CA$, $C_a, C_b \in AB$, let the line $AA_b, AA_c$ meet the circumconic again at $A'_b, A'_c$ define $B'_c, B'_a$ and $C'_a, C'_b$ cyclically, let $A'_bA'_c \cap BC = A'$ define $B', C'$ cyclically.


Question: I am looking for a proof that $A', B', C'$ are collinear if and only if $A_b, A_c, B_c, B_a, C_a, C_b$ lie on a conic and Is it a new discovery on conic section?

See also:

*

*Carnot's theorem (conics)

*Dao's theorem (conics)

*Reuschle's theorem

 A: It suffices to consider the case when $\Omega$ is a circumcircle, so let it be.
At first, the points $A_b, A_c, B_c, B_a, C_a, C_b$ lie on a conic if and only if
$$
\frac{AB_a\cdot AB_c}{AC_a\cdot AC_b}\cdot \frac{BC_a\cdot BC_b}{BA_b\cdot BA_c}\cdot
\frac{CA_b\cdot CA_c}{CB_a\cdot CB_c}=1\quad\quad\quad\quad(\heartsuit)
$$
(by Pascal theorem, they lie on a conic if and only if $X:=B_aC_a\cap BC$ and two analogous points are collinear. Applying Menelaus theorem we get $XB:XC=(BC_a\cdot AB_a):(AC_a\cdot CB_a)$. Multiplying three such expressions we get $(\heartsuit)$.)
For finding $B'A:B'C$ (we look at collinearity of $A',B',C'$ via Menelaus too) we project the quadruple $(B',A,B_a,C)$ to $\Omega$ from the point $B_a'$. We get
$$
\frac{B'A}{B'C}:\frac{B_aA}{B_aC}=\frac{B'_cA}{B'_cC}:\frac{BA}{BC}=
\frac{S(B'_cAB)/BA}{S(B'_cCB)/BC}:\frac{BA}{BC}=\frac{BC^2}{BA^2}\cdot \frac{AB_c}{CB_c}.
$$
Substituting this to Menelaus we get condition $(\heartsuit)$.
A: Application of the theorem in post #1. I give a special case and give a proof as follows:
Generalization of conjugate of a point: Let $ABC$ be a triangle, and $\Omega$ is arbitrary circumconic of $ABC$, $P$ be arbitrary point in the plane. Let $AP$, $BP$, $CP$ meet the circumconic at $A'$, $B'$, $C'$. Three lines through $A'$, $B'$, $C'$ and parallel to $BC$, $CA$, $AB$ meets the circumconic again at $A''$, $B''$, $C''$ then $AA'', BB'', CC''$ are concurrent
Proof: Three lines through $A'$, $B'$, $C'$ and parallel to $BC$, $CA$, $AB$ meet $BC$, $CA$, $AB$ at three collinear points at $\infty$.
Let the lines $AA'$, $AA''$ meet $BC$ again at $A_{b}$, $A_{c}$ define $B_{c}$, $B_{a}$ and $C_{a}$, $C_{b}$ are cyclically, apply theorem #1 then six points  $A_{b}$, $A_{c}$, $B_{c}$, $B_{a}$, $C_{a}$, $C_{b}$ lie on a conic.
By Carnot theorem we have:
$$
\frac{AB_a\cdot AB_c}{AC_a\cdot AC_b}\cdot \frac{BC_a\cdot BC_b}{BA_b\cdot BA_c}\cdot
\frac{CA_b\cdot CA_c}{CB_a\cdot CB_c}=1\quad\quad\quad\quad(1)
$$
But $AA'$, $BB'$, $CC'$ are concurrent at $P$, other word $AA_b$, $BB_c$, $CC_a$ are concurrent at $P$, thus:
$$\frac{BA_b}{CA_b}.\frac{CB_b}{AB_b}.\frac{AC_a}{BC_a}=1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)
$$
Since $(1)$ and $(2)$ we get:
$$\frac{BA_c}{CA_c}.\frac{CB_a}{AB_a}.\frac{AC_b}{BC_b}=1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(3)
$$
Since $(3)$ by converse of the Ceva's theorem we get: Three lines $AA_c$, $BB_a$, $CC_b$ are concurrent, other words $AA'', BB'', CC''$ are concurrent
Professor César Lozada let me know that (via email):
Among all known conjugations, I found your conjugation is:

*

*Isogonal conjugate, for circumconic=circumcircle


*Isotomic conjugate, for circumconic=Steiner circumellipse


*Polar conjugate, for circumconic with center X(1249)
In general, for the circumconic with center $O=x : y : z$,  your transformation for a point $P=u : v : w$ is as simple as:
$Q(O,P) = x*(-x+y+z)*v*w : :$
