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:
Best Answer
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 $(\heartsuit)$.