I was trying to prove Pascal's theorem by using homogeneous coordinates with the following configurations (interactive graph at Desmos):
- A,B,C,D,E,F (homogeneous coordinates) are on a conic.
- G,H,I are the intersection points of pairs of lines (AB,DE), (CD,FA), (EF,BC) respectively.
Because G is the intersection of lines AB, DE, we can write (by using vector quadruple prodcut):
$$ \begin{array}{rcl} G &= & (A \times B) \times (D \times E) \\ &= & (A \cdot (D \times E))B - (B \cdot (D \times E))A \\ &= & \begin{vmatrix} A & D & E \end{vmatrix} B - \begin{vmatrix} B & D & E \end{vmatrix} A \\ &= & bB - aA \end{array} $$
where $b=\begin{vmatrix} A & D & E \end{vmatrix}$, $a=\begin{vmatrix} B & D & E \end{vmatrix}$
Note: $\begin{vmatrix} A & D & E \end{vmatrix}$ means the determinant determined by the three points A,D,E.
Similarly, we have:
$$ \begin{array}{rcl} H &= & (C \times D) \times (F \times A) \\ &= & (C \cdot (F \times A))D - (D \cdot (F \times A))C \\ &= & \begin{vmatrix} C & F & A \end{vmatrix} D - \begin{vmatrix} D & F & A \end{vmatrix} C \\ &= & dD - cC \end{array} $$
$$ \begin{array}{rcl} I &= & (E \times F) \times (B \times C) \\ &= & (E \cdot (B \times C))F - (F \cdot (B \times C))E \\ &= & \begin{vmatrix} E & B & C \end{vmatrix} F - \begin{vmatrix} F & B & C \end{vmatrix} E \\ &= & fF - eE \end{array} $$
where:
$$ d=\begin{vmatrix} C & F & A \end{vmatrix} \\ c=\begin{vmatrix} D & F & A \end{vmatrix} \\ f=\begin{vmatrix} E & B & C \end{vmatrix} \\ e=\begin{vmatrix} F & B & C \end{vmatrix} $$
If we were to prove that G, H, I are collinear, it would be suffice to prove that $\begin{vmatrix} G & H & I \end{vmatrix}=0$, so I went for it.
$$ \require{cancel} \begin{array}{ccl} \begin{vmatrix} G \\ H \\ I \end{vmatrix} & = & \begin{vmatrix} bB - aA \\ dD - cC \\ fF - eE \end{vmatrix} \\ & = & bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix} -bde \begin{vmatrix} B \\ D \\ E \end{vmatrix} -bcf \begin{vmatrix} B \\ C \\ F \end{vmatrix} +bce \begin{vmatrix} B \\ C \\ E \end{vmatrix} \\ & & -adf \begin{vmatrix} A \\ D \\ F \end{vmatrix} +ade \begin{vmatrix} A \\ D \\ E \end{vmatrix} +acf \begin{vmatrix} A \\ C \\ F \end{vmatrix} -ace \begin{vmatrix} A \\ C \\ E \end{vmatrix} \\ & = & bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix} \color{red}{\cancel{-bdea}} \color{blue}{\cancel{-bcfe}} \color{blue}{\cancel{+bcef}} \color{olive}{\cancel{-adfc}} \color{red}{\cancel{+adeb}} \color{olive}{\cancel{+acfd}} -ace \begin{vmatrix} A \\ C \\ E \end{vmatrix} \\ & = & bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix} -ace \begin{vmatrix} A \\ C \\ E \end{vmatrix} \end{array} $$
At this point, if we wanted to prove the theorem of Pappus, it would be obvious:
Since A,C,E and B,D,F are collinear respectively, it is obvious that:
$$ \begin{vmatrix} A & C & E \end{vmatrix}=0 \\ \begin{vmatrix} B & D & F \end{vmatrix}=0 $$
and hence we can conclude immediately that $\begin{vmatrix} G & H & I \end{vmatrix}=0$
But for the case of Pascal's theorem, how can we prove that:
$$ bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix} -ace \begin{vmatrix} A \\ C \\ E \end{vmatrix} = 0 $$
i.e.
$$ \begin{vmatrix} A \\ D \\ E \end{vmatrix} \begin{vmatrix} C \\ F \\ A \end{vmatrix} \begin{vmatrix} E \\ B \\ C \end{vmatrix} \begin{vmatrix} B \\ D \\ F \end{vmatrix} = \begin{vmatrix} B \\ D \\ E \end{vmatrix} \begin{vmatrix} D \\ F \\ A \end{vmatrix} \begin{vmatrix} F \\ B \\ C \end{vmatrix} \begin{vmatrix} A \\ C \\ E \end{vmatrix} $$
where A,B,C,D,E,F are points on a conic?
Define $$\phi: X \longmapsto \begin{vmatrix} X \\ D\\E\end{vmatrix} \begin{vmatrix} C \\ F\\X\end{vmatrix} \begin{vmatrix} E \\ B \\C\end{vmatrix} \begin{vmatrix} B \\ D \\F\end{vmatrix}-\begin{vmatrix} B \\ D \\E\end{vmatrix} \begin{vmatrix} D \\ F \\X\end{vmatrix} \begin{vmatrix} F \\ B \\C\end{vmatrix} \begin{vmatrix} X \\ C \\E\end{vmatrix}.$$
We can assume $\phi$ is a nonzero quadratic polynomial (else we are done). It is easy to check that $\phi$ vanishes at $X=C,D,E,F$, and not much harder to see that $\phi$ vanishes at $B$.
Assume that $B,C,D,E,F$ are pairwise distinct. Then the vanishing set of $\phi$ is a conic going through $B,C,D,E,F$: now, there is only one projective conic (up to a scalar factor) going through any five given pairwise distinct points. Thus $\phi$ must vanish on the conic containing $A,B,C,D,E,F$, hence $\phi(A)=0$.