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For six distinct points on an irreducible conic in the complex projective plane, the three intersections of opposite sides of the resulting hexagon are collinear. Form the two cubics given by the products of alternate side equations. They meet at the six conic points and the three opposite-side intersections. Applying residual intersection through a curve pencil with degrees three and two places the last three points on a line. This is the conic counterpart of Pappus's hexagon theorem for two lines.

The curves have no common component, since their intersection is finite. Let $R=0$ define the irreducible degree-$m$ curve $E$. Choose $p\in E$ outside $C\cap D$, and select $(\lambda ,\mu )\ne (0,0)$ with $\lambda P(p)+\mu Q(p)=0$. The polynomial $H=\lambda P+\mu Q$ is not identically zero, since $P,Q$ are not proportional.

The degree-$n$ curve $H=0$ meets $E$ at the prescribed $nm$ base points and also at $p$. If $R$ did not divide $H$, irreducibility of $R$ and the Bézout theorem would bound the number of distinct intersections by $nm$, contradicting these $nm+1$ points. ThusAt every remaining common zero of $P,Q$ lying off $E$, $R$ is nonzero and hence $S=0$. ThereforeThis is residual intersection through a curve pencil. The existence of a point $p$ outside the finite base locus is automatic on the projective curve $E$.

Let the six distinct conic points be $p_1,\dots ,p_6$, and write ${\ell }_{ij}$ for their joining line. Form two reducible cubicsNo three distinct points of an irreducible conic are collinear, by the Bézout theorem applied to a line and the conic. Thus all six side lines are distinct, the cubics have no common component, and six of their intersections are precisely the vertices. The remaining pairings giveThese three points are off the conic and distinct. For example, equality of $q_1$ and $q_2$ would put that point on both ${\ell }_{34}$ and ${\ell }_{45}$, making it $p_4$ and then placing $p_4$ on ${\ell }_{12}$, impossible. The other cases are identical. Hence the two cubics meet in nine distinct points.

Apply residual intersection through a curve pencil with $n=3$, $m=2$ and $E$ the conic: six intersections lie on $E$, so the other three lie on a curve of degree at most one. Therefore $q_1,q_2,q_3$ are collinear, proving Pascal's theorem.