= Solution
Let the six distinct <conic> points be $p_1,\ldots,p_6$, and write $\ell_{ij}$ for their joining line. Form two reducible cubics
$$
C=\ell_{12}\cup\ell_{34}\cup\ell_{56},\qquad D=\ell_{23}\cup\ell_{45}\cup\ell_{61}.
$$
No 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 give
$$
q_1=\ell_{12}\cap\ell_{45},\qquad q_2=\ell_{34}\cap\ell_{61},\qquad q_3=\ell_{56}\cap\ell_{23}.
$$
These 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 \b[$q_1,q_2,q_3$ are collinear], proving <Pascal's theorem>.
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