Pascal's theorem

Remarks

Here $AB$ denotes the line through $A$ and $B$. No three points of a nondegenerate conic are collinear, so the lines in each intersection are distinct and the three points are well defined. The line through them is the Pascal line of the hexagon $ABCDEF$.

One proof over $\C$, which extends to any field by passing to an algebraic closure, uses cubic curves. The cubics $AB \cdot CD \cdot EF$ and $BC \cdot DE \cdot FA$, products of three lines each, meet in the six points and in the three points of the statement. A suitable combination of them vanishes on a seventh point of $\mathcal{C}$, so by Bézout's theorem it contains $\mathcal{C}$, and the remaining factor is a line through the three points.

Example: on the unit circle $x^2 + y^2 = 1$ in the real plane, take a regular hexagon. Opposite sides are then parallel, so the three points lie on the line at infinity.

When the conic degenerates into two lines, the statement becomes Pappus's hexagon theorem. The dual statement is Brianchon's theorem.

Going deeper

For a student

Pascal's theorem is a basic tool for constructions with conics: given five points of a conic, it lets one construct any number of further points with a straightedge alone. Its converse also holds: if the three points are collinear, the six points lie on a conic, which may be degenerate. It is the projective form of a statement that, in the real plane, covers ellipses, parabolas and hyperbolas at once, since these are all the same in projective geometry. Through duality it gives Brianchon's theorem.