Editing Pascal's theorem (section)

==Proof using Bézout's theorem==

Suppose {{math|''f''}} is the cubic polynomial vanishing on the three lines through {{math|''AB, CD, EF''}} and {{math|''g''}} is the cubic vanishing on the other three lines {{math|''BC, DE, FA''}}. Pick a generic point {{math|''P''}} on the conic and choose {{math|''λ''}} so that the cubic {{math|''h'' {{=}} ''f'' + ''λg''}} vanishes on {{math|''P''}}. Then {{math|''h'' {{=}} 0}} is a cubic that has 7 points {{math|''A, B, C, D, E, F, P''}} in common with the conic. But by [[Bézout's theorem]] a cubic and a conic have at most 3&nbsp;&times;&nbsp;2&nbsp;=&nbsp;6 points in common, unless they have a common component. So the cubic {{math|''h'' {{=}} 0}} has a component in common with the conic which must be the conic itself, so {{math|''h'' {{=}} 0}} is the union of the conic and a line. It is now easy to check that this line is the Pascal line.