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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 × 2 = 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.
== A
Again given the hexagon on a conic of Pascal's theorem with the above notation for points (in the first figure), we have<ref>{{cite web|url=http://www.cut-the-knot.org/Generalization/OverlookedPascal.shtml|title=A Property of Pascal's Hexagon Pascal May Have Overlooked|date= 2014-02-03}}</ref>
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