Pascal's Theorem may be restated as

Suppose that no three of the points A,B,C,D,E, are collinear.

If the point F lies on the conic determined by A,B,C,D and E,

then X = ACnDB, Y = CFnBE, Z = FDnEA are collinear.

Then the converse is plainly

Suppose that no three of the points A,B,C,D,E, are collinear.

If X = ACnDB, Y = CFnBE, Z = FDnEA are collinear,

then the point F lies on the conic determined by A,B,C,D and E.

The figure below was drawn with *Cabri* and displayed using *CabriJava*.

If you drag the red line XY, F moves, but always lies on the yellow curve.

This is the conic determined by A,B,C,D,E.

Here, you may also move A,B,C,D or E to vary the conic.

**Notes**

(1) The condition on A,B,C,D,E is to ensure that the conic is non-degenerate.

(2) The proof is quite short, though tricky. You can find it here.

(3) Pascal's Theorem and its Converse give rise to an interesting construction.

Pascal's Theorem | Main Geometry Page |