We are given five points *P*, *P'*, *Q*, *R*, and
*S*, and can show that the conic lying on
these five points was given by the locus of blue points.

Now let us define *N* as the intersection of *x* and
*z*. We see in the diagram that *N* is on the conic, and
can verify that our construction would send *PN* to *P'N*.
We can state this as a theorem:

If *PR.QN*, *RP'.NS*, and *P'Q.SP* are collinear, then
*N* lies on the conic determined by *PP'QRS*.

Rather than saying that *N* lies on the conic determined by
*PP'QRS*, we could simply say that *NPP'QRS* lie on a conic. It
will also simplify things to speak about the hexagon *PRP'QNS*;
then the points lie on a conic if and only if the hexagon is
inscribed in that conic. Making these modifications and some
changes of labelling, we have the theorem:

If opposite sides of a hexagon *(ABCDEF)* intersect in three points
*(AB.DE, BC.EF, CD.FA)* which are collinear, then the hexagon may
be inscribed in a conic.

This is known as the *converse of Pascal's theorem*.

So *Pascal's theorem* says:

If a hexagon *(ABCDEF)* is inscribed in a conic, then opposite
sides intersect in three points *(AB.DE, BC.EF, CD.FA)* which are
collinear.

Proof: Define *L=BC.EF*, *M=CD.FA*, *N=AB.DE*; we want
to show *L*, *M*, *N* collinear.
To do this, we also define *J=AB.CD*, *K=FA.BC*. Then

Thus the projectivity sending( A, N; J, B) = (DA,DE;DC,DB) = (FA,FE;FC,FB) = ( K, L; C, B).

Note that Pascal's theorem is true regardless of where the points lie on the conic. The diagram above shows a very non-convex hexagon, but since projective geometry does not deal with convexity, a convex hexagon would do just as well. However, a convex hexagon might have a Pascal line too far off the diagram to be seen.

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Created: Nov 30 1995 ---
Last modified: Thu Nov 30 15:31:14 1995