Pascal's Theorem states that if A,..,F lie on a conic, then X,Y and Z are collinear.
The converse is quite easy to deduce from the theorem.
Outline of a proof:
Consider the point F' where CY cuts the conic.
Apply Pascal's Theorem to A,B,C,D,E,F' to see that DF' must pass through Z
(the intersection of AE and XY).
Then conclude that F=F'.
This allows us to obtain any number of points on the conic
defined by five fixed points A,B,C,D,E.
The BraikenridgeMaclaurin Construction
Let AC and BD meet in X.
Draw any line L through meeting BE and AE.
Suppose L cuts BE at Y and AE at Z.
Let CY and DZ meet at F.
By the Converse of Pascal's Theorem, F lies on the conic.
Of course the conic is the locus of points F so obtained.
The Cabri figure conpascal2.fig illustrates this.
If you don't have Cabri, there is a screen shot here.
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