A Proof of Pascal's Theorem

We will present one of the many proofs of Pascal's Theorem.
This one uses homogeneous coordinates in projective geometry.

We will choose the standard projective conic C0 : xy+yz+xz = 0.
This has the advantage that it an equation symmetric in x, y and z.
It also contains the standard points [1,0,0], [0,1,0] and [0,0,1].

Two preliminary results will be used. We begin by proving these.

The Three Point Theorem
If P,Q and R are points on the projective conic C, then there is a unique
projective transformation which maps C to C0 and maps the points P,Q,R
to X = [1,0,0], Y = [0,1,0], Z = [0,0,1] respectively.

This result has the consequence that any theorem in projective geometry
involving three (or more) points on a conic may be reduced to a problem
about the standard conic and the standard points X,Y,Z.

If the problem involves further points on the conic, the following result
allows us to describe the corresponding points on the standard conic.

The Parametrisation Theorem
If the point P lies on the conic C0 : xy+yz+zx = 0, and P ≠ Z,
then P has the form [t,1-t,t(t-1)] for some real t.