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.

proof of the three point theorem

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.

proof of the parametrization theorem