Pascal's Theorem
Suppose that the points A,B,C,D,E,F lie on a non-degenerate conic.
Suppose also that AC, DB meet in X, CF, BE in Y and FD, EA in Z.
Then X, Y and Z are collinear.
Proof
As collinearity is preserved by projective transformation, it is enough
to prove the result for a suitably transformed figure.
By the Three Points Theorem, we may transform C to the conic C0
with equation xy+yz+zx=0, and A to [1,0,0], B to [0,1,0], F to [0,0,1].
By the Parametrization Lemma, we have constants r,s,t such that
D = [r,1-r,r(r-1)], C = [s,1-s,s(s-1)], E = [t,1-t,t(t-1)].
Note that, since we have distinct points, r,s,t ≠ 0,1, and r,s,t distinct.
By inspection :
AC is sy+z=0, DB is (1-r)x+z=0, so X = [s,1-r,s(r-1)],
BE is (1-t)x+z=0, CF is (s-1)x+sy=0, so Y = [s,1-s,s(t-1)],
FD is (r-1)x+ry=0, EA is ty+z=0, so Z = [r,1-r,t(r-1)].
Then the determinant method gives the following equation for XZ :
(1-r)2(s-t)x + (1-r)s(t-r)y + (1-r)(s-r)z = 0.
Since r ≠ 1, this becomes (1-r)(s-t)x + s(t-r)y + (s-r)z = 0.
It is now routine to check that [s,1-s,s(t-1)], the coordinates of Y,
satisfy this equation. Thus X,Y,Z are collinear.