Pascal's Theorem
If A,B,C,A',B',C' are distinct ppoints on a pconic C, and
AB',A'B meet in P, BC',B'C in Q and CA',C'A in R, then
P,Q,R 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 pconic C_{0}
with equation xy+yz+zx=0, and A to [1,0,0], B to [0,1,0], C to [0,0,1].
By the Parametrization Lemma, we have constants r,s,t such that
A' = [r,1r,r(r1)], B' = [s,1s,s(s1)], C' = [t,1t,t(t1)].
Note that, since we have distinct ppoints, r,s,t ≠ 0,1, and r,s,t distinct.
By inspection :
AB' is sy+z=0, A'B is (1r)x+z=0, so P = [s,1r,s(r1)],
BC' is (1t)x+z=0, B'C is (s1)x+sy=0, so Q = [s,1s,s(t1)],
CA' is (r1)x+ry=0, C'A is ty+z=0, so R = [r,1r,t(r1)].
Then the determinant method gives the following equation for PR :
(1r)^{2}(st)x + (1r)s(tr)y + (1r)(sr)z = 0.
Since r ≠ 1, this becomes (1r)(st)x + s(tr)y + (sr)z = 0.
It is now routine to check that [s,1s,s(t1)], the coordinates of Q,
satisfy this equation. Thus P,Q,R are collinear.

