(** Lemma 14.28 is something like exists f, prod O E X A B C -> prod O' E' f(X) f(A') f(B') f(C') ? *)
(** Lemma 14.29 *)
(** From Pappus-Pascal we can derive that the product is symmetric, hence we have a field. *)
(* already done : prod_comm *)

Lemma eq_squares_eq_or_opp : forall O E E' A B A2,
Prod O E E' A A A2 -> Prod O E E' B B A2 -> A = B \/ Opp O E E' A B.