(** 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 diff_2_prod : forall O E E' A B AMB BMA ME,
Opp O E E' E ME -> Diff O E E' A B AMB -> Diff O E E' B A BMA ->
Prod O E E' AMB ME BMA.