(** 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_of_squares : forall O E E' A B A2 B2 A2MB2 APB AMB F,
Prod O E E' A A A2 -> Prod O E E' B B B2 -> Diff O E E' A2 B2 A2MB2 ->
Sum O E E' A B APB -> Diff O E E' A B AMB -> Prod O E E' APB AMB F ->
A2MB2 = F.