(** We propose here a more complex proof to simplify the ndg. If we know that a quadrilateral has its pairs of opposite side congruent and parallel then it is a parallelogram. *)

Lemma varignon_aux :
forall A B C D I J K L,
(A<>C \/ B<>D) ->
J<>L ->
Midpoint I A B ->
Midpoint J B C ->
Midpoint K C D ->
Midpoint L A D ->
Parallelogram I J K L.