(* C6 (SAS): Given triangles ABC and DEF, suppose that AB = DE and AC = DF and ^BAC = ^EDF, then the two triangles are congruent, namely, BC = EF, ^ABC = ^DEF and ^ACB = ^DFE. *)

(* Specifications: *)
Lemma C6a  : forall A B C D E F : Point,
A <> B ->
A <> C ->
Distance A B = Distance D E ->
Distance A C = Distance D F ->
Angle B A C = Angle E D F ->
Distance B C = Distance E F.
Lemma C6b  : forall A B C D E F : Point,
A <> B ->
A <> C ->
B <> C ->
Distance A B = Distance D E ->
Distance A C = Distance D F ->
Angle B A C = Angle E D F ->
Angle A B C = Angle D E F.
Lemma C6c  : forall A B C D E F : Point,
A <> B ->
A <> C ->
B <> C ->
Distance A B = Distance D E ->
Distance A C = Distance D F ->
Angle B A C = Angle E D F ->
Angle A C B = Angle D F E.