reserve SAS for Semi_Affine_Space;
reserve a,a9,a1,a2,a3,a4,b,b9,c,c9,d,d9,d1,d2,o,p,p1,p2,q,r,r1,r2,s,x, y,t,z
  for Element of SAS;

theorem Th38:
  parallelogram a1,a2,a3,a4 implies not a1,a2,a3 are_collinear &
  not a1,a3,a2 are_collinear & not a1,a2,a4 are_collinear & not a1,a4,a2
are_collinear & not a1,a3,a4 are_collinear & not a1,a4,a3 are_collinear &
not a2,
  a1,a3 are_collinear & not a2,a3,a1 are_collinear &
  not a2,a1,a4 are_collinear &
  not a2,a4,a1 are_collinear & not a2,a3,a4 are_collinear & not a2,a4,a3
are_collinear & not a3,a1,a2 are_collinear &
not a3,a2,a1 are_collinear & not a3,
  a1,a4 are_collinear & not a3,a4,a1 are_collinear &
not a3,a2,a4 are_collinear &
  not a3,a4,a2 are_collinear & not a4,a1,a2 are_collinear & not a4,a2,a1
are_collinear & not a4,a1,a3 are_collinear & not a4,a3,a1 are_collinear &
  not a4, a2,a3 are_collinear & not a4,a3,a2 are_collinear
proof
  assume
A1: parallelogram a1,a2,a3,a4;
  then
A2: ( not a3,a4,a1 are_collinear)& not a4,a3,a2 are_collinear by Th37;
  ( not a1,a2,a3 are_collinear)& not a2,a1,a4 are_collinear by A1,Th37;
  hence thesis by A2,Th22;
end;
