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 Th36:
  parallelogram a,b,c,d implies a<>b & a<>c & c <>b & a<>d & b<>d & c <>d
proof
  assume
A1: parallelogram a,b,c,d;
  then not a,b,c are_collinear;
  hence a<>b & a<>c & c <>b by Th24;
A2: now
    assume a=d;
    then a,b // c,a by A1;
    then
A3: a,b // a,c by Th6;
    not a,b,c are_collinear by A1;
    hence contradiction by A3;
  end;
A4: now
    assume c =d;
    then a,c // b,c by A1;
    then c,a // c,b by Th6;
    then
A5: c,a,b are_collinear;
    not a,b,c are_collinear by A1;
    hence contradiction by A5,Th22;
  end;
A6: now
    assume b=d;
    then a,b // c,b by A1;
    then b,a // b,c by Th6;
    then
A7: b,a,c are_collinear;
    not a,b,c are_collinear by A1;
    hence contradiction by A7,Th22;
  end;
  assume not thesis;
  hence contradiction by A2,A6,A4;
end;
