theorem Th36:
  parallelogram a,b,c,d implies not a,d '||' b,c
proof
  assume
A1: parallelogram a,b,c,d;
  then not a,b,c are_collinear;
  then
A2: not a,b '||' a,c;
  a,b '||' c,d & a,c '||' b,d by A1;
  then not b,c '||' a,d by A2,Def1;
  hence thesis by PARSP_1:19;
end;
