theorem Th32:
  L1 <> L2 & L1 meets L2 implies
  (ex x st L1 = {x} or L2 = {x})
  or
  (L1 is being_line & L2 is being_line & ex x st L1 /\ L2 = {x})
  proof
    assume that
A1: L1<>L2 and
A2: L1 meets L2;
    not L1 // L2 by A1,A2,EUCLIDLP:71;
    hence thesis by Th16;
  end;
