theorem Th34:
  A=[X,1] & X is being_line & p on A & p is not Element of AS implies p=LDir(X)
proof
  assume that
A1: A=[X,1] and
A2: X is being_line and
A3: p on A and
A4: not p is Element of AS;
  consider Xp being Subset of AS such that
A5: p=LDir(Xp) and
A6: Xp is being_line by A4,Th20;
  Xp '||' X by A1,A2,A3,A5,A6,Th28;
  hence thesis by A2,A5,A6,Th12;
end;
