theorem Th01:
  a <> b & a <> p & right_angle b,a,p & right_angle a,b,q implies
    not Collinear p,a,q
  proof
    assume that
A1: a <> b and
A2: a <> p and
A3: right_angle b,a,p and
A4: right_angle a,b,q;
    assume
A5: Collinear p,a,q;
    now
A6:   right_angle p,a,b by A3,Satz8p2;
      right_angle a,b,q & a <> p & Collinear a,p,q by A2,A4,A5,GTARSKI3:45;
      hence right_angle q,a,b by A6,Satz8p3;
      thus right_angle q,b,a by A4,Satz8p2;
    end;
    hence contradiction by A1,Satz8p7;
  end;
