theorem Th26:
  i in Seg n & j in Seg n & nt.i = nt.j & i<>j implies Det Segm(M,
  nt,nt1) = 0.K
proof
  assume that
A1: i in Seg n and
A2: j in Seg n and
A3: nt.i = nt.j and
A4: i<>j;
A5: i<j or j<i by A4,XXREAL_0:1;
  Line(Segm(M,nt,nt1),i) = Line(Segm(M,nt,nt1),j) by A1,A2,A3,Th25;
  hence thesis by A1,A2,A5,MATRIX11:50;
end;
