theorem Th25:
  i in Seg n & j in Seg n & nt.i = nt.j implies Line(Segm(A,nt,mt)
  ,i) = Line(Segm(A,nt,mt),j)
proof
  set S=Segm(A,nt,mt);
  set Li=Line(S,i);
  set Lj=Line(S,j);
  assume that
A1: i in Seg n and
A2: j in Seg n and
A3: nt.i=nt.j;
A4: now
    let k such that
A5: 1<=k and
A6: k<=width S;
A7: k in Seg width S by A5,A6;
    then [i,k] in [:Seg n,Seg width S:] by A1,ZFMISC_1:87;
    then [i,k] in Indices S by MATRIX_0:25;
    then
A8: S*(i,k)=A*(nt.i,mt.k) by Def1;
    [j,k] in [:Seg n,Seg width S:] by A2,A7,ZFMISC_1:87;
    then [j,k] in Indices S by MATRIX_0:25;
    then
A9: S*(j,k)=A*(nt.j,mt.k) by Def1;
    S*(i,k) =Li.k by A7,MATRIX_0:def 7;
    hence Li.k=Lj.k by A3,A7,A8,A9,MATRIX_0:def 7;
  end;
A10: len Lj=width S by MATRIX_0:def 7;
  len Li=width S by MATRIX_0:def 7;
  hence thesis by A10,A4;
end;
