theorem Th25:
  MD = LineVec2Mx bD iff Line(MD,1) = bD & len MD = 1
proof
  thus MD = LineVec2Mx bD implies Line(MD,1) = bD & len MD = 1
  proof
    1 in Seg 1;
    then
A1: Line(LineVec2Mx bD,1)=(LineVec2Mx bD).1 by MATRIX_0:52;
    assume MD = LineVec2Mx bD;
    hence thesis by A1,FINSEQ_1:40;
  end;
  assume that
A2: Line(MD,1) = bD and
A3: len MD = 1;
  reconsider md=MD as Matrix of 1,width MD,D by A3,MATRIX_0:51;
  1 in Seg 1;
  then md.1=bD by A2,MATRIX_0:52;
  hence thesis by A3,FINSEQ_1:40;
end;
