theorem Th6:
  p1 in LSeg(p,q) & p`2 = q`2 implies p1`2 = q`2
proof
  assume p1 in LSeg(p,q);
  then consider r such that
A1: p1 = (1-r)*p+r*q and
  0<=r and
  r<=1;
  assume
A2: p`2 = q`2;
  p1`2 = ((1-r)*p)`2+(r*q)`2 by A1,TOPREAL3:2
    .= ((1-r)*p)`2+r*q`2 by TOPREAL3:4
    .= (1-r)*p`2+r*q`2 by TOPREAL3:4;
  hence thesis by A2;
end;
