# PSTricks: Perpendicular bisectors of line segments using pst-eucl

First: This is a crosspost from the LaTeX Community since I haven't gotten any answer there.

Consider the following MWE:

\documentclass{article}

\usepackage{pst-eucl}

\begin{document}

\begin{figure}
\centering
\begin{pspicture}(10.5,9.8)
\pnode(0,2.3){A}
\pnode(3.7,7.8){B}
\pnode(10.5,9.8){C}
\psdot(A)
\psdot(B)
\psdot(C)
\psline(A)(B)(C)
\psset{
PointSymbol=none,
PointNameB=none,
CodeFig=true,
CodeFigColor=black
}
\pstMediatorAB{B}{A}{M}{M''}
\pstMediatorAB{C}{B}{M'}{M'''}
\end{pspicture}
\end{figure}

\end{document}


Is it possible to

(1) remove M and M' from the figure and

(2) extend the lines in order to make them intersect?

Ad (1): I tried to make the third argument blank but that is not legal.

P.S. I know that it can be achieved with the following but I would like to do it with pst-eucl only:

\documentclass{article}

\begin{document}

\begin{figure}
\centering
\begin{pspicture}(10.5,9.8)
\pnode(0,2.3){A}
\pnode(3.7,7.8){B}
\pnode(10.5,9.8){C}
\psdot(A)
\psdot(B)
\psdot(C)
\psline(A)(B)(C)
\psset{
PointSymbol=none,
PointNameB=none,
CodeFig=true,
CodeFigColor=black
}
\psRelLine(A)(B){1 2 div}{I'}
\psRelLine[angle=90](I')(A){3}{I''}
\psRelLine(B)(C){1 2 div}{J'}
\psRelLine[angle=90](J')(B){2.8}{J''}
\psIntersectionPoint(I')(I'')(J')(J''){X}
\pcline(I')(X)
\pcline(J')(X)
\pstRightAngle{A}{I'}{I''}
\pstRightAngle{B}{J'}{J''}
\end{pspicture}
\end{figure}

\end{document}

-

\documentclass{article}
\usepackage{pst-eucl}
\begin{document}

\begin{pspicture}(10.5,9.8)
\pstGeonode[PointName=none](0,2.3){A}(3.7,7.8){B}(10.5,9.8){C}
\psline(A)(B)(C)
\psset{
PointSymbol=none,
PointName=none,
CodeFig=true,
CodeFigColor=black}
\pstMediatorAB[nodesepB=-6.5cm]{B}{A}{M}{M''}
\pstMediatorAB[nodesepB=-6.5cm]{C}{B}{M'}{M'''}
\end{pspicture}

\end{document}


-