Why does my RPN expression to find a distance of 2 nodes produce a wrong output?

Question

I want to find a variant for the following,

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\addtopsstyle{gridstyle}{gridlabels=0pt,griddots=0}
\begin{document}
\begin{pspicture}[showgrid](-2,-2)(2,2)
    \pstGeonode[PosAngle={180,45,-90}]
        (-1.5,-1){A}
        (1.5,1){B}
        (1.5,-1){C}
    \pstRightAngle[RightAngleSize=0.2,fillstyle=solid,fillcolor=green!50]{A}{C}{B}
    \pstArcOAB[arcsepB=-1,arcsepA=-2.5]{A}{B}{B}
    \pstCurvAbsNode[CurvAbsNeg=true,PointName=none]{A}{B}{D}{\pstDistVal{.75}}
    \psline[linestyle=dashed](A)(D)(D|C)(C)
    \psset{shortput=nab,labelsep=-3pt}
    \ncline{A}{B}^{$g$}
    \ncline{B}{C}^{$a$}
    \ncline{C}{A}^{$b$}
\end{pspicture}
\end{document}

by calculating the distance with Pyth2 operator.

The following is my attempt.

\documentclass[pstricks,border={12pt 12pt 2cm 12pt}]{standalone}
\usepackage{pst-eucl}
\addtopsstyle{gridstyle}{gridlabels=0pt,griddots=0}
\psset{saveNodeCoors}
\begin{document}
\begin{pspicture}[showgrid](-2,-2)(2,2)
    \pstGeonode[PosAngle={180,45,-90}]
        (-1.5,-1){A}
        (1.5,1){B}
        (1.5,-1){C}
    \pstRightAngle[RightAngleSize=0.2,fillstyle=solid,fillcolor=green!50]{A}{C}{B}
        \pstVerb{/Dist {N-A.x N-A.y N-B.x N-B.y Pyth2} def}%
        \pnode(!Dist 10 PtoC){D}
    \psarc[arcsepB=-1,arcsepA=-2.5](A){!Dist}{(D)}{(B)}
    \psline[linestyle=dashed](A)(D)(D|C)(C)
    \psset{shortput=nab,labelsep=-3pt}
    \ncline{A}{B}^{$g$}
    \ncline{B}{C}^{$a$}
    \ncline{C}{A}^{$b$}
\end{pspicture}
\end{document}

And its output below is different from the first version shown above.

Could you find the glitch?

Answer 1

A is not the origin of the coordinate system!

\pnode[A](!Dist 10 PtoC ){D}

sets the node relative to A