# pst-eucl symmetry

I'm trying to draw a diagonal using the tags for symmetry but I really can't understand the meaning of the parameters and how to make the lines long just to fit the figure. For this source I have to use only the pst-eucl package. I really do search here for something similar but may be this package is not so often used. I hope someone can explain me or help me with the code.

The picture is the result I'm trying to reach:

\begin{pspicture}(0,0)(3,3)
\pstGeonode[PosAngle=-100,PointSymbol=none](0,0){A}
\psset{CodeFig=true}
\pstGeonode[PointSymbol=none,PosAngle=-100](4.4,0){B}
\pstGeonode[PosAngle=90,PointSymbol=none](7,3.4){C}
\pstGeonode[PosAngle=90,PointSymbol=none](2.6,3.4){D}
\pspolygon(A)(B)(C)(D)
\pstMarkAngle[]{C}{A}{D}{$\alpha /2$}
\pstMarkAngle[]{B}{A}{C}{$\alpha /2$}
\pstLineAB[linestyle=dashed]{A}{C}
\pstLineAB[linestyle=dashed]{B}{D}
\pstInterLL[PointSymbol=+,PosAngle=60]{A}{C}{B}{D}{O}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german]{A}{O}{B}
\pstGeonode[PointSymbol=none,PosAngle=-100](3.5,0){P}
\pstLineAB[linestyle=solid]{O}{P}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german,CodeFigColor=black,RightAngleSize=.3]{O}{P}{A}
\pstGeonode[PointSymbol=none,PosAngle=-100](2.6,0){M}
\pstLineAB[linestyle=solid]{D}{M}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german,CodeFigColor=black,RightAngleSize=.3]{D}{M}{A}
\pstGeonode[PointSymbol=none,PosAngle=100](3.5,3.4){Q}
\pstLineAB[linestyle=solid]{O}{Q}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german,CodeFigColor=black,RightAngleSize=.3]{D}{Q}{O}
\pstGeonode[PointSymbol=none,PointNameSep=0.3em](5,3.6){a}
\pstGeonode[PointSymbol=none,PointNameSep=0.3em](1.3,2.3){a}
\pstGeonode[PointSymbol=none,PointNameSep=0.3em](2.8,2){2r}
\pstGeonode[PointSymbol=none,PointNameSep=0.3em](3.6,2.5){r}
\end{pspicture}


Well if someone is interested I think this could be the code for this. I'll be happy if you have some advices how to make it on a better way.

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Is there any way to draw a DM and the line for P. I try to use a pstProjection but the line I get is dashed and the type of the angle is square. Do you have some ideas how to draw this lines? – nemo May 10 '12 at 9:24
what exactly do you mean with "draw a DM and the line for P"? – Herbert May 30 '12 at 17:36
Your code snippet doesn't work. Please provide a complete minimal example. Otherwise this question will be closed. – Marco Daniel Jun 6 '12 at 19:56

As far as I can see, the lower left angle annotations is the only difference between your code and the image you provided. So adding a few options fixes that problem:

\documentclass[12pt]{article}
\usepackage{pstricks,pst-node,pst-eucl}

\begin{document}

\begin{pspicture}(0,0)(3,3)
\pstGeonode[PosAngle=-100,PointSymbol=none](0,0){A}
\psset{CodeFig=true}
\pstGeonode[PointSymbol=none,PosAngle=-100](4.4,0){B}
\pstGeonode[PosAngle=90,PointSymbol=none](7,3.4){C}
\pstGeonode[PosAngle=90,PointSymbol=none](2.6,3.4){D}
\pspolygon(A)(B)(C)(D)
\pstMarkAngle[LabelSep=1.3,MarkAngleRadius=1]{C}{A}{D}{$\scriptstyle\alpha /2$}
\pstMarkAngle[MarkAngleRadius=0.7]{B}{A}{C}{$\scriptstyle\alpha /2$}
\pstLineAB[linestyle=dashed]{A}{C}
\pstLineAB[linestyle=dashed]{B}{D}
\pstInterLL[PointSymbol=+,PosAngle=60]{A}{C}{B}{D}{O}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german]{A}{O}{B}
\pstGeonode[PointSymbol=none,PosAngle=-100](3.5,0){P}
\pstLineAB[linestyle=solid]{O}{P}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german,CodeFigColor=black,RightAngleSize=.3]{O}{P}{A}
\pstGeonode[PointSymbol=none,PosAngle=-100](2.6,0){M}
\pstLineAB[linestyle=solid]{D}{M}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german,CodeFigColor=black,RightAngleSize=.3]{D}{M}{A}
\pstGeonode[PointSymbol=none,PosAngle=100](3.5,3.4){Q}
\pstLineAB[linestyle=solid]{O}{Q}
\pstRightAngle[PointSymbol=none,linecolor=black,RightAngleType=german,CodeFigColor=black,RightAngleSize=.3]{D}{Q}{O}