# How to rotate the label such that its base line is parallel to the bisector?

How to rotate the label counter clockwise such that its base line is parallel to the bisector?

\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
\pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
\pstBissectBAC[linestyle=dotted,linecolor=red]{A}{B}{C}{C'}
\pstMarkAngle[MarkAngleRadius=1.5,LabelSep=0.75]{A}{B}{C}{\tiny$180^\circ-\theta$}
\end{pspicture}

\end{document}

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## 2 Answers

\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
\pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
\pstBissectBAC[linestyle=dotted,linecolor=red]{A}{B}{C}{B'}
\pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{}
\pcline[linestyle=none](B')(B)\ncput[nrot=:U,npos=0.84]{\tiny$180^\circ-\theta$}
\end{pspicture}

\end{document}


a solution without knowing B'

\documentclass{article}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}[showgrid=false](6,6)
\pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
\pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{}
\pcline[linestyle=none](!
\psGetNodeCenter{A} \psGetNodeCenter{B} \psGetNodeCenter{C}
/LengthBA B.x A.x sub dup mul B.y A.y sub dup mul add sqrt def
/LengthBC C.x A.x sub dup mul C.y A.y sub dup mul add sqrt def
/Factor LengthBC LengthBA div def
A.x B.x sub Factor mul B.x add /A.x ED
A.y B.y sub Factor mul B.y add /A.y ED
A.x C.x add 2 div A.y C.y add 2 div )(B)\ncput[nrot=:U,npos=0.84]{\tiny$180^\circ-\theta$}
\end{pspicture}

\end{document}

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It becomes more complicated :-) – kiss my armpit Aug 10 '12 at 14:27
no, because you do not need any coodinate – Herbert Aug 10 '12 at 14:30
I meant that your solution depends on B' while my solution does not. However, I cannot rotate the label clockwise. – kiss my armpit Aug 10 '12 at 14:35
no I do not need the B'. I used it only while it was defined. See edit for another solution – Herbert Aug 10 '12 at 15:15
The calculation in the second example will put the label on the median rather than on the angle bisector. – kiss my armpit Aug 10 '12 at 17:49

I got the solution even though it is not elegant enough.

\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
\pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
\pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{\rput{(B)}(0,0){\tiny$180^\circ-\theta$}}
\end{pspicture}

\end{document}


Clockwise rotation:

\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
\pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
\pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{\rput{!\psGetNodeCenter{B} B.y B.x atan 180 sub}(0,0){\tiny$180^\circ-\theta$}}
\end{pspicture}

\end{document}

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