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¿Por qué \tkzLabelAngle no coloca la etiqueta en el ángulo correcto?

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¿Por qué \tkzLabelAngle no coloca la etiqueta en el ángulo correcto?

En la siguiente figura, estoy intentando etiquetar angle (lA,A,oA), pero no importa lo que intente, la etiqueta termina estando en angle (rA,A,B). ¿Que está pasando aqui?

\documentclass[border=10pt]{standalone}
\usepackage{tkz-euclide}
\usetikzlibrary{arrows.meta}
\usetkzobj{all}

\newlength\aehlength
\setlength\aehlength{1in}

\begin{document}

\begin{tikzpicture}[my dot/.style={fill,circle,inner sep=1.5pt},>={To[scale=2]}]
  \coordinate (A) at (0,0);
  \coordinate (B) at (55:\aehlength);
  \coordinate (lA) at ($(A)+(180:\aehlength)$);
  \coordinate (rA) at ($(A)+(0:\aehlength)$);
  \coordinate (lB) at ($(B)+(180:\aehlength)$);
  \coordinate (rB) at ($(B)+(0:\aehlength)$);
  \coordinate (oA) at ($(A)!-1cm!(B)$);
  \coordinate (oB) at ($(B)!-1cm!(A)$);

  \foreach \mynA/\mypA/\mynB/\mypB in {A/2cm/B/2cm,lA/1cm/rA/1cm,lB/1cm/rB/1cm}
  {
    \draw[arrows=<->] ($(\mynA)!-\mypA!(\mynB)$) -- ($(\mynB)!-\mypB!(\mynA)$);
  }

  \foreach \myn/\myp in {A/90,B/90,lA/90,rA/90,lB/90,rB/90,oA/90,oB/90 }
  {
      \node[my dot] at (\myn) {};
      \node at ($(\myn)+(\myp:8pt)$) {\myn};
  }

  \tkzLabelAngle[pos=0.45](oA,A,lA) {$x$}
  \tkzLabelAngle[pos=0.45](lA,A,oA) {$x$}
  \tkzLabelAngle[pos=0.75](oB,B,rB) {$60^\circ$}

\end{tikzpicture}
\end{document}

ingrese la descripción de la imagen aquí

Respuesta1

Con nueva versión pura de tkz-euclide.CTAN v 3.01

\documentclass[border=10pt]{standalone}
\usepackage{tkz-euclide}

\def\aehlength{3}
\begin{document}

\begin{tikzpicture}
  \tkzDefPoint(0,0){A}
  \tkzDefPoint(55:\aehlength){B}
  \tkzDefShiftPoint[A](180:\aehlength){lA}
  \tkzDefShiftPoint[A](0:\aehlength){rA}
  \tkzDefShiftPoint[B](180:\aehlength){lB}
  \tkzDefShiftPoint[B](0:\aehlength){rB}
  \tkzDefPointWith[linear normed,K=-1](A,B) \tkzGetPoint{oA}
  \tkzDefPointWith[linear normed,K=-1](B,A) \tkzGetPoint{oB}
  \tkzDrawLines[add=.5 and .5,<->](oA,oB rA,lA rB,lB)
  \tkzDrawPoints(A,B,lA,rA,oA,rB,lB,oB)
  \tkzLabelPoints[above=3pt](A,B,lA,rA,rB,lB)
  \tkzLabelPoints[right=3pt](oA,oB)
  \tkzLabelAngle[pos=0.75](oA,A,lA) {$x$}
  \tkzLabelAngle[pos=0.75](lA,A,oA) {$x$}     
  \tkzLabelAngle[pos=0.75](oB,B,rB) {$60^\circ$}

\end{tikzpicture}
\end{document}

ingrese la descripción de la imagen aquí

Respuesta2

No tengo idea si es un error o una característica, pero uso \tkzLabelAngle[pos=-0.45](lA,A,oA) {$x$}. Tenga en cuenta el signo menos pos.

\documentclass[border=10pt]{standalone}
\usepackage{tkz-euclide}
\usetikzlibrary{arrows.meta}
\usetkzobj{all}

\newlength\aehlength
\setlength\aehlength{1in}

\begin{document}

\begin{tikzpicture}[my dot/.style={fill,circle,inner sep=1.5pt},>={To[scale=2]}]
  \coordinate (A) at (0,0);
  \coordinate (B) at (55:\aehlength);
  \coordinate (lA) at ($(A)+(180:\aehlength)$);
  \coordinate (rA) at ($(A)+(0:\aehlength)$);
  \coordinate (lB) at ($(B)+(180:\aehlength)$);
  \coordinate (rB) at ($(B)+(0:\aehlength)$);
  \coordinate (oA) at ($(A)!-1cm!(B)$);
  \coordinate (oB) at ($(B)!-1cm!(A)$);

  \foreach \mynA/\mypA/\mynB/\mypB in {A/2cm/B/2cm,lA/1cm/rA/1cm,lB/1cm/rB/1cm}
  {
    \draw[arrows=<->] ($(\mynA)!-\mypA!(\mynB)$) -- ($(\mynB)!-\mypB!(\mynA)$);
  }

  \foreach \myn/\myp in {A/90,B/90,lA/90,rA/90,lB/90,rB/90,oA/90,oB/90 }
  {
      \node[my dot] at (\myn) {};
      \node at ($(\myn)+(\myp:8pt)$) {\myn};
  }

  \tkzLabelAngle[pos=0.45](oA,A,lA) {$x$}
  \tkzLabelAngle[pos=-0.45](lA,A,oA) {$x$}         %%% <=== here
  \tkzLabelAngle[pos=0.75](oB,B,rB) {$60^\circ$}

\end{tikzpicture}
\end{document}

ingrese la descripción de la imagen aquí

Respuesta3

Aquí está mi solución (para definir mi propia bisectriz de ángulo):

\makeatletter
\newcommand\aeFindAngleBisector{\ae@find@angle@bisector}
\def\ae@find@angle@bisector(#1)(#2)(#3)[#4]{%%
  \bgroup
  \pgfmathanglebetweenpoints{\pgfpointanchor{#2}{center}}%%
                            {\pgfpointanchor{#1}{center}}%%
  \edef\ae@angle@start@position{\pgfmathresult}%%
  \pgfmathanglebetweenlines{\pgfpointanchor{#2}{center}}%%
                           {\pgfpointanchor{#1}{center}}%%
                           {\pgfpointanchor{#2}{center}}%%
                           {\pgfpointanchor{#3}{center}}%%
  \edef\ae@angle@between{\pgfmathresult}%%
  \pgfmathparse{(\ae@angle@start@position+\ae@angle@between/2}%%
  \xdef\ae@half@angle@bisector{\pgfmathresult}%%
  \coordinate (#4) at ($(#2)+(\ae@half@angle@bisector:10pt)$);
  \egroup
}%%
\makeatother

que toma un ángulo ABCy define un punto a lo largo de la bisectriz del ángulo dentro del ángulo definido en sentido antihorario alrededor del vértice B.

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