如何在 tkz-euclide 中反转 tkz 箭头方向

Question

使用该选项时如何反转箭头的方向tkz arrow？我在\tkzDrawArc命令中使用该选项，它似乎总是想沿正方向绘制圆弧。

\documentclass{standalone}
\usepackage{tkz-euclide}
\begin{document}

\begin{tikzpicture}
  \draw (-2,-2) rectangle (5,2);
  \tkzDefPoints{0/0/O,3/0/U}
  \tkzDefPoint(10:1){A}
  \tkzDefPoint(90:1){B}
  \tkzLabelPoints(A,B)
  \tkzDrawArc[tkz arrow](O,A)(B)
  \tkzDrawPoints(A,B,O)
  
  \begin{scope}[shift=(U)]
    \tkzDefPoints{0/0/O}
    \tkzDefPoint(10:1){A}
    \tkzDefPoint(90:1){B}
    \tkzDrawArc[tkz arrow](O,B)(A)
    \tkzLabelPoints(A,B)
    \tkzDrawPoints(A,B,O)
  \end{scope}
  
\end{tikzpicture}

\end{document}

Answer 1

如果是选项tkz arrow，箭头的类型、大小和方向都是默认的，所以tkz arrow相当于tkz arrow={Latex[scale=1]}。如果要反转箭头的方向， 的值scale必须为负数，例如tkz arrow={Latex[scale=-1]}。

\documentclass{standalone}
\usepackage{tkz-euclide}
\begin{document}

\begin{tikzpicture}
  \draw (-2,-2) rectangle (5,2);
  \tkzDefPoints{0/0/O,3/0/U}
  \tkzDefPoint(10:1){A}
  \tkzDefPoint(90:1){B}
  \tkzLabelPoints(A,B)
  \tkzDrawArc[tkz arrow={Latex[scale=-1]}](O,A)(B)
  \tkzDrawPoints(A,B,O)
  
  \begin{scope}[shift=(U)]
    \tkzDefPoints{0/0/O}
    \tkzDefPoint(10:1){A}
    \tkzDefPoint(90:1){B}
    \tkzDrawArc[tkz arrow={Latex[scale=-1]}](O,B)(A)
    \tkzLabelPoints(A,B)
    \tkzDrawPoints(A,B,O)
  \end{scope}
  
\end{tikzpicture}

\end{document}

Answer 2

好的解决方案实际上来自一丝不苟。你的问题很有趣，因为它让我想到了修改宏的代码/tkzDrawArc。我需要一些测试来检查是否没有不良副作用。

现在可以使用该选项reverse来绘制弧线。

\documentclass{standalone}
\usepackage{tkz-euclide}

\makeatletter

\newif\iftkz@reverse
\gdef\tkz@numa{0}
\pgfkeys{/tkzdrawarc/.cd,
      type/.is choice,
      type/towards/.code               = \def\tkz@numa{0},
      type/rotate/.code                = \def\tkz@numa{1},
      type/angles/.code                = \def\tkz@numa{2}, 
      type/R/.code                     = \def\tkz@numa{3},
      type/R with nodes/.code          = \def\tkz@numa{4},
      towards/.style                   = {type=towards},
      rotate/.style                    = {type=rotate},
      R/.style                         = {type=R},
      angles/.style                    = {type=angles},
      R with nodes/.style              = {type=R with nodes},
      diameter/.code                   = {},
      arc/.code                        = {},
      size/.code                       = {},
      mark/.code                       = {},
      mkpos/.code                      = {},
      mksize/.code                     = {},
      mkcolor/.code                    = {},
      type/.default                    =  towards,
      delta/.store in                  = \tkz@delta,
      delta                            = 0,
      reverse/.is if                   = tkz@reverse,
      reverse/.default                 = true,
      reverse                          = false,
      /tkzdrawarc/.search also         = {/tikz}
}  
\def\tkzDrawArcRAngles{\pgfutil@ifnextchar[{\tkz@DrawArcRAngles}{%
                                            \tkz@DrawArcRAngles[]}} 
\def\tkz@DrawArcRAngles[#1](#2,#3)(#4,#5){% 
 \begingroup
 \tkzNormalizeAngle(#4,#5)    
 \pgfmathsubtract{\tkz@FirstAngle}{\tkz@delta}
 \edef\tkz@FirstAngle{\pgfmathresult}%
 \pgfmathadd{\tkz@SecondAngle}{\tkz@delta}
 \edef\tkz@SecondAngle{\pgfmathresult} 
 \iftkz@reverse 
   \let\tkztemp\tkz@FirstAngle 
   \let\tkz@FirstAngle\tkz@SecondAngle
    \let\tkz@SecondAngle\tkztemp
    \fi
  \draw[shift = {(#2)},arc style,/tkzdrawarc/.cd,#1]%
  (\tkz@FirstAngle:#3) arc (\tkz@FirstAngle:\tkz@SecondAngle:#3);
\endgroup  
}
\makeatother
\begin{document}
  

\begin{tikzpicture}
  \draw (-2,-2) rectangle (5,2);
  \tkzDefPoints{0/0/O,3/0/U}
  \tkzDefPoint(10:1){A}
  \tkzDefPoint(90:1){B}
  \tkzLabelPoints(A,B)
  \tkzDrawArc[reverse,tkz arrow={Stealth}](O,A)(B)
  \tkzDrawPoints(A,B,O)

  \begin{scope}[shift=(U)]
    \tkzDefPoints{0/0/O}
    \tkzDefPoint(10:1){A}
    \tkzDefPoint(90:1){B}
    \tkzDrawArc[tkz arrow={Stealth}](O,A)(B)
    \tkzLabelPoints(A,B)
    \tkzDrawPoints(A,B,O)
  \end{scope}

\end{tikzpicture}
\end{document}