用虚线段绘制角的平分线

Question

（1）AX角C^{'}AC的角平分线必定是虚线线段。

（2）C^{'}X必须画出一条线，而且那条线也必须是虚线。

 \documentclass[11pt,a4paper]{article}
 \usepackage{blindtext}
 \usepackage{tikz}
 \usepackage{tkz-euclide}
 \usetkzobj{all}
 \usepackage{color}

 \begin{document}
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(12,0){B}
\tkzDefPoint(3,7){C}
\tkzDefPoint(10,4){C^{'}}

\tkzLabelPoints[below](A)
\tkzLabelPoints[below](B)
\tkzLabelPoints[above](C)
\tkzLabelPoints[right](C^{'})

\tkzDrawSegment(A,B)
\tkzDrawSegment(A,C^{'})
\tkzDrawSegment(B,C^{'})
\tkzDrawSegment(A,C)
\tkzDrawSegment(B,C)

\tkzMarkSegment[color=red,pos=0.5,mark=|](A,B)
\tkzMarkSegment[color=red,pos=0.5,mark=||](A,C^{'})
\tkzMarkSegment[color=red,pos=0.5,mark=||](A,C)

\tkzDefLine[bisector](C^{'},A,C)
\tkzGetPoint{X}
\tkzDrawSegment[add=0pt and -273pt](A,X)
\tkzLabelLine[pos=0.45](A,X){X}
\tkzDrawSegment[dashed](C^{'},X)
\end{tikzpicture}
\end{document}

Answer 1

以下是相关内容：

\tkzDefLine[bisector](C^{'},A,C)
\tkzGetPoint{x}

\tkzInterLL(A,x)(C,B)
\tkzGetPoint{X}

求得的角平分线C^{'}AC并得到其端点x；然后求得和的X交点。AxCB

代码：

\documentclass[11pt,a4paper]{article}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(12,0){B}
\tkzDefPoint(3,7){C}
\tkzDefPoint(10,4){C^{'}}

\tkzLabelPoints[below](A)
\tkzLabelPoints[below](B)
\tkzLabelPoints[above](C)
\tkzLabelPoints[right](C^{'})

\tkzDrawSegment(A,B)
\tkzDrawSegment(A,C^{'})
\tkzDrawSegment(B,C^{'})
\tkzDrawSegment(A,C)
\tkzDrawSegment(B,C)

\tkzMarkSegment[color=red,pos=0.5,mark=|](A,B)
\tkzMarkSegment[color=red,pos=0.5,mark=||](A,C^{'})
\tkzMarkSegment[color=red,pos=0.5,mark=||](A,C)

\tkzDefLine[bisector](C^{'},A,C)
\tkzGetPoint{x}

\tkzInterLL(A,x)(C,B)
\tkzGetPoint{X}
\tkzLabelPoints[above](X)

\tkzDrawSegment[dashed](A,X)

\tkzDrawSegment[dashed](C^{'},X)
\end{tikzpicture}

\end{document}