从三角形的三个顶点画三条角平分线

从三角形的三个顶点画三条角平分线

在此处输入图片描述

(1)必须从两个顶点A和处画出两条角平分线,B使它们相交于P。(2)但从顶点画出的另一条角平分线C必须是终止在相交之前P

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

 \begin{document}
 \normalsize{\textbf{Theorem 1.24.} \textit{The bisectors of the angles of a triangle meet in a point which is equally distant from the sides.}}
\begin{center}
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(12,0){B}
\tkzLabelPoints[below](A)
\tkzLabelPoints[below](B)
\tkzDrawSegment(A,B)
\tkzDefPoint(6,7){C}
\tkzLabelPoints[above](C)
\tkzDrawSegment(A,C)
\tkzDrawSegment(B,C)
\end{tikzpicture}
\end{center}
\end{document}

答案1

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

\begin{document}

\normalsize{\textbf{Theorem 1.24.} \textit{The bisectors of the angles of a triangle meet in a point which is equally distant from the sides.}}
\begin{center}
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(12,0){B}
\tkzLabelPoints[below](A)
\tkzLabelPoints[below](B)
\tkzDrawSegment(A,B)
\tkzDefPoint(6,7){C}
\tkzLabelPoints[above](C)
\tkzDrawSegment(A,C)
\tkzDrawSegment(B,C)

\tkzDefLine[bisector](C,B,A)
\tkzGetPoint{i}
\tkzDefLine[bisector](B,A,C)
\tkzGetPoint{j}
\tkzInterLL(A,j)(B,i) 
\tkzGetPoint{P}
\tkzLabelPoints[below](P)
\tkzDrawBisector(C,B,A)(P)
\tkzDrawBisector(C,A,B)(Q)
\tkzDrawSegment[add=0pt and -30pt](C,P)
\end{tikzpicture}
\end{center}

\end{document}

在此处输入图片描述

评论:

  • -30pt在某一方面的变化

    \tkzDrawSegment[add=0pt and -30pt](C,P)
    

    得到从C到段所需的缩短量P

  • 你不应该手动标记你的定理,而应该考虑使用专用包,例如amsthmntheorem:下面是一个小例子amsthm

    \documentclass[11pt,a4paper]{article}
    \usepackage{blindtext}
    \usepackage{tikz}
    \usepackage{tkz-euclide}
    \usetkzobj{all}
    \usepackage{color}
    \usepackage{amsthm}
    
    \newtheorem{theo}{Theorem}
    
    \begin{document}
    
    \begin{theo}
    The bisectors of the angles of a triangle meet in a point which is equally distant from the sides.
    \end{theo}
    \begin{center}
    \begin{tikzpicture}
    \tkzDefPoint(0,0){A}
    \tkzDefPoint(12,0){B}
    \tkzLabelPoints[below](A)
    \tkzLabelPoints[below](B)
    \tkzDrawSegment(A,B)
    \tkzDefPoint(6,7){C}
    \tkzLabelPoints[above](C)
    \tkzDrawSegment(A,C)
    \tkzDrawSegment(B,C)
    
    \tkzDefLine[bisector](C,B,A)
    \tkzGetPoint{i}
    \tkzDefLine[bisector](B,A,C)
    \tkzGetPoint{j}
    \tkzInterLL(A,j)(B,i) 
    \tkzGetPoint{P}
    \tkzLabelPoints[below](P)
    \tkzDrawBisector(C,B,A)(P)
    \tkzDrawBisector(C,A,B)(Q)
    \tkzDrawSegment[add=0pt and -30pt](C,P)
    \end{tikzpicture}
    \end{center}
    
    \end{document}
    

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