삼각형의 꼭지점 3개로 이등분선 3개 그리기

Question

(1). 두 꼭지점에서 두 개의 이등분선을 그려야 하며 두 정점 A에서 B교차해야 합니다.P . (2). 그러나 꼭지점에서 그려지는 다른 이등분선은 다음 C과 같아야 합니다.종료됨교차하기 전에 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}

Answer 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}

비고:

• 변경-30pt​

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

  C에서 까지의 세그먼트에 대해 원하는 단축을 얻으려면 P.

• 정리에 대한 수동 마크업 대신 amsthm또는 다음 과 같은 전용 패키지 사용을 고려해야 합니다 ntheorem.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}