标记三角形中的角度

标记三角形中的角度

我尝试查看该问题的过去的答案但似乎无法解决我的问题。

我想标记这个三角形中的角度 ABD 和 DBC:

\documentclass{article}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\begin{document}
\begin{tikzpicture}

    \tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5]
    \tkzClip[space=0.25]

    \coordinate (A) at (0,0); \tkzLabelPoints[left](A)
    \coordinate (B) at (2.25,3.320718914); \tkzLabelPoints[above](B)
    \coordinate (C) at (6,0); \tkzLabelPoints[right](C)
    \coordinate (E) at (3,0); \tkzLabelPoints[below](E)
    \draw (A)--(B)--(C)--cycle;
    \draw (B)--(E);

    \tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
    \tkzInterLL(A,C)(B,a) \tkzGetPoint{D} \tkzLabelPoints[below](D)
    \draw (B)--(D);

    \tkzDefMidPoint(A,B) \tkzGetPoint{12}
    \tkzLabelPoints[left](12)
    \tkzDefMidPoint(B,C) \tkzGetPoint{15}
    \tkzLabelPoints[right=0.1cm](15)
    \tkzDefMidPoint(A,C) \tkzGetPoint{18}
    \tkzLabelPoints[below=0.4cm](18)

    \tkzMarkSegment[mark=|](A,E)
    \tkzMarkSegment[mark=|](C,E)

    \tkzLabelSegment(A,B)
    \tkzLabelSegment(B,C)
    \tkzLabelSegment(C,A)

\end{tikzpicture}
\end{document}

答案1

您可以angles为此使用库。

在此处输入图片描述

\documentclass{article}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\usetikzlibrary{angles}

\begin{document}
\begin{tikzpicture}

    \tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5]
    \tkzClip[space=0.25]

    \coordinate (A) at (0,0); \tkzLabelPoints[left](A)
    \coordinate (B) at (2.25,3.320718914); \tkzLabelPoints[above](B)
    \coordinate (C) at (6,0); \tkzLabelPoints[right](C)
    \coordinate (E) at (3,0); \tkzLabelPoints[below](E)
    \draw (A)--(B)--(C)--cycle;
    \draw (B)--(E);

    \tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
    \tkzInterLL(A,C)(B,a) \tkzGetPoint{D} \tkzLabelPoints[below](D)
    \draw (B)--(D);

    \tkzDefMidPoint(A,B) \tkzGetPoint{12}
    \tkzLabelPoints[left](12)
    \tkzDefMidPoint(B,C) \tkzGetPoint{15}
    \tkzLabelPoints[right=0.1cm](15)
    \tkzDefMidPoint(A,C) \tkzGetPoint{18}
    \tkzLabelPoints[below=0.4cm](18)

    \tkzMarkSegment[mark=|](A,E)
    \tkzMarkSegment[mark=|](C,E)

%    \tkzLabelSegment(A,B)
%    \tkzLabelSegment(B,C)
%    \tkzLabelSegment(C,A)
    \draw pic[draw=blue, angle radius=5mm] {angle = A--B--D};
    \draw pic[draw=red, angle radius=7mm] {angle = D--B--C};

\end{tikzpicture}
\end{document}

答案2

您可以使用\tkzFindAngle、、tkzGetAngletkzMarkAngletkzLabelAngle

顺便说一下,

\tkzDefMidPoint(A,B) \tkzGetPoint{12}
\tkzLabelPoints[left](12)

并且做

\tkzLabelSegment[left](A,B){12}

是相同的(至少在这种情况下)。但正如您所见,命令要短得多。如果我是你,我会使用它。我已经从您的代码中更正了它,因为\tkzLabelSegment(A,B)缺少花括号之间的名称。

输出

图1

代码

\documentclass[margin=10pt]{standalone}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\usetkzobj{all}

\begin{document}
\begin{tikzpicture}

    \tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5]
    \tkzClip[space=0.25]

    \coordinate (A) at (0,0); \tkzLabelPoints[left](A)
    \coordinate (B) at (2.25,3.320718914); \tkzLabelPoints[above](B)
    \coordinate (C) at (6,0); \tkzLabelPoints[right](C)
    \coordinate (E) at (3,0); \tkzLabelPoints[below](E)
    \draw (A)--(B)--(C)--cycle;
    \draw (B)--(E);

    \tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
    \tkzInterLL(A,C)(B,a) \tkzGetPoint{D} \tkzLabelPoints[below](D)
    \draw (B)--(D);

    \tkzDefMidPoint(A,B) \tkzGetPoint{12}
    \tkzLabelPoints[left](12)
    \tkzDefMidPoint(B,C) \tkzGetPoint{15}
    \tkzLabelPoints[right=0.1cm](15)
    \tkzDefMidPoint(A,C) \tkzGetPoint{18}
    \tkzLabelPoints[below=0.4cm](18)

    \tkzMarkSegment[mark=|](A,E)
    \tkzMarkSegment[mark=|](C,E)

    \tkzFindAngle(A,B,D) 
    \tkzFindAngle(D,B,C)
    \tkzGetAngle{angleABD}; \FPround\angleABD\angleABD{0}
    \tkzGetAngle{angleDBC}; \FPround\angleDBC\angleDBC{0}

    \tkzMarkAngle[draw=black,opacity=.5,fill=green!10](A,B,D)
    \tkzMarkAngle[draw=black,opacity=.5,fill=red!10](D,B,C)
    \tkzLabelAngle[dist=1.2](A,B,D){\tiny $\angleABD^\circ$}
    \tkzLabelAngle[dist=1.2](D,B,C){\tiny $\angleDBC^\circ$}

    %\tkzLabelSegment[left](A,B){AB}
    %\tkzLabelSegment[right](B,C){BC}
    %\tkzLabelSegment[below](C,A){CA}

\end{tikzpicture}
\end{document}

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