我想获取三角形 ABC 下的角 alpha,我该怎么做?现在角部分在顶部可见。我还想将角的外部设为黑色。
\documentclass[10pt]{article}
\usepackage{pgfplots}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{textcomp}
\usepackage{mathtools}
\usepackage{xcolor}
\usepackage{tabularx}
\usepackage{pstricks-add}
\usepackage{pgfplots}
\usepackage{background}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}[scale=0.65]
\tkzDefPoint(0,0){A},
\tkzDefPoint(4,2){B},
\tkzDefPoint(5,0){C},
\tkzDefPoint(-1,1.8){D},
\tkzDefPoint(-5,0){E},
\tkzDefPoint(1.5,1.9){O},
\tkzDrawPoints(A,B,C,D,E,O)
\tkzLabelPoints[below](A)
\tkzLabelPoints[right](B)
\tkzLabelPoints[right](C)
\tkzLabelPoints[below](E)
\tkzLabelPoints[above](O)
\tkzLabelSegment[above,font=\footnotesize](A,B){$4$}
\tkzLabelSegment[above,font=\footnotesize](A,C){$5$}
\tkzLabelSegment[right,font=\footnotesize](A,D){$2$}
\tkzLabelSegment[below,font=\footnotesize](A,E){$5$}
\draw[thick,black](0,0)--(-1,1.8)--(-5,0)--cycle;
\draw[thick,black](D)--(B)--cycle;
\draw[thick,black](0,0)--(4,2)--(5,0)--cycle;
\tkzDefCircle[circum](A,B,D)
\tkzGetPoint{O} \tkzGetLength{rayon}
\tkzDrawCircle[R,thick](O,\rayon pt)
\tkzMarkAngle[size=1cm,opacity=.7,black](C,A,B)
\tkzFillAngle[fill= blue!50,size=1cm,opacity=.7](C,A,B)
\tkzLabelAngle[pos=1.4](C,A,B){$\alpha$}
\node at (-1.4, 2.15) {$D$};
\end{tikzpicture}
\end{document}
答案1
只需在绘制所有内容之前放置\tkzFillAngle
命令即可(但在定义点之后)。
\documentclass[10pt]{article}
\usepackage{pgfplots}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{textcomp}
\usepackage{mathtools}
\usepackage{xcolor}
\usepackage{tabularx}
\usepackage{pstricks-add}
\usepackage{pgfplots}
\usepackage{background}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}[scale=0.65]
\tkzDefPoint(0,0){A},
\tkzDefPoint(4,2){B},
\tkzDefPoint(5,0){C},
\tkzDefPoint(-1,1.8){D},
\tkzDefPoint(-5,0){E},
\tkzDefPoint(1.5,1.9){O},
\tkzFillAngle[fill= blue!50,size=1cm,opacity=.7](C,A,B) % <--- Moved here
\tkzDrawPoints(A,B,C,D,E,O)
\tkzLabelPoints[below](A)
\tkzLabelPoints[right](B)
\tkzLabelPoints[right](C)
\tkzLabelPoints[below](E)
\tkzLabelPoints[above](O)
\tkzLabelSegment[above,font=\footnotesize](A,B){$4$}
\tkzLabelSegment[above,font=\footnotesize](A,C){$5$}
\tkzLabelSegment[right,font=\footnotesize](A,D){$2$}
\tkzLabelSegment[below,font=\footnotesize](A,E){$5$}
\draw[thick,black](0,0)--(-1,1.8)--(-5,0)--cycle;
\draw[thick,black](D)--(B)--cycle;
\draw[thick,black](0,0)--(4,2)--(5,0)--cycle;
\tkzDefCircle[circum](A,B,D)
\tkzGetPoint{O} \tkzGetLength{rayon}
\tkzDrawCircle[R,thick](O,\rayon pt)
\tkzMarkAngle[size=1cm,opacity=.7,black](C,A,B)
% Deleted from here
\tkzLabelAngle[pos=1.4](C,A,B){$\alpha$}
\node at (-1.4, 2.15) {$D$};
\end{tikzpicture}
\end{document}