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三角形。改变顶点。画出三角形的高

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三角形。改变顶点。画出三角形的高

1) 我在乳胶中绘制三角形，但我希望顶点 A 和 B 是水平的。在下面的代码中，顶点 A 和 C 是水平的。我在更改代码时遇到了问题。

2) 我想在线段 AB 上从 C 到 D 点添加一条虚线，表示三角形的高度。请帮忙。这是我的代码：

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

\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings,shapes}

\tikzset{
    myangle/.style={fill=green!20!white, draw=green!50!black,size=.3,opacity=.3},
    intnode/.style={circle,fill=black,inner sep=1pt,outer sep=0pt}
}

\begin{document}
\begin{tikzpicture}

% Drawing the triangle and the coordinates
\draw coordinate[label=left:A] (a) --++(60:6) coordinate[label=above:B] (b);

\path[name path=ac] (a)--++(0:8.5);
\path[name path=bc] (b)--++(-45:8);
\path[name intersections={of = ac and bc, by=c}];
\node[anchor=west] at (c) {C};

\draw[use as bounding box] (a)--(b)--(c)--cycle;

% Drawing the coordinates S and T
%\coordinate (s) at ($(a)!0.75!(b)$);

%\path[name path=incls] (s) --++ (-10:5);
\path[name path=altbc] (b) -- (c);
%\path[name intersections={of = incls and altbc, by=t}]; 

%\draw[dashed] (s) -- (t) node[intnode,label={right:{\color{black}\scriptsize $T$}}] (t) {};

% Angles
\tkzFindAngle(a,b,c)
\tkzGetAngle{angleABC};
\FPround\angleABC\angleABC{0}
\tkzFindAngle(c,a,b)
\tkzGetAngle{angleCAB};
\FPround\angleCAB\angleCAB{0}
\tkzFindAngle(b,c,a)
\tkzGetAngle{angleBCA};
\FPround\angleBCA\angleBCA{0}
%\tkzFindAngle(a,s,t)
%\tkzGetAngle{angleAST};
%\FPround\angleAST\angleAST{0}

\tkzMarkAngle[myangle](a,b,c)
\tkzLabelAngle[pos=.4](a,b,c){\tiny $\angleABC^\circ$}

\tkzMarkAngle[myangle](c,a,b)
\tkzLabelAngle[pos=0.5](c,a,b){\tiny $\angleCAB^\circ$}

\tkzMarkAngle[myangle](b,c,a)
\tkzLabelAngle[pos=0.45](b,c,a){\tiny $\angleBCA^\circ$}

%\tkzMarkAngle[myangle](a,s,t)
%\tkzLabelAngle[pos=0.4](a,s,t){\tiny $\angleAST^\circ$}

%\node[intnode,label={left:\scriptsize $S$}] at (s) {};
\end{tikzpicture}
\end{document}

答案1

在评论中你实际上提出了新的问题...如果我理解正确的话，你会问这样的问题：

在此处输入图片描述

如果我是对的，那么请提出新的问题，我将能够提供上面图片的代码，首先网站上的其他成员有可能帮助您。

让我说一下，对于你最初的问题，你会收到两个答案，现在是时候选择其中一个来接受。对于后续问题，请在你的评论中提出新问题。

代码现已可用这里。

答案2

仅供比较，以下是元帖子包含在内luamplib，因此您需要用它进行编译lualatex或研究如何使其适应普通 MP，或者良好生产规范+ pdflatex。

在此处输入图片描述

\RequirePackage{luatex85}
\documentclass[border=5mm]{standalone}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
vardef mark_angle(expr a, b, c, s, tint) =
    save arc; path arc; 
    arc = fullcircle scaled 2s rotated angle (c-b) shifted b cutafter (b--a);
    fill b--arc--cycle withcolor 7/8[tint,white];
    draw arc;
    label("$" & decimal floor(0.5 + (angle (a-b) - angle (c-b)) mod 360) & "^\circ$", 
    b shifted ((s+9) * unitvector(point arctime 1/2 arclength arc of arc of arc - b)));
enddef;

beginfig(1);
    pair A, B, C, D;
    A = origin;
    B = 300 right; % rotated 15; % <- try this just to test generality
    C-A = whatever * (A-B) rotated 60;
    C-B = whatever * (A-B) rotated -45;
    C-D = whatever * (A-B) rotated 90;
    D = whatever[A,B];

    mark_angle(A,B,C,16,blue);
    mark_angle(B,C,A,16,red);
    mark_angle(C,A,B,16,blue);

    draw A--B--C--cycle;
    draw C--D dashed evenly withcolor 1/2 white;

    label.rt(decimal (abs(C-D)/abs(A-B)) , 1/2[C,D]);

    dotlabel.llft("A", A);
    dotlabel.lrt ("B", B);
    dotlabel.top ("C", C);
    dotlabel.bot ("D", D);
endfig;
\end{mplibcode}
\end{document}

答案3

纯tikz溶液：

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{angles, intersections, positioning, quotes}
\usepackage{siunitx}

\begin{document}
    \begin{tikzpicture}[
myangle/.style={angle radius=5mm, angle eccentricity=1.6,
                draw=green!50!black, fill=green!20!white}
                        ]
% Drawing the triangle and the coordinates
\coordinate[label=left:A]  (a);
\coordinate[label=right:B,right=8.5 of a] (b);
\path[name path=ac] (a) -- ++ ( 60:6.5);
\path[name path=bc] (b) -- ++ (135:8);
\path[name intersections={of = ac and bc, by=c}] node[label=C] at (c)  {};;
\draw (a)--(c)--(b)--cycle;
% Angles
\pic [myangle, "\ang{60}"] {angle = b--a--c};
\pic [myangle, "\ang{45}"] {angle = c--b--a};
\pic [myangle, "\ang{75}"] {angle = a--c--b};
\draw[dashed] (c) -- (c |- a) coordinate[label=below:D] (d);
    \end{tikzpicture}
\end{document}

在此处输入图片描述

答案4

你的意思是像这样吗？

在此处输入图片描述

代码：

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

\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings,shapes}

\tikzset{
    myangle/.style={fill=green!20!white, draw=green!50!black,size=.3,opacity=.3},
    intnode/.style={circle,fill=black,inner sep=1pt,outer sep=0pt}
}

\begin{document}
\begin{tikzpicture}

% Drawing the triangle and the coordinates
\draw coordinate[label=left:A] (a) --++(60:6) coordinate[label=above:C] (c);

\path[name path=ac] (a)--++(0:8.5);
\path[name path=bc] (c)--++(-45:8);
\path[name intersections={of = ac and bc, by=b}];
\node[anchor=west] at (b) {B};

\draw[use as bounding box] (a)--(c)--(b)--cycle;

\draw[dashed] (c) |- (b);%%%% DASHED LINE

% Drawing the coordinates S and T
%\coordinate (s) at ($(a)!0.75!(c)$);

%\path[name path=incls] (s) --++ (-10:5);
\path[name path=altbc] (c) -- (b);
%\path[name intersections={of = incls and altbc, by=t}]; 

%\draw[dashed] (s) -- (t) node[intnode,label={right:{\color{black}\scriptsize $T$}}] (t) {};

% Angles
\tkzFindAngle(a,c,b)
\tkzGetAngle{angleABC};
\FPround\angleABC\angleABC{0}
\tkzFindAngle(b,a,c)
\tkzGetAngle{angleCAB};
\FPround\angleCAB\angleCAB{0}
\tkzFindAngle(c,b,a)
\tkzGetAngle{angleBCA};
\FPround\angleBCA\angleBCA{0}
%\tkzFindAngle(a,s,t)
%\tkzGetAngle{angleAST};
%\FPround\angleAST\angleAST{0}

\tkzMarkAngle[myangle](a,c,b)
\tkzLabelAngle[pos=.4](a,c,b){\tiny $\angleABC^\circ$}

\tkzMarkAngle[myangle](b,a,c)
\tkzLabelAngle[pos=0.5](b,a,c){\tiny $\angleCAB^\circ$}

\tkzMarkAngle[myangle](c,b,a)
\tkzLabelAngle[pos=0.45](c,b,a){\tiny $\angleBCA^\circ$}

%\tkzMarkAngle[myangle](a,s,t)
%\tkzLabelAngle[pos=0.4](a,s,t){\tiny $\angleAST^\circ$}

%\node[intnode,label={left:\scriptsize $S$}] at (s) {};
\end{tikzpicture}
\end{document}

我只是将所有出现的BbyC和Cby交换B，使该段成为AB水平的。

为了绘制垂直虚线，我使用了一对垂直线，它们从 开始C，向下延伸到 相同的y坐标，B然后向右延伸到B，如下所示：\draw[dashed] (c) |- (b);。由于该段AB是连续的，因此虚线的水平部分会消失。

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