Tikz：與直線垂直

Question

使用tikz-pgf，我可以繪製從點到線的垂直線。有沒有一種可靠的方法來繪製垂直線從一個點？例如，在下面的程式碼中，假設我想從線上的特定點BC向外提出垂直線。

\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
    \coordinate (A) at (0,0);
    \coordinate (B) at (2,4);
    \coordinate (C) at (8,0);

    \draw(A)--(B)--(C)--cycle;
    \draw[red] (B) -- ($(A)!(B)!(C)$);

    \node[label={below left:$A$}] at (A) {};
    \node[label={above:$B$}] at (B) {};
    \node[label={below right:$C$}] at (C) {};
\end{tikzpicture}
\end{document}

Answer 1

我剛剛寫了這樣的風格這個答案。我稍微改變了語法，所以你需要說

\draw[blue,vert={of {(B)--(C)} at (3,0)}];

在 處畫一條垂直線(3,0)，一直延伸到 處BC。我添加了vert outwards它只是距離修飾符的包裝（請參閱 pgf 手冊的第 13.5.4 節距離修飾符的語法），並且可以用作

    \draw[blue,vert outwards={from {($(B)!0.3!(C)$)} by 3cm on line to {(C)}}];

微量元素

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[vert/.style args={of #1 at #2}{insert path={%
#2 -- (intersection cs:first
  line={#1}, second line={#2--($#2+(0,10)$)}) }},
vert outwards/.style args={from #1 by #2 on line to #3}{insert path={
#1 -- ($#1!#2!90:#3$)
}}]
    \coordinate (A) at (0,0);
    \coordinate (B) at (2,4);
    \coordinate (C) at (8,0);

    \draw(A)--(B)--(C)--cycle;
    \draw[red] (B) -- ($(A)!(B)!(C)$);

    \node[label={below left:$A$}] at (A) {};
    \node[label={above:$B$}] at (B) {};
    \node[label={below right:$C$}] at (C) {};
    \draw[blue,vert={of {(B)--(C)} at (3,0)}];
    \draw[blue,vert outwards={from {($(B)!0.3!(C)$)} by 3cm on line to {(C)}}];
\end{tikzpicture}
\end{document}

Answer 2

如果您願意在線定義特定點(B)--(C)及其相對位置，那麼您可以寫下您的姆韋如下簡單解決方案：

\documentclass[tikz,border=10pt]{standalone}

\begin{document}
\begin{tikzpicture}
    \coordinate[label=below left:$A$]   (A) at (0,0);
    \coordinate[label=above:$B$]        (B) at (2,4);
    \coordinate[label=below right:$C$]  (C) at (8,0);

    \draw(A)--(B)-- coordinate[pos=0.3] (aux) % <--- coordinate of the point
                    (C)--cycle;
    \draw[red] (aux) -- (aux |- A);
\end{tikzpicture}
\end{document}

Answer 3

抱歉，不是 tikz。我理解@hpekris 的想法。

\documentclass[pstricks,border=10pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\foreach \i in {.3,.5,.7}{
\begin{pspicture}[PointSymbol=none,linejoin=1](0,-1)(8,4)
\pnodes(0,0){A}(2,4){B}(8,0){C}(4,0){I}
\psline(A)(B)(C)(A)
\pstHomO[HomCoef=\i,PosAngle=75]{B}{C}[M]
\pstProjection[PosAngle=-90]{A}{C}{B}[H]
\pstProjection[PosAngle=-90]{A}{C}{M}[M']
\pcline(M)(M')
\pcline(B)(H)
\end{pspicture}}
\end{document}

Answer 4

另一個 PSTricks 解決方案僅用於比較目的。

我提供了一些可能的技巧，但您可以刪除不需要的部分。

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\foreach \i in {1,2,3}{%
\begin{pspicture}(8,5)
\pstTriangle(1,1){A}(7,1){B}(3,4){C}
\psline(C)(C|A)
\pnode([nodesep=\i]{B}C){P}
\psline(P)(P|A)
\pnode([nodesep=\i,offset=\i]{B}C){Q}
\psline[linecolor=red](P)(Q)
\end{pspicture}}
\end{document}