如何用 Tikz 绘制一条通过一点并平行于 AC 的弦？

Question 1

定义一条（overlay使得它不会增加边界框）路径，该路径具有与相同的斜率A--C（这就是let \p1=($(C)-(A)$),\n1={atan2(\y1,\x1)} in所做的，它计算线的角度）并穿过F，并计算它与圆的交点。

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{arrows,calc,shapes.geometric,intersections}
\begin{document}
\begin{tikzpicture}[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
 % point O
 \coordinate (O) at (0,0);
 % ABC triangle
 \node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) {};
 \coordinate [label=above:$A$] (A) at (t.corner 1);
 \coordinate [label=left:$B$]  (B) at (t.corner 2);
 \coordinate [label=right:$C$] (C) at (t.corner 3);
 % O's label
 \node [above left] at (O) {$O$};
 % circle with 2.25cm radius and centre at O
 \draw[name path=circle] (O) circle (2.25cm);
 % point D: the point in the circumference whose angle is 50° with OC
 \coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
 % radius OD
 \draw [name path=OD] (O) -- (D);
 % point E: intersection between radius OD and the triangle
 \path [name intersections={of=OD and tri,by=E}];
 \node [below] at (E) {$E$};
 % point F: point 33% the way from O to E
 \coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
 \path[overlay,name path=line] let \p1=($(C)-(A)$),\n1={atan2(\y1,\x1)} in % computes the slope of A--C
 ($(F)+(\n1:2*2.25cm)$) -- ($(F)+(180+\n1:2*2.25cm)$);
 \draw[name intersections={of=line and circle,by={G,H}}] (G) node[above left]{$G$}
  -- (H) node[below right]{$H$};
 % draw bullets at each point
 \foreach \point in {A,...,H}
   \fill [black] (\point) circle (1.5pt);
\end{tikzpicture}
\end{document}

或者，您也可以“并行传输”路径A--C，这会产生相同的结果并且不涉及任何内容atan2。

 \path[overlay,name path=line]
 ($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);

Answer

定义一条（overlay使得它不会增加边界框）路径，该路径具有与相同的斜率A--C（这就是let \p1=($(C)-(A)$),\n1={atan2(\y1,\x1)} in所做的，它计算线的角度）并穿过F，并计算它与圆的交点。

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{arrows,calc,shapes.geometric,intersections}
\begin{document}
\begin{tikzpicture}[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
 % point O
 \coordinate (O) at (0,0);
 % ABC triangle
 \node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) {};
 \coordinate [label=above:$A$] (A) at (t.corner 1);
 \coordinate [label=left:$B$]  (B) at (t.corner 2);
 \coordinate [label=right:$C$] (C) at (t.corner 3);
 % O's label
 \node [above left] at (O) {$O$};
 % circle with 2.25cm radius and centre at O
 \draw[name path=circle] (O) circle (2.25cm);
 % point D: the point in the circumference whose angle is 50° with OC
 \coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
 % radius OD
 \draw [name path=OD] (O) -- (D);
 % point E: intersection between radius OD and the triangle
 \path [name intersections={of=OD and tri,by=E}];
 \node [below] at (E) {$E$};
 % point F: point 33% the way from O to E
 \coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
 \path[overlay,name path=line] let \p1=($(C)-(A)$),\n1={atan2(\y1,\x1)} in % computes the slope of A--C
 ($(F)+(\n1:2*2.25cm)$) -- ($(F)+(180+\n1:2*2.25cm)$);
 \draw[name intersections={of=line and circle,by={G,H}}] (G) node[above left]{$G$}
  -- (H) node[below right]{$H$};
 % draw bullets at each point
 \foreach \point in {A,...,H}
   \fill [black] (\point) circle (1.5pt);
\end{tikzpicture}
\end{document}

或者，您也可以“并行传输”路径A--C，这会产生相同的结果并且不涉及任何内容atan2。

 \path[overlay,name path=line]
 ($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);

Question 2

只是为了完整性。你提到tkz-euclide而且我也不会说法语。但是，如果您只搜索诸如“中点”、“平行”等关键字，文档就很容易使用。此外，以下包含来自 tkz-euclide 的命令的速查表非常有用

tkz-euclid-cheatsheet.en.md

要回答你的问题如何定义一条F平行线AC，很简单

\tkzDefLine[parallel=through F](A,C) \tkzGetPoint{f}

在tkz-euclide。假设点已经定义。类似地，可以通过以下方式找到与圆的交点

\tkzInterLC(F,f)(O,A) \tkzGetPoints{G}{H}

其中tkzInterLC可以读作线Ff与以为中心O、为半径的圆的交点r = |OA|。

请参阅下面的完整代码=)

\documentclass{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all} % on charge tous les objets

\usepackage[utf8]{inputenc}

\begin{document}

\begin{tikzpicture}
    \tkzInit[xmin=-0.5,xmax=4.5,ymin=-1.4,ymax=4]
    \tkzClip
    % Defines where F is placed on the segment OE.
    % 0 = O and 1 = E
    \edef\OF{0.5}

    \tkzDefPoint(0,0){B} \tkzDefPoint(4,0){C}

    \tkzDefEquilateral(B,C)\tkzGetPoint{A};
    \tkzDefBarycentricPoint(A=1,B=1,C=1) \tkzGetPoint{O}
    \tkzDefMidPoint(A,C) \tkzGetPoint{E}

    % Finds the intersection between the line OE and the
    % circle with center O, and radius r = |OA|.
    \tkzInterLC(O,E)(O,A) \tkzGetPoints{D2}{D}

    % Calculates the length |OE| multiplies it with scaling
    \tkzCalcLength[cm](O,E)\tkzGetLength{rOE}
    \pgfmathsetmacro{\pointF}{\OF*\rOE}

    \tkzInterLC[R](O,D)(O,\pointF cm) \tkzGetPoints{F2}{F}

    % Calculates the line parallell to AC through F
    \tkzDefLine[parallel=through F](A,C) \tkzGetPoint{f}
    \tkzInterLC(F,f)(O,A) \tkzGetPoints{G}{H}

    \tkzDrawPolygon[fill=black!10](A,B,C)
    \tkzDrawSegments(G,H O,D)
    \tkzDrawPoints[fill=black,size=10](A,B,C,O,E,D,F,G,H)
    \tkzDrawCircle(O,A)

    \tkzLabelPoint[above](A){$A$}
    \tkzLabelPoint[below left](B){$B$}
    \tkzLabelPoint[below right](C){$C$}
    \tkzLabelPoint[right](D){$D$}
    \tkzLabelPoints[above](E,F,O)

    \tkzLabelPoint[above left](G){$G$}
    \tkzLabelPoint[below](H){$H$}
\end{tikzpicture}

\end{document}

Answer