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未画出直角

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未画出直角

我画了一个直角三角形，其直角位于C。顶点C通过直角三角形的高公式从三角形直角所在的顶点开始定位。由于P是 的高的底边C，并且|AP|=6和|BP|=8，高的长度为\sqrt{(6)(8)} = 4\sqrt{3}。（该图缩小了 4 倍。）

我担心的是，我画的直角标记C不是正方形。（用 替换 可以更明显地看出这一点3mm。10mm）要么TikZ是定位不准确C，要么是绘制直角标记的代码有错误。

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}

\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings,decorations.pathreplacing}


\begin{document}

\begin{tikzpicture}

%The vertices A, B, and C for a right triangle are located.
\coordinate (A) at (0,0);
\coordinate (B) at (30:3.5);
\coordinate (P) at (30:1.5);

%P is the foot of the altitude of a right triangle from C to to $\overline{AB}$.
%Since the length of $\overline{AP}$ is 6 and the length of $\overline{BP}$ is 8,
%the length of the altitude is $\sqrt{(6)(8)} = 4\sqrt{3}$.
\coordinate (C) at ($(P) +(120:{sqrt(3)})$);
\draw[dashed] (P) -- (C);


\draw let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(A)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2)}, inner sep=0] at ($(A) + ({0.5*(\n1+\n2)+180}:0.15)$){$A$};
\coordinate (B) at (30:4);
\draw let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2)+180}, inner sep=0] at ($(B) + ({0.5*(\n1+\n2)}:0.15)$){$B$};
\draw let \p1=($(P)-(C)$), \n1={atan(\y1/\x1)} in node[anchor=\n1, inner sep=0] at ($(C)!-0.15cm!(P)$){$C$};
\draw let \p1=($(P)-(C)$), \n1={atan(\y1/\x1)} in node[anchor={\n1-180}, inner sep=0] at ($(P)!-0.15cm!(C)$){$P$};

%The right-angle mark at C is drawn.
\coordinate (U) at ($(C)!3mm!45:(A)$);
\draw (U) -- ($(C)!(U)!(A)$);
\draw (U) -- ($(C)!(U)!(B)$);

%The right-angle mark at P is drawn.
\coordinate (V) at ($(P)!3mm!-45:(A)$);
\draw (V) -- ($(P)!(V)!(A)$);
\draw (V) -- ($(P)!(V)!(C)$);


%$\triangle{ABC}$ is drawn.
\draw (C) -- (A) -- (B) -- (C);



%The angle mark for $\angle{ABC}$ is drawn. It is marked with "|".
\draw[draw=blue] let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in ($(B)!0.55cm!(A)$) arc ({\n1+180}:{\n2+180}:0.55);
\draw[draw=blue] let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in ($(B) +({0.5*(\n1+\n2)+180}:{0.55cm-3pt})$) -- ($(B) +({0.5*(\n1+\n2)+180}:{0.55cm+3pt})$);

%The angle mark for $\angle{ACP}$ is drawn. It is marked with "|".
\draw[draw=blue] let \p1=($(C)-(P)$), \n1={atan(\y1/\x1)}, \p2=($(C)-(A)$), \n2={atan(\y2/\x2)} in ($(C)!0.55cm!(P)$) arc (\n1:{\n2-180}:0.55);
\draw[draw=blue] let \p1=($(C)-(P)$), \n1={atan(\y1/\x1)}, \p2=($(C)-(A)$), \n2={atan(\y2/\x2)} in ($(C) +({0.5*(\n1+\n2-180)}:{0.55cm-3pt})$) -- ($(C) +({0.5*(\n1+\n2-180)}:{0.55cm+3pt})$);

\end{tikzpicture}


\end{document}

答案1

您的图片包含两个B在不同位置命名的不同坐标。如果您注释掉第二个定义，据我所知，三角形看起来没问题：

定义 1

\documentclass[border=10pt,tikz]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}

  %The vertices A, B, and C for a right triangle are located.
  \coordinate (A) at (0,0);
  \coordinate (B) at (30:3.5);
  \coordinate (P) at (30:1.5);

  %P is the foot of the altitude of a right triangle from C to to $\overline{AB}$.
  %Since the length of $\overline{AP}$ is 6 and the length of $\overline{BP}$ is 8,
  %the length of the altitude is $\sqrt{(6)(8)} = 4\sqrt{3}$.
  \coordinate (C) at ($(P) +(120:{sqrt(3)})$);
  \draw[dashed] (P) -- (C);

  \draw let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(A)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2)}, inner sep=0] at ($(A) + ({0.5*(\n1+\n2)+180}:0.15)$){$A$};
%   \coordinate (B) at (30:4);
  \draw let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2)+180}, inner sep=0] at ($(B) + ({0.5*(\n1+\n2)}:0.15)$){$B$};
  \draw let \p1=($(P)-(C)$), \n1={atan(\y1/\x1)} in node[anchor=\n1, inner sep=0] at ($(C)!-0.15cm!(P)$){$C$};
  \draw let \p1=($(P)-(C)$), \n1={atan(\y1/\x1)} in node[anchor={\n1-180}, inner sep=0] at ($(P)!-0.15cm!(C)$){$P$};

  %The right-angle mark at C is drawn.
  \coordinate (U) at ($(C)!3mm!45:(A)$);
  \draw (U) -- ($(C)!(U)!(A)$);
  \draw (U) -- ($(C)!(U)!(B)$);

  %The right-angle mark at P is drawn.
  \coordinate (V) at ($(P)!3mm!-45:(A)$);
  \draw (V) -- ($(P)!(V)!(A)$);
  \draw (V) -- ($(P)!(V)!(C)$);

  %$\triangle{ABC}$ is drawn.
  \draw (C) -- (A) -- (B) -- (C);

  %The angle mark for $\angle{ABC}$ is drawn. It is marked with "|".
  \draw[draw=blue] let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in ($(B)!0.55cm!(A)$) arc ({\n1+180}:{\n2+180}:0.55);
  \draw[draw=blue] let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in ($(B) +({0.5*(\n1+\n2)+180}:{0.55cm-3pt})$) -- ($(B) +({0.5*(\n1+\n2)+180}:{0.55cm+3pt})$);

  %The angle mark for $\angle{ACP}$ is drawn. It is marked with "|".
  \draw[draw=blue] let \p1=($(C)-(P)$), \n1={atan(\y1/\x1)}, \p2=($(C)-(A)$), \n2={atan(\y2/\x2)} in ($(C)!0.55cm!(P)$) arc (\n1:{\n2-180}:0.55);
  \draw[draw=blue] let \p1=($(C)-(P)$), \n1={atan(\y1/\x1)}, \p2=($(C)-(A)$), \n2={atan(\y2/\x2)} in ($(C) +({0.5*(\n1+\n2-180)}:{0.55cm-3pt})$) -- ($(C) +({0.5*(\n1+\n2-180)}:{0.55cm+3pt})$);

\end{tikzpicture}
\end{document}

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