如何将文字对齐到三角形图形上方

Question

我设法将斜边文本与三角形的斜边对齐，但我觉得~~~~~~在这一行中使用了很多东西，效率很低node[above] {$\sqrt{1+x^2}$~~~~~~~} (B) --。

有没有更好的方法来获得与我现在相同的对齐，而无需过度使用~？

\documentclass[hidelinks,14pt, letterpaper]{extarticle}
\usepackage{amsmath, amssymb, tikz}

\newcommand{\pythagwidth}{3cm}
\newcommand{\pythagheight}{2cm}

\begin{document}
\begin{figure}[h]
\centering

\begin{tikzpicture}[scale=1.25]
\coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
\coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
\coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
\draw 
  (A) -- 
  node[above] {$\sqrt{1+x^2}$~~~~~~~} (B) -- 
  node[right] {?} (C) -- 
  node[below] {?} 
  (A);
\draw 
  (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

\end{tikzpicture}
\caption{Caption}
\label{fig:my_label}
\end{figure}
\end{document}

Answer 1

像这样吗？我使用node[midway,above left=0pt,inner sep=0.5pt] {$\sqrt{1+x^2}$}，其中inner sep=0.5pt控制距离。

\documentclass[hidelinks,14pt, letterpaper]{extarticle}
\usepackage{amsmath, amssymb, tikz}

\newcommand{\pythagwidth}{3cm}
\newcommand{\pythagheight}{2cm}

\begin{document}
\begin{figure}[h]
\centering

\begin{tikzpicture}[scale=1.25]
\coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
\coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
\coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
\draw 
  (A) -- 
  node[midway,above left=0pt,inner sep=0.5pt] {$\sqrt{1+x^2}$} (B) -- 
  node[right] {?} (C) -- 
  node[below] {?} 
  (A);
\draw 
  (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

\end{tikzpicture}
\caption{Caption}
\label{fig:my_label}
\end{figure}
\end{document}

附录：只是为了好玩：使用 Ti 编写更简单、更短的代码钾是……

\documentclass[tikz,border=3.14mm]{standalone}
\begin{document}
\begin{tikzpicture}[scale=1.25]
\draw  (-1.5,-1) coordinate [label=left:$A$] (A) -- 
  node[midway,above,sloped] {$\sqrt{1+x^2}$} 
  (1.5,1) coordinate [label=above:$B$] (B) --
  node[right] {?} 
  (1.5,-1)coordinate [label=below right:$C$] (C) -- 
  node[below] {?} cycle;
\draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
\end{tikzpicture}
\end{document}

Answer 2

只是为了好玩：用pstricks一个非常短的代码来得到这个数字：

 \documentclass{article}
\usepackage{pst-eucl}%,
\usepackage{auto-pst-pdf}

\begin{document}

    \begin{postscript}
    \psset{unit=2, linejoin=1, PointSymbol=none,}
    \pstTriangle(-1.5,-1){A}(1.5,1){B}(1.5,-1){C}
    \ncline[linestyle=none]{A}{B}\naput*[nrot=:U]{$ \sqrt{1 + x^2}$}
    \psset{PointName=none}
    \pstMiddleAB{A}{C}{I}\uput[d](I){?}
    \pstMiddleAB{B}{C}{J}\uput[r](J){?}
    \end{postscript}

\end{document}