编辑

Question

您想要这样的东西吗？请注意，B_1除非您希望标签打印在线上，否则您必须将标签向上移动。因此，\smash如果您还要应用所要求的水平调整，那么使用这里就没有什么意义了。

水平调整是通过测量宽度 $_1$ 并使用该值的一半作为标记节点的偏移量来完成的。

\documentclass[tikz,border=5pt]{standalone}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings}
\usepackage{pgfplots}
\pgfplotsset{compat=1.11}

\begin{document}
\newlength\myshift
\settowidth{\myshift}{$_1$}
  \begin{tikzpicture}

    %Rays k and $\ell$ have a common endpoint at the origin, which is labeled A. With respect to
    %the horizontal line through A, k is inclined at an angle of 50 degrees, and $\ell$ is
    %inclined at an angle of 15 degrees. To place the label for A, the rays are extended 15pt by an
    %invisible path command "before" A. The endpoints of these extensions are called "label_A_below"
    %and "label_A_above" and the midpoint of the line segment between label_A_below and label_A_above
    %is called "label_A".  A node command typesets "A" at the point 7.5pt from A on the invisible
    %line segment between A and label_A .
    \draw[fill] coordinate (A) circle (1.5pt);

    \coordinate (label_A_below) at (-130:15pt);
    \coordinate (label_A_above) at (-165:15pt);
    \coordinate (label_A) at ($(label_A_below)!0.5!(label_A_above)$);
    \node[blue] at ($(A)!7.5pt!(label_A)$){$A$};

    %B' is a point on ray k and C' is a point on $\ell$.
    \draw[name path=ray_k, -latex,] (A) -- (50:7) coordinate (B');
    \node (label_ray_k) at ($(B') + (50:7pt)$){$k$};
    \draw[name path=ray_ell, -latex,] (A) -- (15:5.5) coordinate (C');
    \node (label_ray_ell) at ($(C') + (15:6pt)$){$\ell$};

    %This command labels a point "B" 3.75cm from A on ray $k$.
    \draw (A) -- (50:{15/4}) coordinate (B);

    %These commands label the projection of AB onto \ell "C."  A line segment is drawn between
    %B and C, and a right-angle mark is placed at C.
    \draw (B) -- ($(A)!(B)!(C')$) coordinate (C) node [midway,right]{$y$};

    \draw ($(C)!3mm!-45:(A)$) coordinate (U) -- ($(C)!(U)!(A)$);
    \draw (U) -- ($(C)!(U)!(B)$);

    %This command labels the length of AC.
    \draw (A) -- (C) node [midway, below]{$x$};

    %These commands draw an invisible line segment perpendicular to ray k from B to
    %ray $\ell$. The intersection is labeled C_1. (The label for C_1 is placed after
    %the point B_1 is defined.)
    \path[name path=perpendicular_line_segment_from_B] (B) -- ($(B)!3.5cm!90:(A)$);
    \coordinate[name intersections={of=perpendicular_line_segment_from_B and ray_ell,by={C_1}}];

    %The following commands define B_1 as the intersection of k and a line segment
    %perpendicular to $\ell$ through C_1. A line segment is drawn between
    %B_1 and C_1, and a right-angle mark is placed at C_1. The labels "$B_{1}$" and "$C_{1}$"
    %are typeset either directly above or directly below this line segment.
    \path[name path=perpendicular_line_segment_from_C_1] (C_1) -- ($(C_1)!4cm!90:(C')$);
    \coordinate[name intersections={of=perpendicular_line_segment_from_C_1 and ray_k, by={B_1}}];
    \draw (C_1) -- (B_1) node [midway,right]{$y_{1}$};

    \draw ($(C_1)!3mm!-45:(A)$) coordinate (U_1) -- ($(C_1)!(U_1)!(A)$);
    \draw (U_1) -- ($(C_1)!(U_1)!(B_1)$);

    %These commands put a brace below line segment AC_1 and label the length `$x_{1}$`.  This brace
    %needs to be shifted downward from AC_1 to avoid interference with the labels A, C, and C_1.
    %First, coordinates for A_shift and C_1_shift are defined to be 5mm from AC_1. A brace is drawn
    %between A_shift  and D_shift.
    \draw[decorate,decoration={brace,raise=5pt,amplitude=5pt,mirror}] ($(A)!5.5mm!-90:(C_1)$) coordinate (A_shift) --
    node[below right=4mm and 0.0625mm, fill=white, inner sep=1pt] {$x_{1}$} ($(C_1)!5.5mm!90:(A)$) coordinate (C_1_shift);

    %This command draws the two sides of the triangle with a common vertex A.
    \draw (A) -- (B_1);
    \draw (A) -- (C_1);

    %These commands draw an invisible line segment 2.5mm above AB_1 for labels B and B_1.
    \coordinate (A_shift_for_labels_version_1) at ($(A)!2.5mm!90:(B_1)$);
    \coordinate (B_1_shift_for_labels) at ($(B_1)!2.5mm!-90:(A)$);
    \coordinate (B_1_shift_for_labels_extension) at ($(B_1_shift_for_labels)!-0.25cm!(A_shift_for_labels_version_1)$);
    \path[name path=path_for_labels_B_and_B_1] (A_shift_for_labels_version_1) -- (B_1_shift_for_labels_extension);
    \path[name path=path_from_C_to_B_above] (C) -- ($(B)!-0.75cm!(C)$) coordinate (B_above);
    \path[name path=path_from_C_1_to_B_1_above] (C_1) -- ($(B_1)!-0.75cm!(C_1)$) coordinate (B_1_above);
    \coordinate[name intersections={of=path_for_labels_B_and_B_1 and path_from_C_to_B_above, by={B_label}}];
    \coordinate[name intersections={of=path_for_labels_B_and_B_1 and path_from_C_1_to_B_1_above, by={B_1_label}}];
    \node[green] at (B_label){$B$};
    \node[blue, xshift=.5\myshift, yshift=.5em] at (B_1_label){$B_{1}$};

    %These commands draw an invisible line segment 2.5mm below AC_1 for labels C and C_1. The center of the node
    %containing $C_{1}$ is displaced upwards by the subscript "1" in  $C_{1}$. To avoid this upward vertical
    %displacement, a \smash command is issued.
    \coordinate (A_shift_for_labels_version_2) at ($(A)!2.5mm!-90:(C_1)$);
    \coordinate (C_1_shift_for_labels) at ($(C_1)!2.5mm!90:(A)$);
    \path[name path=path_for_labels_C_and_C_1] (A_shift_for_labels_version_2) -- (C_1_shift_for_labels);
    \path[name path=path_from_B_to_C_below] (B) -- ($(C)!-0.75cm!(B)$) coordinate (C_below);
    \path[name path=path_from_B_1_to_C_1_below] (B_1) -- ($(C_1)!-0.75cm!(B_1)$) coordinate (C_1_below);
    \coordinate[name intersections={of=path_for_labels_C_and_C_1 and path_from_B_to_C_below, by={C_label}}];
    \coordinate[name intersections={of=path_for_labels_C_and_C_1 and path_from_B_1_to_C_1_below, by={C_1_label}}];
    \node[green] at (C_label){$C$};
    \node[blue, xshift=.5\myshift] at (C_1_label){$C\smash{_{1}}$};

  \end{tikzpicture}

\end{document}

移位标签

如果您不介意标签写在线上，您当然可以删除yshift并恢复\smash。

编辑

这里证明的C排版C_1好像_1没有任何面积。

C_1我在标签排版之前添加了以下行：

   \node[red] at (C_1_label){$C$};

由于此节点为红色，而实际节点为蓝色，因此测试是红色C在最终结果中是否可见。如下所示，红色不可见：

如果将_1视为具有面积，则C中的蓝色C_1将相对于其下方的红色发生偏移C。但事实并非如此。

Answer 1