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在 TikZ 中获取错误的交点

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在 TikZ 中获取错误的交点

目的是画出三角形的所有三个角平分线。这里 ABC 是一个三角形。对 (a1,a2)、(b1,b2) 和 (c1,c2) 分别表示顶点 A、B 和 C 处圆弧的起点和终点。我试图做的是取直线 A -- ($(a1​​)!0.5!(a2)$) 和相应的边 BC 并得到它们的交点（称为 k1）。然后我将使用顶点和交点 k1 绘制角平分线。并对剩下的两个顶点重复此操作。但我的问题是，当我试图获取直线的交点时，我得到的是点 c1。

这是我的代码：

\documentclass[11pt,a4paper]{article}

\usepackage[margin=0.75in,marginparsep=0pt]{geometry}

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}


\begin{document}
\begin{tikzpicture}
% vertices of the triangle
\coordinate[label=below:{$A(x_1,y_1)$}] (A) at (0,0);
\coordinate[label=below:{$B(x_2,y_2)$}] (B) at (4.5cm, 0);
\coordinate[label=above:{$C(x_3,y_3)$}] (C) at (6cm, 5cm);

% drawing triangle
\draw[name path=trg] (A) -- (B) -- (C) -- cycle;

% circle at each vertex A, B and C to get intersection
% points with triangle
\path[name path=circa] (A) circle (6mm);
\path[name path=circb] (B) circle (6mm);
\path[name path=circc] (C) circle (6mm);

% labeling intersections of circles and each vertex angle
\path [name intersections={of=trg and circa, by={a1,a2}}]; % at vertex A
\path [name intersections={of=trg and circb, by={b1,b2}}]; % at vertex B
\path [name intersections={of=trg and circc, by={c1,c2}}]; % at vertex C


% drawing arc at each vertex
\draw[bend right] (a1) to (a2);
\draw[bend right] (b2) to (b1);
\draw[bend right] (c2) to (c1);

\path[name path=AB] (A) -- (B);
\path[name path=BC] (B) -- (C);
\path[name path=CA] (C) -- (A);

% determining intersection of angle bisector and 
% corresponding side of the triangles
\path[name path=abs1] (A) -- ($(a1)!0.5!(a2)$);
\path[name intersections={of=abs1 and BC, by={k1}}];
\fill[red] (k1) circle [radius=2pt];

\path[name path=abs2] (B) -- ($(b1)!0.5!(b2)$);
\path[name intersections={of=abs2 and CA, by={k2}}];
\fill[red] (k2) circle [radius=2pt];

\path[name path=abs3] (C) -- ($(c1)!0.5!(c2)$);
\path[name intersections={of=abs3 and AB, by={k3}}];
\fill[red] (k3) circle [radius=2pt];

% drawing angle bisectors
% the problem is that k1, k2, and k3 are at the same place
\draw [red] (A) -- (k1); 
\draw [red] (B) -- (k2);
\draw [red] (C) -- (k3);
\end{tikzpicture}


\end{document}

答案1

线段abs1与线段BC没有交点...您可以abs1通过以下方式扩大您的线段：

\path[overlay,name path=abs1] (A) -- ($(A)!20!($(a1)!0.5!(a2)$)$);

我任意选择值20并添加overlay选项，以免干扰边界框的计算。

在此处输入图片描述

\documentclass[]{standalone}

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}


\begin{document}
\begin{tikzpicture}
% vertices of the triangle
\coordinate[label=below:{$A(x_1,y_1)$}] (A) at (0,0);
\coordinate[label=below:{$B(x_2,y_2)$}] (B) at (4.5cm, 0);
\coordinate[label=above:{$C(x_3,y_3)$}] (C) at (6cm, 5cm);

% drawing triangle
\draw[name path=trg] (A) -- (B) -- (C) -- cycle;

% circle at each vertex A, B and C to get intersection
% points with triangle
\path[name path=circa] (A) circle (6mm);
\path[name path=circb] (B) circle (6mm);
\path[name path=circc] (C) circle (6mm);

% labeling intersections of circles and each vertex angle
\path [name intersections={of=trg and circa, by={a1,a2}}]; % at vertex A
\path [name intersections={of=trg and circb, by={b1,b2}}]; % at vertex B
\path [name intersections={of=trg and circc, by={c1,c2}}]; % at vertex C


% drawing arc at each vertex
\draw[bend right] (a1) to (a2);
\draw[bend right] (b2) to (b1);
\draw[bend right] (c2) to (c1);

\path[name path=AB] (A) -- (B);
\path[name path=BC] (B) -- (C);
\path[name path=CA] (C) -- (A);

% determining intersection of angle bisector and 
% corresponding side of the triangles
\path[overlay,name path=abs1] (A) -- ($(A)!20!($(a1)!0.5!(a2)$)$);
\path[name intersections={of=abs1 and BC, by={k1}}];
\fill[red] (k1) circle [radius=2pt];

\path[overlay,name path=abs2] (B) -- ($(B)!20!($(b1)!0.5!(b2)$)$);
\path[name intersections={of=abs2 and CA, by={k2}}];
\fill[red] (k2) circle [radius=2pt];

\path[overlay,name path=abs3] (C) -- ($(C)!15!($(c1)!0.5!(c2)$)$);
\path[name intersections={of=abs3 and AB, by={k3}}];
\fill[red] (k3) circle [radius=2pt];

% drawing angle bisectors
% the problem is that k1, k2, and k3 are at the same place
\draw [red] (A) -- (k1); 
\draw [red] (B) -- (k2);
\draw [red] (C) -- (k3);
\end{tikzpicture}


\end{document}

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