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TikZ で誤った交差点を取得する

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TikZ で誤った交差点を取得する

目的は、三角形の 3 つの角の二等分線をすべて描くことです。ここで、ABC は三角形です。ペア (a1,a2)、(b1,b2)、(c1,c2) は、それぞれ頂点 A、B、C の円弧の始点と終点を表します。私がやろうとしているのは、線分 A -- ($(a1​​)!0.5!(a2)$) と対応する辺 BC を取り、それらの交点 (k1 とします) を取得することです。次に、頂点と交点 k1 を使用して角の二等分線を描きます。残りの 2 つの頂点についてもこれを繰り返します。しかし、問題は、線の交点を取得しようとすると、点 c1 を取得してしまうことです。

これが私のコードです:

\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}

答え1

セグメントabs1とセグメントにはBC交差がありません...次のabs1方法でセグメントを拡大できます:

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

境界ボックスの計算に干渉しないように、値を任意に選択してオプション20を追加します。overlay

ここに画像の説明を入力してください

\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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