O objetivo é desenhar todas as três bissetrizes de um triângulo. Aqui ABC é um triângulo. Os pares (a1,a2), (b1,b2) e (c1,c2) representam o ponto inicial e o ponto final do arco no vértice A, B e C respectivamente. O que estou tentando fazer é pegar as linhas A -- ($(a1)!0.5!(a2)$) e o lado correspondente BC e obter seu ponto de interseção (digamos k1). Então desenharei a bissetriz do ângulo usando o vértice e o ponto de interseção k1. E repetindo isso para os dois vértices restantes. Mas meu problema é que quando estou tentando obter o ponto de intersecção das linhas, estou obtendo o ponto c1.
Aqui está o meu código:
\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}
Responder1
O segmento abs1
e o segmento BC
não têm interseção... Você pode ampliar seu abs1
segmento:
\path[overlay,name path=abs1] (A) -- ($(A)!20!($(a1)!0.5!(a2)$)$);
Escolho arbitrariamente o valor 20
e adiciono a overlay
opção para não interferir no cálculo da caixa delimitadora.
\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}