A perpendicular bisector of a line segment is a line which is perpendicular to this line and passes through its midpoint. This drawing shows perpendicular bisectors of a triangle. They meet in the center of the circumcircle of the triangle.
This example was written by Sam Britt answering a question on TeX.SE.
Edit and compile if you like:
% Perpendicular bisectors of a triangle % Author: Sam Britt \documentclass[tikz,border=10pt]{standalone} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture} [ scale=3, >=stealth, point/.style = {draw, circle, fill = black, inner sep = 1pt}, dot/.style = {draw, circle, fill = black, inner sep = .2pt}, ] % the circle \def\rad{1} \node (origin) at (0,0) [point, label = {below right:$P_c$}]{}; \draw (origin) circle (\rad); % triangle nodes: just points on the circle \node (n1) at +(60:\rad) [point, label = above:$1$] {}; \node (n2) at +(-145:\rad) [point, label = below:$2$] {}; \node (n3) at +(-45:\rad) [point, label = {below right:$3$ $(0, 0, 0)$}] {}; % triangle edges: connect the vertices, and leave a node at the midpoint \draw[->] (n3) -- node (a) [label = {above right:$\vec{v}_1$}] {} (n1); \draw[->] (n3) -- node (b) [label = {below right:$\vec{v}_2$}] {} (n2); \draw[dashed] (n2) -- (n1); % Bisectors % start at the point lying on the line from (origin) to (a), at % twice that distance, and then draw a path going to the point on % the line lying on the line from (a) to the (origin), at 3 times % that distance. \draw[dotted] ($ (origin) ! 2 ! (a) $) node [right] {Bisector 1} -- ($(a) ! 3 ! (origin)$ ); % similarly for origin and b \draw[dotted] ($ (origin) ! 2 ! (b) $) -- ($(b) ! 3 ! (origin)$ ) node [right] {Bisector 2}; % short vectors \draw[->] ($ (origin) ! -.7 ! (a) $) -- node [below] {$\vec{u}_4$} ($ (origin) ! -.1 ! (a) $); \draw[->] ($ (origin) ! -.1 ! (b) $) -- node [right] {$\vec{u}_3$} ($ (origin) ! -.7 ! (b) $); % Right angle symbols \def\ralen{.5ex} % length of the short segment \foreach \inter/\first/\last in {a/n3/origin, b/n2/origin} { \draw let \p1 = ($(\inter)!\ralen!(\first)$), % point along first path \p2 = ($(\inter)!\ralen!(\last)$), % point along second path \p3 = ($(\p1)+(\p2)-(\inter)$) % corner point in (\p1) -- (\p3) -- (\p2) % path ($(\inter)!.5!(\p3)$) node [dot] {}; % center dot } \end{tikzpicture} \end{document}
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