# Example: Perpendicular bisectors of a triangle

Published 2014-07-29 | Author: Sam Britt

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.

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% 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
\node (origin) at (0,0) [point, label = {below right:$P_c$}]{};

% 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} 