Perpendicular bisectors of a triangle

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:

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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
  \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) $)
 
 
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