Anonymous 1123
I am trying to node m is midpoint of the arc AB. I tried it by hand

\documentclass[tikz,border=3mm]{standalone}
\usepackage{calc}
\begin{document}
\begin{tikzpicture}[declare function={R=2;},c/.style={circle,fill,inner sep=1pt}]
\path
(0,0)	coordinate (O)
({R*cos(60)},{R*sin(60)})	coordinate (A)
({R*cos(-30)},{R*sin(-30)})	coordinate (C)
({R*cos(210)},{R*sin(210)})	coordinate (B)
({R*cos(270/2)},{R*sin(270/2)})	 coordinate[label=left:{$m$}] (M);
\path foreach \p/\g in {O/0,A/90,B/0,C/0}
{(\p)node[c]{}+(\g:2.5mm) node{$\p$}};
\end{tikzpicture}
\end{document}

![ScreenHunter 239.png](/image?hash=ceaed93c2d6353f561224dd83e5342848b9baae3b21227f0ba8f53c03126da3c)

How to node midpoint of an arc automatically?

user 3.14159
An arc between two points on a circle is not uniquely defined, there are two options. In the case at hand, there is also a midpoint close to C of an arc between A and B on a circle with center O. This is why the following code has two angles, \n1 and \n2, which differ by 180. While in this example the angles of the points are known, in general they may not be, which is why they get determined with the calc library. In more detail, after saying inside the let ... in block

\p{A}=(A)

the two lengths \x{A} and \y{A} contain the x and y coordinates on the screen. Then

\n{A}={scalar(atan2(\y{A},\x{A}))}

contains the polar angle of A, and likewise for B. Note also that using polar coordinates is convenient in this case.

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[declare function={R=2;},c/.style={circle,fill,inner sep=1pt}]
\path
(0,0)	coordinate (O)
(60:R)	coordinate (A)
(-30:R)	coordinate (C)
(210:R)	coordinate (B)
let \p{A}=(A),\p{B}=(B),
\n{A}={scalar(atan2(\y{A},\x{A}))},\n{B}={scalar(atan2(\y{B},\x{B}))},
\n1={0.5*\n{A}+0.5*\n{B}},\n2={\n1+180}
in
(\n2:R) coordinate[label=left:{$m$}] (M)
;
{(\p)node[c]{}+(\g:2.5mm) node{$\p$}};

![Screen Shot 2021-06-19 at 7.59.12 PM.png](/image?hash=b4e3da90e9751483f1bda23390044564605be812d0ff01c576b64580f3eab9f4)