salve
in altro forum mi hanno proposto due soluzioni
la prima che parla dell’ inversione a–c e c–a
l’altra decisamente più sofisticata che riposto
`
\documentclass{minimal}
\usepackage{etex}
\usepackage{tikz}
\usetikzlibrary{shapes,calc,arrows,through,intersections}
\begin{document}
\begin{tikzpicture}[scale=0.5]
\tikzset{mark coordinate/.style={inner sep=0pt,
outer sep=0pt,
minimum size=3pt,
fill=#1,
circle}
}
\draw[thick] (0,0) coordinate [mark coordinate=black,label=$A$] (a) —
(6,0) coordinate [mark coordinate=black,label=50:$B$] (b) —
(9,6) coordinate [mark coordinate=black,label=$C$] (c) — cycle ;
\node [name path=Circle1,draw,circle through=(a)] at (b) {};
\path [name path=BC] (b) — (c);
\path [name intersections={of=BC and Circle1, name=i}]
(i-1) coordinate [mark coordinate=red];
\node [name path=Circle2,draw,circle through=(a)] at (i-1) {};
\node [name path=Circle3,draw,circle through=(i-1)] at (a) {};
\fill [name intersections={of=Circle2 and Circle3, name=j}]
(j-1) coordinate [mark coordinate=green];
\path [name path=AC] (a) — (c);
\path [name path=Bj-1] (b) — (j-1);
\fill [name intersections={of=Bj-1 and AC, name=k}]
(k-1) coordinate [mark coordinate=blue];
\draw [red] (b)–(k-1);
\node [name path=Circle4,draw,circle through=(b)] at (c) {};
\fill [name intersections={of=Circle4 and AC, name=l}]
(l-1) coordinate [mark coordinate=orange];
\end{tikzpicture}
\end{document}
`
ciao claudio