Re: centering

in effetti non avete tutti i torti
Utilizzando Qtikz ho ottenuto questa semplificazione
`
% Graphic for TeX using PGF
% Title: C:\Users\claudio\Pictures\Diagramma2.dia
% Creator: Dia v0.97.1
% CreationDate: Mon Oct 24 15:56:03 2011
% For: claudio
% \usepackage{tikz}
% The following commands are not supported in PSTricks at present
% We define them conditionally, so when they are implemented,
% this pgf file will use them.
\ifx\du\undefined
\newlength{\du}
\fi
\setlength{\du}{12\unitlength}
\begin{tikzpicture}
\pgftransformxscale{1.000000}
\pgftransformyscale{-1.000000}
\pgfpathellipse{\pgfpoint{7.721636\du}{4.898318\du}}{\pgfpoint{3.028364\du}{0\du}}{\pgfpoint{0\du}{2.401682\du}}
\pgfusepath{stroke}
\node at (7.721636\du,5.093318\du){A};
\definecolor{dialinecolor}{rgb}{1.000000, 1.000000, 1.000000}
\pgfsetfillcolor{dialinecolor}
\fill (7.569670\du,10.044670\du)–(11.350000\du,12.822335\du)–(7.569670\du,15.600000\du)–(3.789340\du,12.822335\du)–cycle;
\pgfsetlinewidth{0.100000\du}
\draw (7.569670\du,10.044670\du)–(11.350000\du,12.822335\du)–(7.569670\du,15.600000\du)–(3.789340\du,12.822335\du)–cycle;
\node at (7.569670\du,13.017335\du){B};
\fill (3.900000\du,18.800000\du)–(3.900000\du,21.850000\du)–(11.200000\du,21.850000\du)–(11.200000\du,18.800000\du)–cycle;
\draw (3.900000\du,18.800000\du)–(3.900000\du,21.850000\du)–(11.200000\du,21.850000\du)–(11.200000\du,18.800000\du)–cycle;
\node at (7.550000\du,20.520000\du){C};
{
\definecolor{dialinecolor}{rgb}{0.000000, 0.000000, 0.000000}
\pgfsetfillcolor{dialinecolor}
% was here!!!
\pgfsetarrowsend{stealth}
\draw (7.674620\du,7.349908\du)–(7.623142\du,10.034130\du);
}
{
\definecolor{dialinecolor}{rgb}{0.000000, 0.000000, 0.000000}
\pgfsetfillcolor{dialinecolor}
% was here!!!
\pgfsetarrowsend{stealth}
\definecolor{dialinecolor}{rgb}{0.000000, 0.000000, 0.000000}
\pgfsetstrokecolor{dialinecolor}
\draw (7.562272\du,15.644077\du)–(7.554120\du,18.753397\du);
}
\end{tikzpicture}
`
Saluti e grazie
Claudio