Inserting the \markangle on tkz-euclide 1.16c

How do i insert \markangle in tkz-euclid 1.16c?

\documentclass{article}
\usepackage{tikz}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{5pt}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}[>=latex]
\tkzInit
\tkzDefPoint(2,2){A}
\tkzDefPoint(0,0){B}
\tkzDefPoint(5,0){C}
%angulos
\tkzMarkAngle[fill=green!50,size=7mm](A,C,B)
\tkzLabelPoints[right](C)
\tkzDrawSegments(B,C C,A)
\draw ($(C)+(165:.5)$) -- ($(C)+(165:.9)$);
\draw[red,<-] ($(C)+(165:.9)$) to[out=180,in=0] +(-1,.5) node[left] {mark};
\end{tikzpicture}

\end{document}


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It isn't clear what you are asking. You have an image that is produced by the quoted code. What do you want to do differently? – Seamus Jan 21 '12 at 11:34

Are you just trying to label the angle? If so just use label={} to specify what you want.

\documentclass{article}
\usepackage{tikz}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{5pt}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}[>=latex]
\tkzInit
\tkzDefPoint(2,2){A}
\tkzDefPoint(0,0){B}
\tkzDefPoint(5,0){C}
%angulos
\tkzMarkAngle[fill=green!50,size=7mm,label={$35^\circ$}](A,C,B)
\tkzLabelPoints[right](C)
\tkzDrawSegments(B,C C,A)
\draw ($(C)+(165:.5)$) -- ($(C)+(165:.9)$);
%\draw[red,<-] ($(C)+(165:.9)$) to[out=180,in=0] +(-1,.5) node[left] {mark};
\end{tikzpicture}
\end{document}

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See my \markangle was done manually using the commands I wanted native tkz-euclide before existed, but now I found the command, nor in the manual. I need \tkzMarkAngle[mark=x] or similar. – Regis da Silva Oct 15 '11 at 1:34
@RegisdaSilva: Still not sure what you wanted, but if you have solved your issue you should post that as an answer here with some explanation. – Peter Grill Oct 15 '11 at 1:41
I thought, is very simple: \tkzMarkAngle[mark=|]. – Regis da Silva Oct 15 '11 at 1:51
@RegisdaSilva: Simple or not, you should post the answer with an explanation. If you had a problem am sure others will too, and since my answer was not what you were looking for, so adding an answer might help others. – Peter Grill Oct 15 '11 at 3:39