# tkz-euclide: How to get opposite angle?

I made this picture with tkz-euclide but now I want to mark the opposite angle that completes alpha to 180 degree. How can I do this?

\documentclass[12pt,a4paper]{article}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B,1/3/C}
\tkzDrawLines(A,B B,C A,C)
\tkzMarkAngle(A,C,B)
\tkzMarkAngle(C,B,A)
\tkzMarkAngle(B,A,C)
\tkzLabelAngles[pos=0.5](A,C,B){$\gamma$}
\tkzLabelAngles[pos=0.7](C,B,A){$\beta$}
\tkzLabelAngles[pos=0.5](B,A,C){$\alpha$}
\end{tikzpicture}

\end{document} • Welcome to TeX.SE! Apr 25 at 20:02
• What is your version of tkz-euclide ? Now usetkzobj{all} is not necessary Apr 25 at 20:02

With the last version 4.05

\documentclass[12pt,a4paper]{article}
\usepackage{tkz-euclide}

\begin{document}

\begin{tikzpicture}
\tkzDefPoints{0/0/A,4/0/B,1/3/C}
\tkzDrawLines(A,B B,C A,C)
\tkzMarkAngle(A,C,B)
\tkzMarkAngle(C,B,A)
\tkzMarkAngle(B,A,C)
\tkzLabelAngles[pos=0.5](A,C,B){$\gamma$}
\tkzLabelAngles[pos=0.7](C,B,A){$\beta$}
\tkzLabelAngles[pos=0.5](B,A,C){$\alpha$}
\tkzDefPointBy[symmetry=center A](B) \tkzGetPoint{B'}
\tkzLabelAngles[pos=0.5](C,A,B'){$\delta$}
\tkzMarkAngle(C,A,B')
\end{tikzpicture}

\end{document} • Thanks! This is exactly what I was looking for. Apr 25 at 20:24