# Alternate colors in a dashed line when using Tkz-Euclide-euclide

Is there a way when using Tikz-Euclide make the lines on the third triangle dashed but with alternating colours, sort of showing the other two triangles are on top of each other.

\documentclass{book}

\usepackage{blindtext}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usepackage{
float,
graphicx
}
% Proposition environment
\newenvironment{proposition}
{\begin{center}\em}
{\end{center}}

%Diagram enviroment
\newenvironment{diagram}
{\begin{center}\vspace*{10pt}\begin{tikzpicture}}
{\end{tikzpicture}\vspace*{-5pt}\end{center}}

\begin{document}
\begin{diagram}
\tkzDefPoint(0,0){A}
\tkzDefPoint(-2,0){B}
\tkzDefPoint(-1.5,2){C}
\tkzDefPoint(4,0){P}
\tkzDefPoint(2,0){Q}
\tkzDefPoint(2.5,2){R}
\tkzDrawPolygon[red](A,B,C)
\tkzDrawPolygon[blue](P,Q,R)
\tkzLabelPoints[right](A,P)
\tkzLabelPoints[left](B,Q)
\tkzLabelPoints[above](C,R)
\tkzMarkAngle[size=0.5](A,B,C)
\tkzMarkAngle[size=0.5](P,Q,R)
\end{diagram}

$\overline{AB} = \overline{PQ}$ given in the proposition\\
$\overline{BC} = \overline{QR}$ given in the proposition\\
$\angle{ABC} = \angle{PQR}$ given in the proposition\\

We can now superimpose  the triangles onto one another with point $A$      matched with $P$, point $B$ matched with $Q$ and $C$ matched with $R$. and and  with $\angle{ABC}$ matched with $\angle{PQR}$. we can say that the two triangles  and \textit{congruent} with each other.

\begin{diagram}

\tkzDefPoint(0,0){A}
\tkzDefPoint(-2,0){B}
\tkzDefPoint(-1.5,2){C}

\tkzDrawPolygon(A,B,C)

\tkzLabelPoints[right](A)
\tkzLabelPoints[left](B)
\tkzLabelPoints[above](C)
\tkzMarkAngle[size=0.5](A,B,C)

\end{diagram}

\end{document}


I doesn't have to be Tkz-Euclide if anyone has another solution. Thanks

I think I have solved it but any improvements are sill always welcome

    \documentclass{book}

\usepackage{blindtext}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usepackage{
float,
graphicx
}
% Proposition environment
\newenvironment{proposition}
{\begin{center}\em}
{\end{center}}

%Diagram enviroment
\newenvironment{diagram}
{\begin{center}\vspace*{10pt}\begin{tikzpicture}}
{\end{tikzpicture}\vspace*{-5pt}\end{center}}

\begin{document}
\begin{diagram}
\tkzDefPoint(0,0){A}
\tkzDefPoint(-2,0){B}
\tkzDefPoint(-1.5,2){C}
\tkzDefPoint(4,0){P}
\tkzDefPoint(2,0){Q}
\tkzDefPoint(2.5,2){R}
\tkzDrawPolygon[red](A,B,C)
\tkzDrawPolygon[blue](P,Q,R)
\tkzLabelPoints[right](A,P)
\tkzLabelPoints[left](B,Q)
\tkzLabelPoints[above](C,R)
\tkzMarkAngle[size=0.5](A,B,C)
\tkzMarkAngle[size=0.5](P,Q,R)
\end{diagram}

$\overline{AB} = \overline{PQ}$ given in the proposition\\
$\overline{BC} = \overline{QR}$ given in the proposition\\
$\angle{ABC} = \angle{PQR}$ given in the proposition\\

We can now superimpose  the triangles onto one another with  point $A$      matched with $P$, point $B$ matched with $Q$ and $C$  matched with $R$. and and  with $\angle{ABC}$ matched with  $\angle{PQR}$. we can say that the two triangles  and      \textit{congruent} with each other.

\begin{diagram}

\tkzDefPoint(0,0){A}
\tkzDefPoint(-2,0){B}
\tkzDefPoint(-1.5,2){C}

\tkzDrawPolygon[red, dash pattern= on 3pt off 5pt](A,B,C) %-------Added
\tkzDrawPolygons[blue, dash pattern = 0n 3pt off 5pt,dash phase=4pt](A,B,C)%--------Added

\tkzLabelPoints[right](A)
\tkzLabelPoints[left](B)
\tkzLabelPoints[above](C)
\tkzMarkAngle[size=0.5](A,B,C)

\end{diagram}

\end{document}