How to find points M and N such that AM=CN=BC only with a straight edge and a compass?

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl,pst-calculate,textcomp}
\begin{document}
    \begin{pspicture}[showgrid](-1,-1)(8,8)
    \pstTriangle[PosAngle={180,90,0}](1,2){B}(2.5,5){C}(6,2){A}
    \pstInterLC[PointSymbolB=none,PointName=none]{A}{C}{C}{B}{N}{N0}\uput[0](N){N}
    \pstInterLC[PointSymbolB=none,PointName=none,
                Radius=\pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)}]{B}{A}{A}{}{M}{M0}\uput[-90](M){M}
    \psarc[origin={C}]{->}(C){0.5}{(B)}{(N)}\uput{5mm}[-10](C){75.9636\textdegree}
    \end{pspicture}
\end{document}

The simpler the code the more challenging it becomes.

\documentclass[pstricks,12pt]{standalone}
\usepackage{pst-eucl}

\begin{document}
\begin{pspicture}[PointSymbolB=none,PointNameB=](7,5)
\pstTriangle(1,1){B}(2.5,4){C}(6,1){A}
\pstTranslation[PointName=none,PointSymbol=none]{B}{A}{C}[C']
\pstInterLC[PosAngle=45]{A}{C}{C}{B}{N}{x}
\pstInterLC[PosAngle=-90]{B}{A}{A}{C'}{M}{y}
\pstMarkAngle[MarkAngleRadius=1.1,LabelSep=.8,arrows=->]{B}{C}{A}{\scriptsize$75.96^\circ$}
\end{pspicture}
\end{document}

Notes

PointNameA and PointNameB cannot be assigned none but empty. It looks like a bug.

An alternative tikz version:

\documentclass[tikz,border=12pt]{standalone}
\usetikzlibrary{calc,bending,arrows.meta,quotes,angles}
\usepackage{textcomp}
\begin{document}

\begin{tikzpicture}[dot/.style={fill,circle,inner sep=1.2pt},>= Stealth]\footnotesize
\pgfmathsetmacro\BC{sqrt(1.5*1.5+3*3)}
\pgfmathsetmacro\AC{sqrt(3.5*3.5+3*3)}
\draw (1,2)coordinate[dot,label=left:$B$](b)--(2.5,5)coordinate[dot,label=above:$C$](c)--(6,2)coordinate[dot,label=right:$A$](a)--cycle;
\node at ([xshift=-\BC cm]a) [dot,label=below:$M$] {};
\node at ($(c)!\BC/\AC!(a)$) [dot,label=right:$N$] {};
\pic["75.96\textdegree",draw,->,angle eccentricity=.7,angle radius=1cm] {angle=b--c--a};
\end{tikzpicture}

\end{document}