The following code is my attempt to find M and N such that AM=CN=BC.

documentclass[pstricks,border=12pt]{standalone}

usepackage{pst-eucl,pst-calculate}

begin{document}

defx{pscalculate{6-3.3541}}

begin{pspicture}[showgrid](-1,-1)(8,8)

pstGeonode[PosAngle={180,90,0,-90}](1,2){B}(2.5,5){C}(6,2){A}(x,2){M}

pspolygon(B)(C)(A)

pstRotation[PosAngle=90,RotAngle=75.9638]{C}{B}[N]



psGetDistanceAB[xShift=-8,yShift=4](B)(C){M}

psGetAngleABC[AngleValue=true,MarkAngleRadius=.5,LabelSep=0.5,%

ShowWedge=false,xShift=-5,yShift=7,arrows=->](B)(C)(A){}



end{pspicture}

end{document}

However, I am trying not to use psGetDistanceAB and psGetAngleABC.

Question

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

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){75.9636textdegree}

    end{pspicture}

end{document}

The shorter the code the more beautiful it becomes.

documentclass[pstricks,12pt]{standalone}

usepackage{pst-eucl}



begin{document}

begin{pspicture}(0,1)(7,6)

pstTriangle[PosAngle={180,90,0}](1,2){B}(2.5,5){C}(6,2){A}

pstInterLC[PointSymbolB=none,PointNameB=,PosAngle=45]{A}{C}{C}{B}{N}{x}

pstTranslation[PointName=none,PointSymbol=none]{B}{A}{C}[C']

pstInterLC[PointNameA=,PointSymbolA=none,PosAngleB=-90]{A}{B}{A}{C'}{y}{M}

pstMarkAngle[arrows=->,MarkAngleRadius=1.1,LabelSep=.8]{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

pgfmathsetmacroBC{sqrt(1.5*1.5+3*3)}

pgfmathsetmacroAC{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["75.96textdegree",draw,->,angle eccentricity=.7,angle radius=1cm] {angle=b--c--a};

end{tikzpicture}



end{document}

A version in which you do not have to compute anything yourself. All distances and angles can be very conveniently computed with TikZ. In more detail, the following code computes the distance between B and C (n1) as well as the "slopes" of all the edges, and then uses polar coordinates to set M n1 away from A in B direction and N n1 away from C in A direction. The angle at B is also computed by tikz and inserted via pgfmathprintnumber which allows you to specify the number of digits and so on, if needed.

documentclass[tikz,border=3.14mm]{standalone}

usetikzlibrary{calc,angles,quotes}

begin{document}

begin{tikzpicture}

draw (1,2) coordinate[label=below:$B$] (B) -- 

(2.5,5) coordinate[label=above:$C$] (C) -- 

(6,2) coordinate[label=right:$A$] (A) -- cycle;

path let p1=($(B)-(C)$),p2=($(B)-(A)$),p3=($(A)-(C)$),

n1={veclen(x1,y1)},n2={atan2(y2,x2)},n3={atan2(y3,x3)},

n4={n3-atan2(y1,x1)} in 

($(A)+(n2:n1)$) coordinate[label=below:$M$] (M)

($(C)+(n3:n1)$) coordinate[label=above right:$N$] (N)

pic["$pgfmathparse{n4}pgfmathprintnumber{pgfmathresult}^circ$",draw,-latex,angle

eccentricity=1.3,angle radius=1cm] 

{angle=B--C--A};

foreach X in {A,B,C,M,N}

{fill (X) circle (1pt);}

end{tikzpicture}

end{document}