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












6














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}


enter image description here



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?










share|improve this question





























    6














    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}


    enter image description here



    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?










    share|improve this question



























      6












      6








      6







      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}


      enter image description here



      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?










      share|improve this question















      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}


      enter image description here



      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?







      pstricks pst-eucl






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 29 '18 at 18:40









      God Must Be Crazy

      5,68711039




      5,68711039










      asked Dec 29 '18 at 15:22









      chishimotoji

      857316




      857316






















          4 Answers
          4






          active

          oldest

          votes


















          5














          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}


          enter image description here






          share|improve this answer



















          • 1




            Radius=pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)} is also possible.
            – Herbert
            Dec 29 '18 at 18:43










          • Why do you replace pscalculate by pstDistCalc ?
            – chishimotoji
            Dec 30 '18 at 11:54






          • 1




            It also does the conversion into the internal screen unit, whereas pscalculate does only the calculation without taken the current screen unit into account.
            – Herbert
            Dec 30 '18 at 12:36





















          4














          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}


          enter image description here



          Notes



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






          share|improve this answer































            3














            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}



            enter image description here







            share|improve this answer























            • Thank you for your answer with TikZ.
              – chishimotoji
              Dec 29 '18 at 18:20





















            3














            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}


            enter image description here






            share|improve this answer





















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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5














              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}


              enter image description here






              share|improve this answer



















              • 1




                Radius=pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)} is also possible.
                – Herbert
                Dec 29 '18 at 18:43










              • Why do you replace pscalculate by pstDistCalc ?
                – chishimotoji
                Dec 30 '18 at 11:54






              • 1




                It also does the conversion into the internal screen unit, whereas pscalculate does only the calculation without taken the current screen unit into account.
                – Herbert
                Dec 30 '18 at 12:36


















              5














              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}


              enter image description here






              share|improve this answer



















              • 1




                Radius=pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)} is also possible.
                – Herbert
                Dec 29 '18 at 18:43










              • Why do you replace pscalculate by pstDistCalc ?
                – chishimotoji
                Dec 30 '18 at 11:54






              • 1




                It also does the conversion into the internal screen unit, whereas pscalculate does only the calculation without taken the current screen unit into account.
                – Herbert
                Dec 30 '18 at 12:36
















              5












              5








              5






              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}


              enter image description here






              share|improve this answer














              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}


              enter image description here







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Dec 29 '18 at 18:45

























              answered Dec 29 '18 at 16:58









              Herbert

              270k24408717




              270k24408717








              • 1




                Radius=pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)} is also possible.
                – Herbert
                Dec 29 '18 at 18:43










              • Why do you replace pscalculate by pstDistCalc ?
                – chishimotoji
                Dec 30 '18 at 11:54






              • 1




                It also does the conversion into the internal screen unit, whereas pscalculate does only the calculation without taken the current screen unit into account.
                – Herbert
                Dec 30 '18 at 12:36
















              • 1




                Radius=pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)} is also possible.
                – Herbert
                Dec 29 '18 at 18:43










              • Why do you replace pscalculate by pstDistCalc ?
                – chishimotoji
                Dec 30 '18 at 11:54






              • 1




                It also does the conversion into the internal screen unit, whereas pscalculate does only the calculation without taken the current screen unit into account.
                – Herbert
                Dec 30 '18 at 12:36










              1




              1




              Radius=pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)} is also possible.
              – Herbert
              Dec 29 '18 at 18:43




              Radius=pstDistCalc{sqrt((2.5-1)^2+(5-2)^2)} is also possible.
              – Herbert
              Dec 29 '18 at 18:43












              Why do you replace pscalculate by pstDistCalc ?
              – chishimotoji
              Dec 30 '18 at 11:54




              Why do you replace pscalculate by pstDistCalc ?
              – chishimotoji
              Dec 30 '18 at 11:54




              1




              1




              It also does the conversion into the internal screen unit, whereas pscalculate does only the calculation without taken the current screen unit into account.
              – Herbert
              Dec 30 '18 at 12:36






              It also does the conversion into the internal screen unit, whereas pscalculate does only the calculation without taken the current screen unit into account.
              – Herbert
              Dec 30 '18 at 12:36













              4














              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}


              enter image description here



              Notes



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






              share|improve this answer




























                4














                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}


                enter image description here



                Notes



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






                share|improve this answer


























                  4












                  4








                  4






                  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}


                  enter image description here



                  Notes



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






                  share|improve this answer














                  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}


                  enter image description here



                  Notes



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







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 29 '18 at 19:24

























                  answered Dec 29 '18 at 17:54









                  God Must Be Crazy

                  5,68711039




                  5,68711039























                      3














                      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}



                      enter image description here







                      share|improve this answer























                      • Thank you for your answer with TikZ.
                        – chishimotoji
                        Dec 29 '18 at 18:20


















                      3














                      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}



                      enter image description here







                      share|improve this answer























                      • Thank you for your answer with TikZ.
                        – chishimotoji
                        Dec 29 '18 at 18:20
















                      3












                      3








                      3






                      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}



                      enter image description here







                      share|improve this answer














                      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}



                      enter image description here








                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Dec 29 '18 at 18:21

























                      answered Dec 29 '18 at 18:19









                      AboAmmar

                      33.3k22882




                      33.3k22882












                      • Thank you for your answer with TikZ.
                        – chishimotoji
                        Dec 29 '18 at 18:20




















                      • Thank you for your answer with TikZ.
                        – chishimotoji
                        Dec 29 '18 at 18:20


















                      Thank you for your answer with TikZ.
                      – chishimotoji
                      Dec 29 '18 at 18:20






                      Thank you for your answer with TikZ.
                      – chishimotoji
                      Dec 29 '18 at 18:20













                      3














                      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}


                      enter image description here






                      share|improve this answer


























                        3














                        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}


                        enter image description here






                        share|improve this answer
























                          3












                          3








                          3






                          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}


                          enter image description here






                          share|improve this answer












                          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}


                          enter image description here







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Dec 29 '18 at 19:08









                          marmot

                          88.5k4102190




                          88.5k4102190






























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