How to find points M and N such that AM=CN=BC only with a straight edge and a compass?
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?
pstricks pst-eucl
add a comment |
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?
pstricks pst-eucl
add a comment |
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?
pstricks pst-eucl
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?
pstricks pst-eucl
pstricks pst-eucl
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
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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}
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, whereaspscalculate
does only the calculation without taken the current screen unit into account.
– Herbert
Dec 30 '18 at 12:36
add a comment |
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.
add a comment |
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}
Thank you for your answer with TikZ.
– chishimotoji
Dec 29 '18 at 18:20
add a comment |
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}
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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}
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, whereaspscalculate
does only the calculation without taken the current screen unit into account.
– Herbert
Dec 30 '18 at 12:36
add a comment |
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}
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, whereaspscalculate
does only the calculation without taken the current screen unit into account.
– Herbert
Dec 30 '18 at 12:36
add a comment |
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}
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}
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, whereaspscalculate
does only the calculation without taken the current screen unit into account.
– Herbert
Dec 30 '18 at 12:36
add a comment |
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, whereaspscalculate
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
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
edited Dec 29 '18 at 19:24
answered Dec 29 '18 at 17:54
God Must Be Crazy
5,68711039
5,68711039
add a comment |
add a comment |
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}
Thank you for your answer with TikZ.
– chishimotoji
Dec 29 '18 at 18:20
add a comment |
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}
Thank you for your answer with TikZ.
– chishimotoji
Dec 29 '18 at 18:20
add a comment |
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}
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}
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
add a comment |
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
add a comment |
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}
add a comment |
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}
add a comment |
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}
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}
answered Dec 29 '18 at 19:08
marmot
88.5k4102190
88.5k4102190
add a comment |
add a comment |
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