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How to draw with Tikz a chord parallel to AC that passes through a point?
TikZ how to draw lines to/from nodes correctly?How to define the default vertical distance between nodes?To wrap the external lines so that it can touch the perimeterHow can I draw tikz arrows on a calculated triangle?Calculate the intersection between a path enclosed by a `scope` and another pathTikz 3d trimetric view coordinate calculation bugTikZ: Drawing an arc from an intersection to an intersectionFill a section between two circles with TikZInclude node at end point tikz decorationsDraw TikZ circles with a specific outer radius
I've drawn the following diagram with Tikz:
The code used:
documentclassstandalone
usepackagepgf,tikz
usetikzlibrarybabel,calc,arrows,shapes.geometric,intersections,through,backgrounds
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
% draw bullets at each point
foreach point in A,B,C,O,D,E,F
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
About my current code:
- I'm aware of
tkz-euclide
and I think it could help me here, but the only documentation I could find (at CTAN) is in French, I language I can't read. - According to the
pgf
documentation, thearrows
library has been deprecated in favor ofarrows.meta
. I only usedarrows
because this code was adapted from Geogebra and I still haven't got into changing that particular aspect of the code.
I want to place 2 points G
and H
in the circle such that the chord GH
:
passes through F and- is parallel to AC
so as to obtain something like this (the following was drawn with Geogebra):
I've found answers regarding drawing parallel lines, but in this case it's not just parallel, it also needs to start and end at the circle, so I couldn't find a good answer by myself.
tikz-pgf circles intersections tikz-angles
add a comment |
I've drawn the following diagram with Tikz:
The code used:
documentclassstandalone
usepackagepgf,tikz
usetikzlibrarybabel,calc,arrows,shapes.geometric,intersections,through,backgrounds
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
% draw bullets at each point
foreach point in A,B,C,O,D,E,F
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
About my current code:
- I'm aware of
tkz-euclide
and I think it could help me here, but the only documentation I could find (at CTAN) is in French, I language I can't read. - According to the
pgf
documentation, thearrows
library has been deprecated in favor ofarrows.meta
. I only usedarrows
because this code was adapted from Geogebra and I still haven't got into changing that particular aspect of the code.
I want to place 2 points G
and H
in the circle such that the chord GH
:
passes through F and- is parallel to AC
so as to obtain something like this (the following was drawn with Geogebra):
I've found answers regarding drawing parallel lines, but in this case it's not just parallel, it also needs to start and end at the circle, so I couldn't find a good answer by myself.
tikz-pgf circles intersections tikz-angles
add a comment |
I've drawn the following diagram with Tikz:
The code used:
documentclassstandalone
usepackagepgf,tikz
usetikzlibrarybabel,calc,arrows,shapes.geometric,intersections,through,backgrounds
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
% draw bullets at each point
foreach point in A,B,C,O,D,E,F
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
About my current code:
- I'm aware of
tkz-euclide
and I think it could help me here, but the only documentation I could find (at CTAN) is in French, I language I can't read. - According to the
pgf
documentation, thearrows
library has been deprecated in favor ofarrows.meta
. I only usedarrows
because this code was adapted from Geogebra and I still haven't got into changing that particular aspect of the code.
I want to place 2 points G
and H
in the circle such that the chord GH
:
passes through F and- is parallel to AC
so as to obtain something like this (the following was drawn with Geogebra):
I've found answers regarding drawing parallel lines, but in this case it's not just parallel, it also needs to start and end at the circle, so I couldn't find a good answer by myself.
tikz-pgf circles intersections tikz-angles
I've drawn the following diagram with Tikz:
The code used:
documentclassstandalone
usepackagepgf,tikz
usetikzlibrarybabel,calc,arrows,shapes.geometric,intersections,through,backgrounds
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
% draw bullets at each point
foreach point in A,B,C,O,D,E,F
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
About my current code:
- I'm aware of
tkz-euclide
and I think it could help me here, but the only documentation I could find (at CTAN) is in French, I language I can't read. - According to the
pgf
documentation, thearrows
library has been deprecated in favor ofarrows.meta
. I only usedarrows
because this code was adapted from Geogebra and I still haven't got into changing that particular aspect of the code.
I want to place 2 points G
and H
in the circle such that the chord GH
:
passes through F and- is parallel to AC
so as to obtain something like this (the following was drawn with Geogebra):
I've found answers regarding drawing parallel lines, but in this case it's not just parallel, it also needs to start and end at the circle, so I couldn't find a good answer by myself.
tikz-pgf circles intersections tikz-angles
tikz-pgf circles intersections tikz-angles
asked May 18 at 20:25
Daniel DinizDaniel Diniz
796
796
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Define an (overlay
such that it does not increase the bounding box) path that has the same slope as A--C
(this is what let p1=($(C)-(A)$),n1=atan2(y1,x1) in
does, which computes the angle of the line) and runs through F
, and compute its intersections with the circle.
documentclass[tikz,border=3.14mm]standalone
usetikzlibraryarrows,calc,shapes.geometric,intersections
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw[name path=circle] (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
path[overlay,name path=line] let p1=($(C)-(A)$),n1=atan2(y1,x1) in % computes the slope of A--C
($(F)+(n1:2*2.25cm)$) -- ($(F)+(180+n1:2*2.25cm)$);
draw[name intersections=of=line and circle,by=G,H] (G) node[above left]$G$
-- (H) node[below right]$H$;
% draw bullets at each point
foreach point in A,...,H
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
Alternatively you may just "parallel transport" the A--C
path, which produces the same result and involves no atan2
.
path[overlay,name path=line]
($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);
thanks a lot! is there by any chance some place that summarizes the job oflet
andp
? I've read about them in the documentation, couldn't understand their purpose and then skipped them. Now that I see they were needed for my problem, I think it's best to stop avoiding them.
– Daniel Diniz
May 18 at 22:04
1
@DanielDinizlet p1=($(C)-(A)$)
stores the vectorA--C
inp1
, after thatx1
will be the x-component of that vector, andy1
its y-component. Thenn1=atan2(y1,x1)
computes the arctan ofy1/x1
, i.e. the angle of the slope. (atan2
also has the information on the quadrant, which is in this very case not needed, but in general good to have.) I also added an alternative that does not make use of thelet ... in
construction.
– marmot
May 18 at 23:56
add a comment |
A PSTricks solution only for fun purposes.
documentclass[pstricks,12pt]standalone
usepackagepst-eucl
begindocument
beginpspicture(-5,-5)(5,5)
pstTriangle[fillcolor=lightgray,fillstyle=solid](4;90)A(4;-150)B(4;-30)C
pstTriangleOCABC
pstGeonode(0,0)O(4;30)D
pstInterLLACODE
pstOIJGeonode[PointName=default,none,PointSymbol=*,none]
(-.5,0)FEDA(-.5,1)T
pstInterLC[PosAngleA=-45,PosAngleB=135]FTODHG
psline(O)(D)
psline(G)(H)
endpspicture
enddocument
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Define an (overlay
such that it does not increase the bounding box) path that has the same slope as A--C
(this is what let p1=($(C)-(A)$),n1=atan2(y1,x1) in
does, which computes the angle of the line) and runs through F
, and compute its intersections with the circle.
documentclass[tikz,border=3.14mm]standalone
usetikzlibraryarrows,calc,shapes.geometric,intersections
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw[name path=circle] (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
path[overlay,name path=line] let p1=($(C)-(A)$),n1=atan2(y1,x1) in % computes the slope of A--C
($(F)+(n1:2*2.25cm)$) -- ($(F)+(180+n1:2*2.25cm)$);
draw[name intersections=of=line and circle,by=G,H] (G) node[above left]$G$
-- (H) node[below right]$H$;
% draw bullets at each point
foreach point in A,...,H
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
Alternatively you may just "parallel transport" the A--C
path, which produces the same result and involves no atan2
.
path[overlay,name path=line]
($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);
thanks a lot! is there by any chance some place that summarizes the job oflet
andp
? I've read about them in the documentation, couldn't understand their purpose and then skipped them. Now that I see they were needed for my problem, I think it's best to stop avoiding them.
– Daniel Diniz
May 18 at 22:04
1
@DanielDinizlet p1=($(C)-(A)$)
stores the vectorA--C
inp1
, after thatx1
will be the x-component of that vector, andy1
its y-component. Thenn1=atan2(y1,x1)
computes the arctan ofy1/x1
, i.e. the angle of the slope. (atan2
also has the information on the quadrant, which is in this very case not needed, but in general good to have.) I also added an alternative that does not make use of thelet ... in
construction.
– marmot
May 18 at 23:56
add a comment |
Define an (overlay
such that it does not increase the bounding box) path that has the same slope as A--C
(this is what let p1=($(C)-(A)$),n1=atan2(y1,x1) in
does, which computes the angle of the line) and runs through F
, and compute its intersections with the circle.
documentclass[tikz,border=3.14mm]standalone
usetikzlibraryarrows,calc,shapes.geometric,intersections
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw[name path=circle] (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
path[overlay,name path=line] let p1=($(C)-(A)$),n1=atan2(y1,x1) in % computes the slope of A--C
($(F)+(n1:2*2.25cm)$) -- ($(F)+(180+n1:2*2.25cm)$);
draw[name intersections=of=line and circle,by=G,H] (G) node[above left]$G$
-- (H) node[below right]$H$;
% draw bullets at each point
foreach point in A,...,H
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
Alternatively you may just "parallel transport" the A--C
path, which produces the same result and involves no atan2
.
path[overlay,name path=line]
($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);
thanks a lot! is there by any chance some place that summarizes the job oflet
andp
? I've read about them in the documentation, couldn't understand their purpose and then skipped them. Now that I see they were needed for my problem, I think it's best to stop avoiding them.
– Daniel Diniz
May 18 at 22:04
1
@DanielDinizlet p1=($(C)-(A)$)
stores the vectorA--C
inp1
, after thatx1
will be the x-component of that vector, andy1
its y-component. Thenn1=atan2(y1,x1)
computes the arctan ofy1/x1
, i.e. the angle of the slope. (atan2
also has the information on the quadrant, which is in this very case not needed, but in general good to have.) I also added an alternative that does not make use of thelet ... in
construction.
– marmot
May 18 at 23:56
add a comment |
Define an (overlay
such that it does not increase the bounding box) path that has the same slope as A--C
(this is what let p1=($(C)-(A)$),n1=atan2(y1,x1) in
does, which computes the angle of the line) and runs through F
, and compute its intersections with the circle.
documentclass[tikz,border=3.14mm]standalone
usetikzlibraryarrows,calc,shapes.geometric,intersections
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw[name path=circle] (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
path[overlay,name path=line] let p1=($(C)-(A)$),n1=atan2(y1,x1) in % computes the slope of A--C
($(F)+(n1:2*2.25cm)$) -- ($(F)+(180+n1:2*2.25cm)$);
draw[name intersections=of=line and circle,by=G,H] (G) node[above left]$G$
-- (H) node[below right]$H$;
% draw bullets at each point
foreach point in A,...,H
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
Alternatively you may just "parallel transport" the A--C
path, which produces the same result and involves no atan2
.
path[overlay,name path=line]
($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);
Define an (overlay
such that it does not increase the bounding box) path that has the same slope as A--C
(this is what let p1=($(C)-(A)$),n1=atan2(y1,x1) in
does, which computes the angle of the line) and runs through F
, and compute its intersections with the circle.
documentclass[tikz,border=3.14mm]standalone
usetikzlibraryarrows,calc,shapes.geometric,intersections
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw[name path=circle] (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
path[overlay,name path=line] let p1=($(C)-(A)$),n1=atan2(y1,x1) in % computes the slope of A--C
($(F)+(n1:2*2.25cm)$) -- ($(F)+(180+n1:2*2.25cm)$);
draw[name intersections=of=line and circle,by=G,H] (G) node[above left]$G$
-- (H) node[below right]$H$;
% draw bullets at each point
foreach point in A,...,H
fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument
Alternatively you may just "parallel transport" the A--C
path, which produces the same result and involves no atan2
.
path[overlay,name path=line]
($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);
edited May 18 at 22:00
answered May 18 at 20:33
marmotmarmot
129k6163310
129k6163310
thanks a lot! is there by any chance some place that summarizes the job oflet
andp
? I've read about them in the documentation, couldn't understand their purpose and then skipped them. Now that I see they were needed for my problem, I think it's best to stop avoiding them.
– Daniel Diniz
May 18 at 22:04
1
@DanielDinizlet p1=($(C)-(A)$)
stores the vectorA--C
inp1
, after thatx1
will be the x-component of that vector, andy1
its y-component. Thenn1=atan2(y1,x1)
computes the arctan ofy1/x1
, i.e. the angle of the slope. (atan2
also has the information on the quadrant, which is in this very case not needed, but in general good to have.) I also added an alternative that does not make use of thelet ... in
construction.
– marmot
May 18 at 23:56
add a comment |
thanks a lot! is there by any chance some place that summarizes the job oflet
andp
? I've read about them in the documentation, couldn't understand their purpose and then skipped them. Now that I see they were needed for my problem, I think it's best to stop avoiding them.
– Daniel Diniz
May 18 at 22:04
1
@DanielDinizlet p1=($(C)-(A)$)
stores the vectorA--C
inp1
, after thatx1
will be the x-component of that vector, andy1
its y-component. Thenn1=atan2(y1,x1)
computes the arctan ofy1/x1
, i.e. the angle of the slope. (atan2
also has the information on the quadrant, which is in this very case not needed, but in general good to have.) I also added an alternative that does not make use of thelet ... in
construction.
– marmot
May 18 at 23:56
thanks a lot! is there by any chance some place that summarizes the job of
let
and p
? I've read about them in the documentation, couldn't understand their purpose and then skipped them. Now that I see they were needed for my problem, I think it's best to stop avoiding them.– Daniel Diniz
May 18 at 22:04
thanks a lot! is there by any chance some place that summarizes the job of
let
and p
? I've read about them in the documentation, couldn't understand their purpose and then skipped them. Now that I see they were needed for my problem, I think it's best to stop avoiding them.– Daniel Diniz
May 18 at 22:04
1
1
@DanielDiniz
let p1=($(C)-(A)$)
stores the vector A--C
in p1
, after that x1
will be the x-component of that vector, and y1
its y-component. Then n1=atan2(y1,x1)
computes the arctan of y1/x1
, i.e. the angle of the slope. (atan2
also has the information on the quadrant, which is in this very case not needed, but in general good to have.) I also added an alternative that does not make use of the let ... in
construction.– marmot
May 18 at 23:56
@DanielDiniz
let p1=($(C)-(A)$)
stores the vector A--C
in p1
, after that x1
will be the x-component of that vector, and y1
its y-component. Then n1=atan2(y1,x1)
computes the arctan of y1/x1
, i.e. the angle of the slope. (atan2
also has the information on the quadrant, which is in this very case not needed, but in general good to have.) I also added an alternative that does not make use of the let ... in
construction.– marmot
May 18 at 23:56
add a comment |
A PSTricks solution only for fun purposes.
documentclass[pstricks,12pt]standalone
usepackagepst-eucl
begindocument
beginpspicture(-5,-5)(5,5)
pstTriangle[fillcolor=lightgray,fillstyle=solid](4;90)A(4;-150)B(4;-30)C
pstTriangleOCABC
pstGeonode(0,0)O(4;30)D
pstInterLLACODE
pstOIJGeonode[PointName=default,none,PointSymbol=*,none]
(-.5,0)FEDA(-.5,1)T
pstInterLC[PosAngleA=-45,PosAngleB=135]FTODHG
psline(O)(D)
psline(G)(H)
endpspicture
enddocument
add a comment |
A PSTricks solution only for fun purposes.
documentclass[pstricks,12pt]standalone
usepackagepst-eucl
begindocument
beginpspicture(-5,-5)(5,5)
pstTriangle[fillcolor=lightgray,fillstyle=solid](4;90)A(4;-150)B(4;-30)C
pstTriangleOCABC
pstGeonode(0,0)O(4;30)D
pstInterLLACODE
pstOIJGeonode[PointName=default,none,PointSymbol=*,none]
(-.5,0)FEDA(-.5,1)T
pstInterLC[PosAngleA=-45,PosAngleB=135]FTODHG
psline(O)(D)
psline(G)(H)
endpspicture
enddocument
add a comment |
A PSTricks solution only for fun purposes.
documentclass[pstricks,12pt]standalone
usepackagepst-eucl
begindocument
beginpspicture(-5,-5)(5,5)
pstTriangle[fillcolor=lightgray,fillstyle=solid](4;90)A(4;-150)B(4;-30)C
pstTriangleOCABC
pstGeonode(0,0)O(4;30)D
pstInterLLACODE
pstOIJGeonode[PointName=default,none,PointSymbol=*,none]
(-.5,0)FEDA(-.5,1)T
pstInterLC[PosAngleA=-45,PosAngleB=135]FTODHG
psline(O)(D)
psline(G)(H)
endpspicture
enddocument
A PSTricks solution only for fun purposes.
documentclass[pstricks,12pt]standalone
usepackagepst-eucl
begindocument
beginpspicture(-5,-5)(5,5)
pstTriangle[fillcolor=lightgray,fillstyle=solid](4;90)A(4;-150)B(4;-30)C
pstTriangleOCABC
pstGeonode(0,0)O(4;30)D
pstInterLLACODE
pstOIJGeonode[PointName=default,none,PointSymbol=*,none]
(-.5,0)FEDA(-.5,1)T
pstInterLC[PosAngleA=-45,PosAngleB=135]FTODHG
psline(O)(D)
psline(G)(H)
endpspicture
enddocument
answered May 19 at 10:59
Artificial Odorless ArmpitArtificial Odorless Armpit
5,51811344
5,51811344
add a comment |
add a comment |
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