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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













5















I've drawn the following diagram with Tikz:
current output



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, the arrows library has been deprecated in favor of arrows.meta. I only used arrows 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):
desired output



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.










share|improve this question


























    5















    I've drawn the following diagram with Tikz:
    current output



    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, the arrows library has been deprecated in favor of arrows.meta. I only used arrows 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):
    desired output



    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.










    share|improve this question
























      5












      5








      5


      0






      I've drawn the following diagram with Tikz:
      current output



      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, the arrows library has been deprecated in favor of arrows.meta. I only used arrows 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):
      desired output



      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.










      share|improve this question














      I've drawn the following diagram with Tikz:
      current output



      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, the arrows library has been deprecated in favor of arrows.meta. I only used arrows 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):
      desired output



      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






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked May 18 at 20:25









      Daniel DinizDaniel Diniz

      796




      796




















          2 Answers
          2






          active

          oldest

          votes


















          7














          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


          enter image description here



          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)$)$);





          share|improve this answer

























          • 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





            @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


















          1














          A PSTricks solution only for fun purposes.



          enter image description here



          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





          share|improve this answer























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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7














            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


            enter image description here



            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)$)$);





            share|improve this answer

























            • 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





              @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















            7














            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


            enter image description here



            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)$)$);





            share|improve this answer

























            • 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





              @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













            7












            7








            7







            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


            enter image description here



            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)$)$);





            share|improve this answer















            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


            enter image description here



            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)$)$);






            share|improve this answer














            share|improve this answer



            share|improve this answer








            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 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





              @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

















            • 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





              @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
















            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











            1














            A PSTricks solution only for fun purposes.



            enter image description here



            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





            share|improve this answer



























              1














              A PSTricks solution only for fun purposes.



              enter image description here



              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





              share|improve this answer

























                1












                1








                1







                A PSTricks solution only for fun purposes.



                enter image description here



                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





                share|improve this answer













                A PSTricks solution only for fun purposes.



                enter image description here



                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






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered May 19 at 10:59









                Artificial Odorless ArmpitArtificial Odorless Armpit

                5,51811344




                5,51811344



























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