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How can I draw overlapping triangles?


How can I draw several polygons on the same base line using TikZ?Triangles inside Triangles, Fractals to an arbitrary depthTikz: How to indicate/display overlapping linesWriting from middle of the page, upside down and from the right hand side?How can I draw two triangles in a row?Is it possible to delete all “data-curves” which are outside from my diagram? (only the overlapping part)How to draw overlapping nodesRemoving lines in overlapping triangles in TikZFilling shape with horizontal linesHow do I draw the following triangles in latex?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2















I need a non-equilateral triangle ABC and an equilateral triangle DEF to be overlapping ABC (with all 6 lines showing). Lines AD, BE, and CF need to be perpendicular bisectors (cut in the middle) of the sides of the triangle DEF. Would really appreciate any help with this thank you!










share|improve this question









New contributor



marmarmar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2





    Hi, welcome. Can you add a hand-drawn drawing of the desired result?

    – AndréC
    8 hours ago











  • someone got it spot on in the comments but will do in the future!

    – marmarmar
    7 hours ago

















2















I need a non-equilateral triangle ABC and an equilateral triangle DEF to be overlapping ABC (with all 6 lines showing). Lines AD, BE, and CF need to be perpendicular bisectors (cut in the middle) of the sides of the triangle DEF. Would really appreciate any help with this thank you!










share|improve this question









New contributor



marmarmar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2





    Hi, welcome. Can you add a hand-drawn drawing of the desired result?

    – AndréC
    8 hours ago











  • someone got it spot on in the comments but will do in the future!

    – marmarmar
    7 hours ago













2












2








2








I need a non-equilateral triangle ABC and an equilateral triangle DEF to be overlapping ABC (with all 6 lines showing). Lines AD, BE, and CF need to be perpendicular bisectors (cut in the middle) of the sides of the triangle DEF. Would really appreciate any help with this thank you!










share|improve this question









New contributor



marmarmar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I need a non-equilateral triangle ABC and an equilateral triangle DEF to be overlapping ABC (with all 6 lines showing). Lines AD, BE, and CF need to be perpendicular bisectors (cut in the middle) of the sides of the triangle DEF. Would really appreciate any help with this thank you!







tikz-pgf geometry shapes






share|improve this question









New contributor



marmarmar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|improve this question









New contributor



marmarmar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|improve this question




share|improve this question








edited 2 hours ago









LianTze Lim

9,7602 gold badges34 silver badges75 bronze badges




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marmarmar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago









marmarmarmarmarmar

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153 bronze badges




New contributor



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marmarmar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 2





    Hi, welcome. Can you add a hand-drawn drawing of the desired result?

    – AndréC
    8 hours ago











  • someone got it spot on in the comments but will do in the future!

    – marmarmar
    7 hours ago












  • 2





    Hi, welcome. Can you add a hand-drawn drawing of the desired result?

    – AndréC
    8 hours ago











  • someone got it spot on in the comments but will do in the future!

    – marmarmar
    7 hours ago







2




2





Hi, welcome. Can you add a hand-drawn drawing of the desired result?

– AndréC
8 hours ago





Hi, welcome. Can you add a hand-drawn drawing of the desired result?

– AndréC
8 hours ago













someone got it spot on in the comments but will do in the future!

– marmarmar
7 hours ago





someone got it spot on in the comments but will do in the future!

– marmarmar
7 hours ago










2 Answers
2






active

oldest

votes


















4
















The fact that A, B, and C should lie on the bisectors makes their location unique up to a real parameter for each of them, for which I chose 1.2, 1.4 and 1.6, respectively.



documentclass[tikz,border=3mm]standalone
begindocument
begintikzpicture[declare function=a=2;]
draw (-30:a) coordinate[label=-30:$D$] (D) --
(90:a) coordinate[label=90:$E$] (E) --
(210:a) coordinate[label=210:$F$] (F) -- cycle;
draw (-30-60:1.4*a) coordinate[label=-30-60:$B$] (B) --
(90-60:1.6*a) coordinate[label=90-60:$C$] (C) --
(210-60:1.2*a) coordinate[label=210-60:$A$] (A) -- cycle;
draw[dashed] (A) -- (D) (B) -- (E) (C) -- (F);
endtikzpicture
enddocument


enter image description here






share|improve this answer






















  • 1





    Bless your heart THANK YOU

    – marmarmar
    7 hours ago



















0
















Using calc TikZ library:



documentclass[tikz,border=3mm]standalone
usetikzlibrarycalc
begindocument
begintikzpicture[declare function=a=2;,
every label/.style = circle, inner sep=0pt
]
draw (0,0) coordinate[label=210:$A$] (A) --
(a,0) coordinate[label=330:$B$] (B) --
(60:a) coordinate[label= 90:$C$] (C) -- cycle;
draw ($(A)!0.5!(B)!a/4!90:(A)$) coordinate[label=270:$D$] (D) --
($(B)!0.5!(C)!a/3!90:(B)$) coordinate[label= 30:$E$] (E) --
($(C)!0.5!(A)!a/2!90:(C)$) coordinate[label=120:$F$] (F) -- cycle;
draw[densely dashed,very thin]
(A) -- (E) (B) -- (F) (C) -- (D);
endtikzpicture
enddocument


enter image description here






share|improve this answer



























    Your Answer








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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4
















    The fact that A, B, and C should lie on the bisectors makes their location unique up to a real parameter for each of them, for which I chose 1.2, 1.4 and 1.6, respectively.



    documentclass[tikz,border=3mm]standalone
    begindocument
    begintikzpicture[declare function=a=2;]
    draw (-30:a) coordinate[label=-30:$D$] (D) --
    (90:a) coordinate[label=90:$E$] (E) --
    (210:a) coordinate[label=210:$F$] (F) -- cycle;
    draw (-30-60:1.4*a) coordinate[label=-30-60:$B$] (B) --
    (90-60:1.6*a) coordinate[label=90-60:$C$] (C) --
    (210-60:1.2*a) coordinate[label=210-60:$A$] (A) -- cycle;
    draw[dashed] (A) -- (D) (B) -- (E) (C) -- (F);
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer






















    • 1





      Bless your heart THANK YOU

      – marmarmar
      7 hours ago
















    4
















    The fact that A, B, and C should lie on the bisectors makes their location unique up to a real parameter for each of them, for which I chose 1.2, 1.4 and 1.6, respectively.



    documentclass[tikz,border=3mm]standalone
    begindocument
    begintikzpicture[declare function=a=2;]
    draw (-30:a) coordinate[label=-30:$D$] (D) --
    (90:a) coordinate[label=90:$E$] (E) --
    (210:a) coordinate[label=210:$F$] (F) -- cycle;
    draw (-30-60:1.4*a) coordinate[label=-30-60:$B$] (B) --
    (90-60:1.6*a) coordinate[label=90-60:$C$] (C) --
    (210-60:1.2*a) coordinate[label=210-60:$A$] (A) -- cycle;
    draw[dashed] (A) -- (D) (B) -- (E) (C) -- (F);
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer






















    • 1





      Bless your heart THANK YOU

      – marmarmar
      7 hours ago














    4














    4










    4









    The fact that A, B, and C should lie on the bisectors makes their location unique up to a real parameter for each of them, for which I chose 1.2, 1.4 and 1.6, respectively.



    documentclass[tikz,border=3mm]standalone
    begindocument
    begintikzpicture[declare function=a=2;]
    draw (-30:a) coordinate[label=-30:$D$] (D) --
    (90:a) coordinate[label=90:$E$] (E) --
    (210:a) coordinate[label=210:$F$] (F) -- cycle;
    draw (-30-60:1.4*a) coordinate[label=-30-60:$B$] (B) --
    (90-60:1.6*a) coordinate[label=90-60:$C$] (C) --
    (210-60:1.2*a) coordinate[label=210-60:$A$] (A) -- cycle;
    draw[dashed] (A) -- (D) (B) -- (E) (C) -- (F);
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer















    The fact that A, B, and C should lie on the bisectors makes their location unique up to a real parameter for each of them, for which I chose 1.2, 1.4 and 1.6, respectively.



    documentclass[tikz,border=3mm]standalone
    begindocument
    begintikzpicture[declare function=a=2;]
    draw (-30:a) coordinate[label=-30:$D$] (D) --
    (90:a) coordinate[label=90:$E$] (E) --
    (210:a) coordinate[label=210:$F$] (F) -- cycle;
    draw (-30-60:1.4*a) coordinate[label=-30-60:$B$] (B) --
    (90-60:1.6*a) coordinate[label=90-60:$C$] (C) --
    (210-60:1.2*a) coordinate[label=210-60:$A$] (A) -- cycle;
    draw[dashed] (A) -- (D) (B) -- (E) (C) -- (F);
    endtikzpicture
    enddocument


    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 7 hours ago

























    answered 8 hours ago









    Schrödinger's catSchrödinger's cat

    7,25011 silver badges22 bronze badges




    7,25011 silver badges22 bronze badges










    • 1





      Bless your heart THANK YOU

      – marmarmar
      7 hours ago













    • 1





      Bless your heart THANK YOU

      – marmarmar
      7 hours ago








    1




    1





    Bless your heart THANK YOU

    – marmarmar
    7 hours ago






    Bless your heart THANK YOU

    – marmarmar
    7 hours ago














    0
















    Using calc TikZ library:



    documentclass[tikz,border=3mm]standalone
    usetikzlibrarycalc
    begindocument
    begintikzpicture[declare function=a=2;,
    every label/.style = circle, inner sep=0pt
    ]
    draw (0,0) coordinate[label=210:$A$] (A) --
    (a,0) coordinate[label=330:$B$] (B) --
    (60:a) coordinate[label= 90:$C$] (C) -- cycle;
    draw ($(A)!0.5!(B)!a/4!90:(A)$) coordinate[label=270:$D$] (D) --
    ($(B)!0.5!(C)!a/3!90:(B)$) coordinate[label= 30:$E$] (E) --
    ($(C)!0.5!(A)!a/2!90:(C)$) coordinate[label=120:$F$] (F) -- cycle;
    draw[densely dashed,very thin]
    (A) -- (E) (B) -- (F) (C) -- (D);
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer





























      0
















      Using calc TikZ library:



      documentclass[tikz,border=3mm]standalone
      usetikzlibrarycalc
      begindocument
      begintikzpicture[declare function=a=2;,
      every label/.style = circle, inner sep=0pt
      ]
      draw (0,0) coordinate[label=210:$A$] (A) --
      (a,0) coordinate[label=330:$B$] (B) --
      (60:a) coordinate[label= 90:$C$] (C) -- cycle;
      draw ($(A)!0.5!(B)!a/4!90:(A)$) coordinate[label=270:$D$] (D) --
      ($(B)!0.5!(C)!a/3!90:(B)$) coordinate[label= 30:$E$] (E) --
      ($(C)!0.5!(A)!a/2!90:(C)$) coordinate[label=120:$F$] (F) -- cycle;
      draw[densely dashed,very thin]
      (A) -- (E) (B) -- (F) (C) -- (D);
      endtikzpicture
      enddocument


      enter image description here






      share|improve this answer



























        0














        0










        0









        Using calc TikZ library:



        documentclass[tikz,border=3mm]standalone
        usetikzlibrarycalc
        begindocument
        begintikzpicture[declare function=a=2;,
        every label/.style = circle, inner sep=0pt
        ]
        draw (0,0) coordinate[label=210:$A$] (A) --
        (a,0) coordinate[label=330:$B$] (B) --
        (60:a) coordinate[label= 90:$C$] (C) -- cycle;
        draw ($(A)!0.5!(B)!a/4!90:(A)$) coordinate[label=270:$D$] (D) --
        ($(B)!0.5!(C)!a/3!90:(B)$) coordinate[label= 30:$E$] (E) --
        ($(C)!0.5!(A)!a/2!90:(C)$) coordinate[label=120:$F$] (F) -- cycle;
        draw[densely dashed,very thin]
        (A) -- (E) (B) -- (F) (C) -- (D);
        endtikzpicture
        enddocument


        enter image description here






        share|improve this answer













        Using calc TikZ library:



        documentclass[tikz,border=3mm]standalone
        usetikzlibrarycalc
        begindocument
        begintikzpicture[declare function=a=2;,
        every label/.style = circle, inner sep=0pt
        ]
        draw (0,0) coordinate[label=210:$A$] (A) --
        (a,0) coordinate[label=330:$B$] (B) --
        (60:a) coordinate[label= 90:$C$] (C) -- cycle;
        draw ($(A)!0.5!(B)!a/4!90:(A)$) coordinate[label=270:$D$] (D) --
        ($(B)!0.5!(C)!a/3!90:(B)$) coordinate[label= 30:$E$] (E) --
        ($(C)!0.5!(A)!a/2!90:(C)$) coordinate[label=120:$F$] (F) -- cycle;
        draw[densely dashed,very thin]
        (A) -- (E) (B) -- (F) (C) -- (D);
        endtikzpicture
        enddocument


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        ZarkoZarko

        147k8 gold badges84 silver badges194 bronze badges




        147k8 gold badges84 silver badges194 bronze badges
























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