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How can I draw overlapping triangles?

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.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!

tikz-pgf geometry shapes

share|improve this question

edited 2 hours ago

LianTze Lim

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asked 8 hours ago

marmarmarmarmarmar

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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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• 
  
  
  

  
  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
  

  
  
  
  

add a comment
 | 

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!

tikz-pgf geometry shapes

share|improve this question

edited 2 hours ago

LianTze Lim

9,76022 gold badges3434 silver badges7575 bronze badges

asked 8 hours ago

marmarmarmarmarmar

1533 bronze badges

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
  

  
  
  
  

add a comment
 | 

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

edited 2 hours ago

LianTze Lim

9,76022 gold badges3434 silver badges7575 bronze badges

asked 8 hours ago

marmarmarmarmarmar

1533 bronze badges

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

edited 2 hours ago

LianTze Lim

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asked 8 hours ago

marmarmarmarmarmar

1533 bronze badges

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

edited 2 hours ago

LianTze Lim

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asked 8 hours ago

marmarmarmarmarmar

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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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edited 2 hours ago

LianTze Lim

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edited 2 hours ago

LianTze Lim

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asked 8 hours ago

marmarmarmarmarmar

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asked 8 hours ago

marmarmarmarmarmar

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2 Answers
2

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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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Schrödinger's catSchrödinger's cat

7,2501111 silver badges2222 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Bless your heart THANK YOU
  
  – marmarmar
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

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

share|improve this answer

answered 2 hours ago

ZarkoZarko

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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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Schrödinger's catSchrödinger's cat

7,2501111 silver badges2222 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Bless your heart THANK YOU
  
  – marmarmar
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Schrödinger's catSchrödinger's cat

7,2501111 silver badges2222 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Bless your heart THANK YOU
  
  – marmarmar
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Schrödinger's catSchrödinger's cat

7,2501111 silver badges2222 bronze badges

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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Schrödinger's catSchrödinger's cat

7,2501111 silver badges2222 bronze badges

answered 8 hours ago

Schrödinger's catSchrödinger's cat

7,2501111 silver badges2222 bronze badges

answered 8 hours ago

Schrödinger's catSchrödinger's cat

7,2501111 silver badges2222 bronze badges

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

share|improve this answer

answered 2 hours ago

ZarkoZarko

147k88 gold badges8484 silver badges194194 bronze badges

add a comment
 | 

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

share|improve this answer

answered 2 hours ago

ZarkoZarko

147k88 gold badges8484 silver badges194194 bronze badges

add a comment
 | 

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

share|improve this answer

answered 2 hours ago

ZarkoZarko

147k88 gold badges8484 silver badges194194 bronze badges

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

share|improve this answer

answered 2 hours ago

ZarkoZarko

147k88 gold badges8484 silver badges194194 bronze badges

answered 2 hours ago

ZarkoZarko

147k88 gold badges8484 silver badges194194 bronze badges

answered 2 hours ago

ZarkoZarko

147k88 gold badges8484 silver badges194194 bronze badges