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!

  • 2
    Hi, welcome. Can you add a hand-drawn drawing of the desired result? – AndréC Sep 15 at 18:52
  • someone got it spot on in the comments but will do in the future! – marmarmar Sep 15 at 20:10

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.

\begin{tikzpicture}[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);

enter image description here

  • 1
    Bless your heart THANK YOU – marmarmar Sep 15 at 19:46

Using calc TikZ library:

\begin{tikzpicture}[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);

enter image description here

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