Equilateral triangle commutative diagram

Question

Here is my LaTeX for a commutative diagram using tikz-cd:

\documentclass{article}
\usepackage{tikz-cd}

\begin{document}

\begin{tikzcd}[column sep=small]
& & Y & &     \\
& A \arrow[ur] \arrow[to=3-1]  & D \arrow[r] \arrow[l] \arrow[d] & C \arrow[ul] \arrow[to=3-5] &    \\
Z & & B \arrow[to=3-1] \arrow[to=3-5] & & X
\end{tikzcd}

\end{document}

Is there a way to make the outer triangle equilateral, such that $A$, $B$, and $C$ lie in the middle of the edges, and $D$ lies in the centre of the triangle? Ideally, the arrows from $D$ to $A$, $B$, and $C$ would be perpendicular to the edges of the triangle.

Answer 1

The tikz-cd manual has an example of a pentagon on p. 13, in which it places the nodes with TikZ methods and uses the styles of the cd library for the rest.

This can be easily adapted to your case.

\documentclass{article}
\usepackage{tikz-cd}

\begin{document}
\begin{tikzpicture}[commutative diagrams/every diagram,
    declare function={R=2;Rs=R*cos(60);}]
 \path 
  (150:Rs) node(A) {$A$} 
  (270:Rs) node(B) {$B$} 
  (30:Rs) node(C) {$C$} 
  (0,0)  node(D) {$D$} 
  (-30:R) node (X) {$X$} 
  (90:R) node (Y) {$Y$}
  (210:R) node (Z) {$Z$};
 \path[commutative diagrams/.cd, every arrow, every label]
 (A) edge (Y) edge(Z)
 (B) edge (Z) edge(X)
 (C) edge (X) edge(Y)
 (D) foreach \X in {A,B,C} {edge (\X)}; 
\end{tikzpicture}
\end{document}