This question is probably not research level that's why I asked it previously on MSE a week ago. Unfortunately it doesn't get much attention there and I thought I would try it here.

Choose a arbitrary discrete group $G$. The classifying space $BG$ of $G$ is classically constructed by forming a certain contractable $\Delta$-complex $EG$ (on concrete construction of $EG$: see below) endowed with an action by $G$. $BG$ is obtained by taking quotient $BG:= EG/G$.

The concrete construction of the $\Delta$-complex $EG$ works as follows: The $n$-simplices of $EG$ are the ordered tuples

$$[g_0,g_1,...,g_n] \cong \Delta_n =\left\{x\in \mathbb {R} ^{n}:x=\sum _{i=0}^{n}t_{i}v_{i}\ {\text{with}}\ 0\leq t_{i}\leq 1\ {\text{and}}\ \sum _{i=0}^{n}t_{i}=1 \right\}$$

with $g_i \in G$. The $v_i$ are spanning $\Delta_n$. We obtain a $\Delta$-complex by attaching $n$-simplices to the $(n − 1)$ simplices $[g_0,g_1,..., \hat{g}_i,...,g_n]$ in standard way as a standard simplex attaches to its faces. Here $\hat{g}_i$ means that this vertex is deleted.

**Question.** Does there exist a *constructive* way to show that $EG$ is contractable. By *constructive* I mean how to construct an explicite homotopy $h_t: EG \times I \to EG$ which contracts $EG$ to a point.
That is $h_0(EG) =EG, h_1(EG) = \{*\}$. How looks it concretely geometrically?

I know some abstract arguments like Whitehead's theorem that also provide a reason for contrability of $EG$ but this is not what I'm looking for.

I tried to define such homotopy as follows: let $p \in [g_0,...,g_n]$. Then $h_t$ "slides" step by step $p$ along $(n+1)$-simplex $[e,g_0,...,g_n]$ to $[e]$ ($e \in G$ the identity element). What I mean by "step by step along $[e,g_0,...,g_n]$"?

If we use again the identification $[g_0,g_1,...,g_n] \cong \Delta_n$ then as long as we sitting "inside" $\Delta_n$ we can interpret the $g_i$ as spanning vectors $v_i$. Let $p = \sum _{i=0}^{n}t_{i}g_{i}$. As $[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$ we can interpret $[g_0,g_1,...,g_n]$ as $x = t_{-1}e + \sum _{i=0}^{n}t_{i}g_{i} \in \Delta_{n+1}$ with $t_{-1}=0$. That is the "point" $p_e:= 1 \cdot e \in [e, g_0,g_1,...,g_n]$ is not contained in $[g_0,g_1,...,g_n]$ and we can define a unique line $l_pe$ which contains $p_e$ and is perpendicular to $[g_0,g_1,...,g_n]$ in our geometric picture $[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$ corresponding to $\Delta_n \subset \Delta_{n+1}$.

This line uniquely intersects $[g_0,g_1,...,g_n]$ in a unique point $p_l$. Then we say that our homotopy slide $p$ along the unique line through $p $ and $p_l$ up to the first contact with a boundary of $[g_0,g_1,...,g_n]$.

Let this boundary be $[g_0,..., \hat{g}_i,...,g_n]$. Then we play the same game with $[g_0,..., \hat{g}_i,...,g_n]$ and $[e, g_0,..., \hat{g}_i,...,g_n]$.

Does this approach work? And is there known a more conventional "textbook" (that I still haven't found) way for the construction of $EG$?