Let $G_1$ and $G_2$ be topological groups. Assume that there exists a continuous homomorphism $f : G_1 \rightarrow G_2$ which (ignoring the group structure) is a homotopy equivalence. If $BG_i$ is a classifying space for $G_i$, then we get an induced map $f_{\ast} : BG_1 \rightarrow BG_2$ which is a homotopy equivalence.

The fact that $f_{\ast}$ is a homotopy equivalence is clear if you construct $BG_i$ via Milnor's construction of classifying spaces. However, the existence of $BG_i$ is also an easy consequence of the (unpointed) Brown representability theorem. This leads me to my question : is there an "abstract nonsense" proof that $f_{\ast}$ is a homotopy equivalence which avoids having to know an explicit description of $BG_i$?

EDIT : As Fernando Muro pointed out, the desired result is equivalent to the fact that if $EG \rightarrow BG$ is the universal $G$-bundle over $BG$, then the space $EG$ is contractible. Of course, this is true, but I don't see how to prove it without knowing an explicit construction of $BG$. My definition of $BG$ and the universal bundle $EG \rightarrow BG$ is that $BG$ is a space such that for any CW complex $X$, the set of principal $G$-bundles on $X$ is naturally in bijection with $[X,BG]$ via the map that takes $\phi : X \rightarrow BG$ to $\phi^{\ast}(EG)$. Is there a way to see that $EG$ has to be contractible without knowing a construction of $BG$?