$\newcommand{\eD}{\overset{\text{D}}\to}
\newcommand{\D}{\overset{\text{D}}=}$
As was noted on the linked MO page Expected value ..., the gaps $G_1,\dots,G_{n-1}$ between the adjacent points are jointly distributed as $\frac{H_1}{H_1+\dots+H_{n+1}},\dots,\frac{H_{n-1}}{H_1+\dots+H_{n+1}}$, where the $H_i$'s are iid standard exponential random variables (r.v.'s); see e.g. Theorem 6.6(c).
So,
\begin{equation*}
B_n\D M_n:=\frac{H_{n-1:n-1}}{S_{n+1}}=\frac1{S_{n+1}}\,\max_{i\le n-1}H_i,
\end{equation*}
where $\D$ means the equality in distribution, and
$S_{n+1}:=H_1+\dots+H_{n+1}$.
Next, for any real $x$ and all large enough natural $n$,
\begin{multline*}
P(H_{n-1:n-1}-\ln n<x)=P(\max_{i\le n-1}H_i<x+\ln n)
=P(H_1<x+\ln n)^{n-1} \\
=(1-e^{-x-\ln n})^{n-1}
\to e^{-e^{-x}}=P(Y<x)
\end{multline*}
for some r.v. $Y$, so that
\begin{equation*}
Y_n:=H_{n-1:n-1}-\ln n\eD Y,
\end{equation*}
where $\eD$ means the convergence in distribution.
Also, by the strong law of large numbers (SLLN) $\frac n{S_{n+1}}\to1$ almost surely and hence in distribution.
So,
\begin{equation*}
\frac n{\ln n}\,B_n\D\frac n{\ln n}\,M_n=\frac n{\ln n}\,\frac{H_{n-1:n-1}}{S_{n+1}}
=\frac{Y_n+\ln n}{\ln n}\,\frac n{S_{n+1}}\eD1. \tag{1}
\end{equation*}
So, by the Fatou lemma,
\begin{equation*}
\liminf_n\frac n{\ln n}\,EB_n\ge1.
\end{equation*}
On the other hand,
\begin{equation*}
A_n\le G_1,
\end{equation*}
and $G_1$ has the beta distribution with parameters $1,n$. So,
\begin{equation*}
EA_n\le EG_1=\frac1{n+1}.
\end{equation*}
So,
\begin{equation*}
\liminf_n\frac{EB_n}{EA_n}\ge\lim_n\frac{\ln n}n\,(n+1)=\infty.
\end{equation*}
Thus, it is confirmed that $EB_n/EA_n\to\infty$.
Also,
\begin{equation}
\frac{B_n}{A_n}\ge \frac{B_n}{G_1}
\D\frac{nB_n}{H_1}\,\frac{S_{n+1}}n\eD\infty,
\end{equation}
because, by (1), $nB_n\eD\infty$ and, by the SLLN, $\frac{S_{n+1}}n\eD1$.
Thus, $\frac{B_n}{A_n}\eD\infty$ and hence indeed $med(B_n/A_n)\to\infty$.
