In what follows $B$ denotes Brownian motion and $M$ its running maximum.
I have two random variables, $\eta_t$ and $\xi_t$, defined as
\begin{align} \eta_t &= \sup\{s \leq t: B_s = 0\} \\ \xi_t &= \sup\{s \leq t: M_s - B_s = 0\} \end{align}
I am trying to prove that $\eta_t \stackrel{d}{=} \xi_t$.
I know that $M_s - B_s \stackrel{d}{=} \lvert B_s\rvert$. If their fdds were also the same, then I would be done but I doubt that is the case. On the other hand, I don't have proof of this either.
I tried applying the time reversal property of $B$, the Markov property of both $B$ and $M-B$ and many other things but I haven't been able to arrive at a proof.
It is the case that $\{M_s-B_s: s\ge 0\}$ has the same distribution as $\{|B_s|: s\ge 0\}$. This result is due to P. Lévy. See p. 97 of the second edition of Brownian Motion and Stochastic Calculus by Karatzas and Shreve.