# Brownian motion and sup of a Brownian motion

I am stuck with the following problem: let $B_t$ be a standard Brownian motion and let $S_{t}:=\sup_{0 \leq s \leq t} B_s$. Prove that for every $\lambda \geq 0$ and $\mu \leq \lambda$, $\mathbb{P}(B_t \leq \mu, S_t \geq \lambda)=\mathbb{P}(B_t \geq 2 \lambda-\mu)$.

I know that for any $a>0,\ \mathbb{P}(S_t \geq a)=2 \mathbb{P}(B_t \geq a)$ but I do not know how I can use this fact.

Thank you very much !!