conditional expectation of brownian motion Let $(B_t)_{t\geq 0}$ be a standard Brownian motion in $\mathbb{R}^d$. It is intuitive that, for fixed $s<t<u$
$$\mathbb{E}[B_t\mid \sigma(B_s,B_u)]=B_s+\frac{t-s}{u-s}(B_u-B_s).$$
However, I cannot think of a way to show this rigorously. 
If first attempted to take $A\in\sigma(B_s,B_u)$ and show that $\mathbb{E}[1_A B_t]=\mathbb{E}[1_A(B_s+\frac{t-s}{u-s}(B_u-B_s))]$. But I cannot manage to show this equality. 
I'd be very thankful for any ideas and suggestions on how to tackle this problem. 

Actually, it just might be that OP is the next Paul Lévy, as the result holds!
Unfortunately, I am far from being as bright as Lévy, so I'll apply the same kind of tricks that I used in this answer.
Set
$$
Z_t=B_t-\frac{u-t}{u-s}B_s-\frac{t-s}{u-s}B_u,
$$
which you can check is a Gaussian random variable independent of $B_s$ and $B_u$.
It follows that
\begin{align*}
E[B_t\,|\,\sigma(B_s,B_u)]
&=E[Z_t\,|\,\sigma(B_s,B_u)]+\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\
&=E[Z_t]+\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\
&=\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\
&=B_s+\frac{t-s}{u-s}\left(B_u-B_s\right).
\end{align*}
