Let $\{W(t)\}_{0 \le t \le 1}$ a Brownian motion. Then $\{B(t)\}_{0 \le t \le 1}$ with $B(t)=W(t)-tW(1)$ is a Brownian brigde. My goal is to prove that $B(t)$ and $W(1)$ are independent. Since their joint distribution is Gaussian, it is enough to see that their covariance is 0. I realize that
$$Cov(B(t),W(1))=E(B(t)W(1))$$
but I'm no sure about how to continue.
We have that $$E(B(t)W(1))=E \left( (W(t)-tW(1))W(1) \right) =E(W(t)W(1)-tW(1)W(1)) $$ Using the linearity of the expectation we also have that this expression is equal to $$ E(W(t)W(1))-E(tW(1)W(1))=E(W(t)W(1))-tE(W(1)W(1))$$ As @PetiteEtincelle said, $E(W(t)W(s))=\min(t,s)$, so finally we conclude that $$ E(B(t)W(1))= \min (t,1)-t\min(1,1)=t-t=0$$