6. Let $O$ be the circumcenter of $\triangle A B C$, and it satisfies
$$
\overrightarrow{O A}+\overrightarrow{O B}=\overrightarrow{O C} \text {. }
$$

Then $\angle A C B=$ $\qquad$
6. $\frac{2 \pi}{3}$.

From $\overrightarrow{O A}+\overrightarrow{O B}=\overrightarrow{O C}$, we know that quadrilateral $O A C B$ is a parallelogram.

Also, $|O A|=|O B|=|O C|$, so, $\square O A C B$ is a rhombus, and $\angle O A C=\angle O B C=\frac{\pi}{3}$.
Thus, $\angle A C B=\frac{2 \pi}{3}$.