8. Given that $A$, $B$, and $C$ are three points on $\odot O$, and
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \text {. }
$$

Then the dot product $\overrightarrow{A B} \cdot \overrightarrow{A C}=$
8. 0 .

Let $AD$ be the diameter of $\odot O$, and connect $DB$, $DC$. Then, from the given equation, we have $\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{AC}$.

Therefore, quadrilateral $ABDC$ is an inscribed parallelogram in the circle, which means it is a rectangle, $\angle BAC=90^{\circ}$, and $\overrightarrow{AB} \cdot \overrightarrow{AC}=0$.