6. $3<A D \leqslant 9$

Solution: Clearly, the vertex of the parabola is $A_{0}(1,9)$, and the axis of symmetry is $x=1$.
$\because$ The parabola intersects the $x$-axis at points
$$
B(-2,0), C(4,0) \text {. }
$$

and passes through points $P(1-2 \sqrt{2}, 1)$,
$$
Q(1+2 \sqrt{2}, 1) \text {. }
$$
$\therefore$ Construct circles $\odot D$ and $\odot E$ with diameters $BC$ and $DA_{0}$, respectively. The two circles intersect the parabola at points $P$ and $Q$ (as shown in the figure).
$\because$ The angle subtended by a diameter at the circumference is a right angle, an angle outside the circle is acute, and an angle inside the circle is obtuse.
$\therefore$ When point $A$ is on the parabolic arc $\overparen{P A_{0} Q}$ excluding the endpoints, $\angle B A C<90^{\circ}$,
$$
\begin{array}{l}
\text { Also, } \because \quad D P=D Q=3, D A_{0}=9, \\
\therefore \quad 3<A \leqslant 9 .
\end{array}
$$
6. Given the parabola $y=-x^{2}+2 x+8$ intersects the $x$-axis at points $B$ and $C$, point $D$ bisects $BC$, if point $A$ on the upper side of the $x$-axis is a moving point on the parabola, and $\angle BAC$ is an acute angle, then the range of $AD$ is $\qquad$