In $\square A B C D$, $A B=8$, the circumradius of $\triangle A B D$ is 5. Find the maximum value of the diagonal $A C$.

The original text has been translated into English, preserving the original formatting and line breaks.
As shown in Figure 4, let the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $P$.
Then $BP = DP$, $AC = 2AP$.
Let the circumcenter of $\triangle ABD$ be $O$, and connect $OB$, $OP$.
Since $P$ is the midpoint of $BD$, we know $\angle OPB = 90^\circ$.
Thus, the locus of point $P$ is the circle $\odot T$ with $OB$ as its diameter (excluding points $B$ and $E$, where $E$ is the midpoint of $AB$).

Therefore, to find the maximum value of $AP$, we only need to find the longest distance from point $A$ to a point on $\odot T$.
Connect $AT$, $TP$, and extend $AT$ to intersect $\odot T$ at point $S$.
It is easy to see that $AP \leq AT + TP = AT + TS = AS$.
Thus, the maximum value of $AP$ is the length of $AS$.
Connect $OE$. Then $OE \perp AB$.
Draw $TF \perp AB$, with the foot of the perpendicular at $F$.
In the right triangle $\triangle OEB$, by the Pythagorean theorem, we have
$OE = \sqrt{OB^2 - BE^2} = 3$.
Since $TF \parallel OE$ and $OT = BT$, we know
$TF = \frac{1}{2} OE = \frac{3}{2}$, $EF = \frac{1}{2} BE = 2$.
In the right triangle $\triangle ATF$, by the Pythagorean theorem, we have
$AT = \sqrt{AF^2 + TF^2} = \frac{3\sqrt{17}}{2}$.
Thus, $AP_{\text{max}} = AS = AT + TS = \frac{3\sqrt{17} + 5}{2}$.
Therefore, $AC_{\max} = 2AP_{\max} = 3\sqrt{17} + 5$.