Example 8 As shown in Figure 12, a square $B C E F$ is constructed on the hypotenuse $B C$ of $\mathrm{Rt} \triangle A B C$ on the same side as $\triangle A B C$. Let the center of the square be $O$, and connect $A O$. If $A B=4, A O=6 \sqrt{2}$, then the length of $A C$ is ( )
(A) 12
(B) 16
(C) $4 \sqrt{3}$
(D) $8 \sqrt{2}$
(2007, Zhejiang Province Junior High School Mathematics Competition)
Analysis 1: Take a point $G$ on $AC$ such that $CG = AB = 4$.
Since $\angle BAC = \angle BOC = 90^{\circ}$, points $A$, $O$, $C$, and $B$ are concyclic. Therefore, $\angle ABO = \angle ACO$.
Because $BO = CO$, $\triangle BAO \cong \triangle CGO$.
From $\angle AOB = \angle GOC$, we know $\angle AOG = 90^{\circ}$.
Since $AO = OG$, $AG = 12$. Therefore, $AC = 16$.
Analysis 2: If we think of constructing a string diagram from the original figure (as shown in Figure 13), it is easy to see that point $O$ is also the center of the smaller square. Thus, $AO$ should be half the diagonal of the smaller square. Hence, its side length is
12. Therefore, $AC = 12 + AB = 16$.