As shown in Figure 7, since $C D / / A F$ and $A B$ intersects $A F$, extending $A B$ and $D C$ will definitely intersect at point $Q$; similarly, we get intersection points $P$ and $S$. Let $X$, $Y$, and $Z$ be the three points of tangency. It is easy to see that
$\triangle P A F \backsim \triangle B Q C$
$\sim \triangle E D S \backsim \triangle P Q S$.
Their perimeters are denoted as $m_{1}$, $m_{2}$, $m_{3}$, and $m$. It is easy to prove that
$$\begin{array}{l} m_{1}+m_{2}+m_{3} \\ =(P X+P Z)+(Q X+Q Y)+(S Y+S Z) \\ =P Q+Q S+S P=m . \end{array}$$

At this point, $1=\frac{m_{1}}{m}+\frac{m_{2}}{m}+\frac{m_{3}}{m}=\frac{A F}{Q S}+\frac{Q C}{Q S}+\frac{D S}{Q S}$, then
$$\frac{A F}{Q S}=1-\frac{Q C+D S}{Q S}=1-\frac{Q S-C D}{Q S}=\frac{C D}{Q S} \text {. }$$

Therefore, $A F=C D$.
Since $A F / / C D$, $A D$ and $C F$ must bisect each other.
Similarly, we can prove that $A D$ and $B E$ bisect each other, and $B E$ and $C F$ bisect each other. Thus, $A D$, $B E$, and $C F$ are concurrent. Let this point be $O$. Clearly, point $O$ is the center of the circle and the symmetry center of the convex hexagon $A B C D E F$.