As shown in the figure, let $C G=t, A G=16-t$.
Let $\angle C B G=\theta, \angle A C B=\alpha$.
Then $\angle C A B=\theta$ (eq. chords eq. $\angle$ s)
Then $\triangle B C G \sim \triangle A C B$ (equiangular)
$t: 8=8: 16$ (ratio of sides, $\sim \Delta \mathrm{s}$ )
$t=4$
It is easy to see that $\triangle A D G \sim \triangle B C G$ (equiangular)
$$\begin{array}{l} (16-t): y=x: t \text { (ratio of sides, } \sim \Delta \mathrm{s} \text { ) } \\ (16-4) \times 4=x y \\ x y=48 \end{array}$$

Assume that $x$ and $y$ are integers, then possible pairs of $(x, y)$ are $(1,48),(2,24), \ldots .,(6,8), \ldots,(48,1)$.
Using triangle inequality $x+t>8$ and $8+t>x$ in $\triangle B C G$, the only possible combinations are:
$$\begin{array}{l} (x, y)=(6,8) \text { or }(8,6) \\ c=x+y=14 \end{array}$$