It is easy to compute the inner angle of a regular octagon is $135^{\circ}$. As what we need to compute is the ratio, let's assume $AB=1$. Then by the law of cosines, we find $$AC^2 = AB^2 + BC^2 - 2 AB \cdot BC \cos 135^{\circ} = 2+\sqrt{2} $$

Because $ACEG$ is a square, therefore its area is $$2+\sqrt{2}$$

The area of $ABCDEFGH$ is simply the area of $ACEG$ plus four congruent isosceles triangles' area each of which equals $$\frac{1}{2}\times 1\times 1\times\sin 135^{\circ}=\frac{\sqrt{2}}{4}$$

Therefore, the desired answer is $$\frac{2+\sqrt{2}}{4\times \frac{\sqrt{2}}{4}+(2+\sqrt{2})}=\boxed{\frac{\sqrt{2}}{2}}$$