Japan Junior Mathematical Olympiad

$$\boxed{\frac{1 + \sqrt{2}}{2}}$$
Since the straight lines $CH$ and $AD$ are symmetrically positioned with respect to the straight line $BF$, the lines $CH$, $AD$, $BF$ intersect at a single point. Therefore, the $3$ points $C$, $I$, $H$ lie on the same straight line. Since the diagonals $DH$ and $CG$ are diameters of the circum-circle of the octagon, we have $\angle DAH = \angle CHG = 90^\circ$. Since $BC \parallel AD$ and $BA \parallel CH$ are satisfied, the quadrilateral $ABCI$ is a parallelogram, and hence we get $AI = BC = 1$. From $\angle IAH = 90^\circ$ we get $IH = \sqrt{AI^2 + AH^2} = \sqrt{2}$. Consequently, the area of the triangle $AIH$ is $\frac{1}{2} \cdot AI \cdot AH = \frac{1}{2}$,

and since $\angle IHG = 90^\circ$, the area of the triangle $HIG = \frac{1}{2} \cdot HI \cdot HG = \frac{\sqrt{2}}{2}$. We then conclude that the area of the quadrilateral $AIGH = \frac{1}{2} + \frac{\sqrt{2}}{2} = \frac{1+\sqrt{2}}{2}$.

Alternate Solution:
Since $AG\parallel BF$, the areas of the triangles $AIG$ and $AFG$ are the same. Hence it is enough to determine the area of the quadrilateral $AFGH$.
Let $J$ be the point of intersection of the straight lines $AH$ and $FG$. Then we have $\angle JHG = \angle JGH = 180^\circ - 135^\circ = 45^\circ$, and we see that the triangle $JHG$ is a right isosceles triangle with $\angle GJH = 90^\circ$. Consequently, we have $JH = JG = \frac{\sqrt{2}}{2}$.
Putting together what we obtained above, we get that the area of the triangle $JAF = \frac{1}{2} \cdot JA \cdot JF = \frac{1}{2} \cdot (1 + \frac{\sqrt{2}}{2})^2 = \frac{3+2\sqrt{2}}{4}$, and the area of the triangle $JHG = \frac{1}{2} \cdot JH \cdot JG = \frac{1}{2} (\frac{\sqrt{2}}{2})^2 = \frac{1}{4}$. And finally, the area of the quadrilateral $AFGH = \frac{3+2\sqrt{2}}{4} - \frac{1}{4} = \frac{1+\sqrt{2}}{2}$, which is the desired answer.