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 $IH=\sqrt{A{I}^2+A{H}^2}=\sqrt{2}$. 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.