The ratio of the area of two regular polygons The polygons in the figure below are all regular polygons(regular heptagon),   share a vertex and the orange line crosses the three vertices of the two regular polygons, the area of the small regular polygon and the large regular polygon is denoted as $S_1$, $S_2$,  what is $\frac{S_1}{S_2}$?

Additional question (regular nine-sided polygon)



Won't go through the calculation, but this is the idea.
First since $\triangle ADE$ and $\triangle BDF$ are similar, we know $AE$ pass through $G$.
Now we can calculate $DG$,$GC$,$AG$ based on the left heptagon and since $AD\parallel CE$ we can calculate $GE=GC\cdot {AD\over DG}$. Also we know $\angle DGE=180^{\circ}-\angle AGD={5\over 7}180^{\circ}$.
Therefore $DE^2=DG^2+GE^2-2\cos({5\over 7}180^{\circ})DG\cdot GE$.
If you let $a=DG,b=DA,c=DB$, there are some identity here
Using the identity, $\cos({5\over 7}180^{\circ})=-{a^2+c^2-b^2\over 2ac}=-{a+b\over 2c}$
New edit:
Actually just realized $\angle GEB=\angle GAD=\angle GBE$ so $GE$ is actually just $b$.
Now the calculation is really simple:
$$ED^2=a^2+b^2+ab\cdot{(a+b)\over c}$$
$$=a^2+b^2+{bc(c-b)+c(c+a)(c-b)\over c}$$
$$=a^2+b^2+bc-b^2+c^2+ac-bc-ab$$
$$=a^2+c^2+ac-ab$$
$$=a^2+c^2+b^2-a^2-c^2+b^2$$
$$=2b^2$$
So the area is exactly twice.
