Intersection in the square

In $\triangle A_1BC$ and $\triangle D_1AB$ we have,

$A_1B = D_1A \newline \angle A_1BC = \angle D_1AB \newline BC = AB$

So, $\triangle A_1BC \cong D_1AB$

Let $\angle CA_1B = \angle AD_1B = \alpha$

Then we have

$\angle D_2BC = \angle AD_1B = \alpha\hspace{20pt} [$ Vertically opposite angles $]$

Also, $\angle A_1CB = 90\degree - \angle CA_1B = 90\degree - \alpha$

So we get, $\angle BGC = 90\degree$ . So $\triangle BGC$ is right angled triangle.

We have,

$\displaystyle\frac{A_1A_2}{A_2C} = \frac{A_1A}{AB} = \frac{\sqrt{2}-1}{1}$

$\displaystyle \Rightarrow A_1A_2 = (\sqrt{2} - 1)A_2C$

Now $A_1C = A_1A_2 + A_2C = (\sqrt{2}-1)A_2C + A_2C = \sqrt{2}A_2C\hspace{20pt}\cdots\text{Eq. 1}$

Let the side of square $ABCD$ be $a$ . Then $A_1B = \sqrt{2}a \Rightarrow A_1C = \sqrt{3}a$ . Now using the above equation we get,

$\displaystyle A_2C = \frac{\sqrt{3}}{\sqrt{2}}a\hspace{20pt}\cdots\text{Eq. 2}$

We have $\triangle A_1BC \sim \triangle BGC$ , so

$\displaystyle\frac{GC}{BC} = \frac{BC}{A_1C} = \frac{a}{\sqrt{3}a}$

$\displaystyle\Rightarrow GC = \frac{1}{\sqrt{3}}a$

Using Eq. 2 and the above equation we get

$\displaystyle A_2G = A_2C - GC\newline \Rightarrow A_2G = \frac{\sqrt{3}}{\sqrt{2}}a - \frac{1}{\sqrt{3}}a\newline \Rightarrow A_2G = \frac{3-\sqrt{2}}{\sqrt{6}}a$

So,

$\displaystyle \frac{A_2G}{GC} = \frac{\frac{3-\sqrt{2}}{\sqrt{6}}a}{\frac{1}{\sqrt{3}}a} = \frac{3 - \sqrt{2}}{\sqrt{2}} = \frac{3}{\sqrt{2}} - 1$

So, $a = 3\,,\,b = 2\,,\,c = 1$

Hence, $a - b - c = 3 - 2 - 1 = \boxed{0}$