Everywhere there's circles!

For red circle with radius $r$ :

For $\overline{AC}: y = x$ and $m_{\overline{ED}} = -2 \implies y = -2x + 2a \implies x = y = \dfrac{2a}{3}$

$\implies G(\dfrac{2a}{3},\dfrac{2a}{3}) \implies \overline{AG} = \dfrac{2\sqrt{2}}{3}a$ and $\overline{GD} = \dfrac{\sqrt{5}}{3}a$

$\implies A_{\triangle{AGD}} = \dfrac{1}{2}(a)(\dfrac{2a}{3}) = \dfrac{a^2}{3} =$ $\dfrac{ar}{2}(\dfrac{3 + 2\sqrt{2} + \sqrt{5}}{3}) \implies$

$2a^2 - (3 + 2\sqrt{2} + \sqrt{5})ar = 0 \implies a(2a - (3 + 2\sqrt{2} + \sqrt{5})r) = 0$ and

$a \neq 0 \implies\boxed{r = \dfrac{2a}{\sqrt{5} + 2\sqrt{2} + 3}}$

For green circle: with radius $r_{2}$ .

$h_{\triangle{EGC}} = a - \dfrac{2a}{2} = \dfrac{a}{3}, \overline{EG} = \dfrac{\sqrt{5}}{6}a$ and $\overline{GC} = \dfrac{\sqrt{2}}{3}a \implies$

$A_{\triangle{EGC}} = \dfrac{1}{2}(\dfrac{a}{2})(\dfrac{a}{3}) = \dfrac{a^2}{12} =$ $\dfrac{1}{2}(a r_{2})(\dfrac{\sqrt{5} + 2\sqrt{2} + 3}{6}) \implies$

$a(a - (\sqrt{5} + 2\sqrt{2} + 3)r_{2}) = 0$ and $a \neq 0 \implies \boxed{r_{2} = \dfrac{a}{\sqrt{5} + 2\sqrt{2} + 3}}$

Similarly using the same method for the pink circle with radius $r_{3}$ we have:

$\boxed{r_{3} = \dfrac{a}{\sqrt{5} + \sqrt{2} + 3}}$

Let $\boxed{R_{1} = r = \dfrac{2a}{\sqrt{5} + 2\sqrt{2} + 3}}$

$\overline{Ow_{0}} = \sqrt{4 + 2\sqrt{2}}R_{1}$

$\triangle{OA_{1}w_{0}} \sim \triangle{w_{1}A_{2}w_{0}} \implies \dfrac{\sqrt{4 + 2\sqrt{2}}R_{1}}{R_{1}} = \dfrac{R_{1} + R_{2}}{R_{1} - R_{2}} \implies$

$(\sqrt{4 + 2\sqrt{2}} - 1)R_{1} = (\sqrt{4 + 2\sqrt{2}} + 1)R_{2} \implies$ $R_{2} = \dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1}R_{1}$

$R_{3} = \dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1}R_{2} = (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2R_{1}$

In General: $R_{n} = (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^{n - 1}R_{1}$

$\implies A_{n} = \pi R_{n}^2 = \pi R_{1}^2((\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2)^{n - 1}$

$\implies A = \sum_{n = 1}^{\infty} A_{n} = \pi R_{1}^2\sum_{n = 1}^{\infty} ((\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2)^{n - 1}$

$= \dfrac{(\sqrt{4 + 2\sqrt{2}} + 1)^2}{4\sqrt{4 + 2\sqrt{2}}}\pi R_{1}^2 \implies$

$\dfrac{A}{A_{ABCD}} = \dfrac{(\sqrt{4 + 2\sqrt{2}} + 1)^2}{\sqrt{4 + 2\sqrt{2}}(\sqrt{5} + 2\sqrt{2} + 3)^2} \approx \boxed{0.2413244723253244}$

and $\dfrac{A^{*}}{A_{ABCD}} = (r_{2}^2 + r_{3}^2)\pi = ((\dfrac{1}{\sqrt{5} + 2\sqrt{2} + 3})^2 + (\dfrac{1}{\sqrt{5} + \sqrt{2} + 3})^2)\pi$

$\approx \boxed{0.1193399593228952}$

$\implies \dfrac{S}{A_{ABCD}} = \dfrac{A + A^{*}}{A_{ABCD}} \approx \boxed{0.3606644316482196}$