What's the Value?

$\triangle{ABD} \sim \triangle{EFD} \implies m = \dfrac{\overline{BD}}{y} \implies \overline{BD} = my$

Using the Pythagorean theorem on $\triangle{ABD} \implies m^2y^2 = m^2r^2 + (r + y)^2 \implies$

$(m^2 - 1)y^2 - 2ry - (m^2 + 1)r^2 = 0 \implies y = \dfrac{m^2 + 1}{m^2 - 1}r \implies$

$\overline{AD} = \dfrac{2m^2}{m^2 - 1}r \implies \overline{AG} = \overline{AD} + r = \dfrac{3m^2 - 1}{m^2 - 1}r$

$\triangle{ABC} \sim \triangle{GHC} \implies m = \dfrac{\overline{BC}}{x} \implies \overline{BC} = mx$

Using the Pythagorean theorem on $\triangle{ABC} \implies m^2x^2 = m^2r^2 + (\dfrac{3m^2 - 1}{m^2 - 1}r + x)^2$

$\implies (m^2 - 1)^3x^2 - 2(3m^2 - 1)(m^2 - 1)rx - ((3m^2 - 1)^2 + m^2(m^2 - 1)^2)r^2 = 0 \implies$

$x = \dfrac{3m^2 - 1 + m^2\sqrt{m^4 + 6m^2 - 3}}{(m^2 - 1)^2}r$ dropping the negative root

$\implies \overline{AC} = \overline{AG} + x = \dfrac{3m^2 - 1 + \sqrt{m^4 + 6m^2 - 3}}{(m^2 - 1)^2}m^2r$

$\implies \dfrac{AC}{r} = \dfrac{3m^2 - 1 + \sqrt{m^4 + 6m^2 - 3}}{(m^2 - 1)^2}m^2 = \dfrac{5m^2}{(m^2 - 1)^2}$

$\implies (*) 3m^2 - 1 + \sqrt{m^4 + 6m^2 - 3} = 5 \implies m^4 + 6m^2 - 3 = 36 - 36m^2 + 9m^4$

$\implies$

$8m^2 - 42m^2 + 39 = 0 \implies m_{1} = \dfrac{\sqrt{2}\sqrt{21 - \sqrt{129}}}{4}$ and $m_{2} = \dfrac{\sqrt{2}\sqrt{21 + \sqrt{129}}}{4}$

dropping the negative roots and since $m_{1} > 1$ satisfies $(*) \implies$

$m = \dfrac{\sqrt{2}\sqrt{21 - \sqrt{129}}}{4} = \dfrac{\sqrt{a}\sqrt{b - \sqrt{c}}}{d} \implies$

$a + b + c + d = \boxed{156}$ .

Notice letting $d(m) = \dfrac{AC}{r} \implies \lim_{m\to\infty} d(m) =$

$\lim_{m\to\infty} \dfrac{3m^2 - 1 + \sqrt{m^4 + 6m^2 - 3}}{m^2(1 - \dfrac{2}{m^2} + \dfrac{1}{m^4})} =$

$\lim_{m\to\infty} \dfrac{3 - \dfrac{1}{m^2} + \sqrt{1 + \dfrac{6}{m^2} - \dfrac{3}{m^4}}}{1 - \dfrac{2}{m^2} + \dfrac{1}{m^4}} = \dfrac{4}{1} = 4$