Perimeters!

Let $\theta_{1} = \angle{OBD}$ and $\theta_{2} = \angle{OBE} \implies \theta_{1} + \theta_{2} = 2\theta_{1}$ , where

$\cos(\theta_{1}) = \dfrac{a}{\sqrt{a^2 + c^2}} \implies \cos(\theta) = \cos(2\theta_{1}) = 2(\dfrac{a^2}{a^2 + c^2}) - 1 = \dfrac{a^2 - c^2}{a^2 + c^2}$

Using law of cosines we obtain:

$(b + x)^2 = (a + x)^2 + (a + b)^2 - 2(a + x)(a + b)(\dfrac{a^2 - c^2}{a^2 + c^2}) \implies$

$b^2 + 2bx + x^2 = a^2 + 2ax + x^2 + a^2 + 2ab + b^2 - 2(a + x)(a + b)(\dfrac{a^2 - c^2}{a^2 + c^2}) \implies$

$(a - b)x + a(a + b) = \dfrac{(a + x)(a + b)(a^2 - c^2)}{a^2 + c^2}$

Let $j = a^2 - c^2$ and $m = a^2 + c^2 \implies$

$m(a - b)x + ma(a + b) = ja(a + b) + j(a + b)x \implies$

$(m(a - b) - j(a + b))x = a(a + b)(j - m) \implies$

$(ab - c^2)x = (a + b)c^2 \implies x = \dfrac{(a + b)c^2}{ab - c^2} \implies$

$AC = x + b = \dfrac{a(b^2 + c^2)}{ab - c^2}, AB = x + a = \dfrac{b(a^2 + c^2)}{ab - c^2}$

and $BC = a + b \implies P = BC + AC + AB = \dfrac{2ab(a + b)}{ab - c^2} \implies$ $\dfrac{(ab - c^2)P}{ab(a + b)} = \boxed{2}$ .

The circle center $O$ is also the incenter of $\triangle ABC$ . $OA$ bisects $\angle A$ and leaving two congruent right triangles $\triangle AEO$ and $\triangle AFO$ , therefore $AE=AF=d$ , similarly, $OB$ and $OC$ bisect $\angle B$ and $\angle C$ respectively. Therefore, the perimeter $P = 2(a+b+d)$ . To find $P$ , we need only to find $d$ .

$\begin{aligned} d & = \frac c{\tan \frac A2} = \frac c{\tan \left(90^\circ - \frac B2 - \frac C2\right)} \\ & = \frac c{\cot \left(\frac B2 + \frac C2 \right)} = c \tan \left(\frac B2 + \frac C2\right) \\ & = \frac {c\left(\tan \frac B2+\tan \frac C2\right)}{1-\tan \frac B2 \tan \frac C2} & \small \blue{\text{Note that }\tan \frac B2 = \frac ca \text{ and }\tan \frac C2 = \frac cb} \\ & = \frac {c\left(\frac ca+\frac cb\right)}{1-\frac {c^2}{ab}} = \frac {c^2(a+b)}{ab-c^2} \end{aligned}$

Therefore,

$\begin{aligned} P & = 2(a+b+d) = 2\left(a+b+\frac {c^2(a+b)}{ab-c^2}\right) = 2\left(\frac {ab(a+b)}{ab-c^2}\right) \\ \implies \frac {(ab-c^2)P}{ab(a+b)} & = \boxed 2 \end{aligned}$

Let $|\overline {AE}|=|\overline {AF}|=d$ . Then $P=2s=2(a+b+d)\implies s=a+b+d\implies$ area of $\triangle {ABC}=\triangle =\sqrt {abd(a+b+d)}$ .

Therefore $c=\dfrac{\triangle }{s}=\sqrt {\frac{abd}{a+b+d}}\implies ab-c^2=\dfrac{ab(a+b)}{a+b+d}$ .

Hence $\dfrac{ab-c^2}{ab(a+b)}=\dfrac{1}{a+b+d}=\dfrac{2}{P}$ .

Therefore $\dfrac{(ab-c^2)P}{ab(a+b)}=\boxed 2$ .