Semicircle inscribed in a Quadrilateral

$\triangle{AEO} \cong \triangle{OGD}$ and $2(m + n + \theta) = 180^{\circ} \implies m + n + \theta = 90^{\circ}$ or $m + n = 90 - \theta$ .

$\triangle{OAB} \sim \triangle{OBC} \implies \dfrac{a}{\overline{OB}} = \dfrac{\overline{OB}}{a + b - 2x} \implies \overline{OB}^2 = a(a + b - 2x)$

and

$\triangle{ODC} \sim \triangle{OBC} \implies \dfrac{b}{\overline{OC}} = \dfrac{\overline{OC}}{a + b - 2x} \implies \overline{OC}^2 = b(a + b - 2x)$

Using the law of cosines on $\triangle{OBC} \implies$

$(a + b - 2x)^2 = (a + b - 2x)(a + b - 2\sqrt{ab}\cos(m + n)) \implies$

$a + b - 2x = a + b - 2\sqrt{ab}\cos(m + n) \implies x = \sqrt{ab}\cos(m + n) = \sqrt{ab}\sin(\theta)$

$= z\sin(\theta) \implies z = \sqrt{ab} \implies \overline{AD} = 2\sqrt{ab} \implies \dfrac{\overline{AD}}{\sqrt{ab}} = \boxed{2}$ .