Semicircle in quadrilateral

$\triangle EAO \cong \triangle GAO$ , so let $\angle EAO = \angle GAO = a$ . Similarly, $\triangle FDO \cong \triangle GDO$ , so let $\angle FDO = \angle GDO = b$ . In addition, $\triangle OBE \cong \triangle OCF$ , so let $\angle OBE = \angle OCF = c$ . Finally, let $OB = OC = \frac{x}{2}$ , so we are solving for $x$ . Looking at the angle measures of $ABCD$ :

$2a + 2b + c + c = 360^{\circ}$ $a + b + c = 180^{\circ}$

By considering the angles of $\triangle AOB$ and $\triangle ODC$ , we find that $\angle AOB = b$ and $\angle COD = a$ , so $\triangle AOB \sim \triangle ODC$ . Therefore:

$\frac{OC}{AB} = \frac{DC}{OB}$ $\frac{\frac{x}{2}}{4} = \frac{5}{\frac{x}{2}}$ $x = \boxed{4\sqrt{5}}$