Trapezium in a circle

According to $Ptolemy's thorem$ here: $AC \cdot BD = AB \cdot CD + AD \cdot BC$ This trapizium is in a circle. Then $AC = BD, AD = BC$ . Therefore, $169 = 88 + AD \cdot AD$ $AD = 9$ Therefore, $\boxed{AD + BC = 9+9 = 18}$

Since the trapezoid is cyclic, it must be isosceles: we have both $\angle A + \angle D = 180^{\circ}$ and $\angle A + \angle C = 180^{\circ},$ leading us to $\angle C = \angle D.$ Thus, $AD = BC,$ and $AD + BC = 2AD.$

Drop a perpendicular from $A$ onto $DC,$ and let the foot of that perpendicular be $X.$ We can easily get $DX = \dfrac{3}{2}$ and $XC = \dfrac{19}{2}$ from the given information. We can apply the Pythagorean Theorem on both right triangles $AXD$ and $AXC$ to find $AD$ :

$\begin{aligned} AD &= \sqrt{AX^2 + DX^2} \\ &= \sqrt{AC^2 - XC^2 + DX^2} \\ &= \sqrt{13^2 - \left ( \dfrac{19}{2} \right )^2 + \left ( \dfrac{3}{2} \right )^2} \\ &= \sqrt{169 - \dfrac{361}{4} + \dfrac{9}{4}} \\ &= \sqrt{169 - 88} \\ &= \sqrt{81} \\ &= 9. \end{aligned}$

Therefore, $AD + BC = 2(9) = \boxed{18}.$