$解:\because 四边形{ABCD}内接于{\odot O}，{\angle A= 90^{{\circ} }}，$
$\therefore {\angle C= 180^{{\circ} }-\angle A= 90^{{\circ} }}，{\angle ABC+ \angle ADC= 180^{{\circ} }}．$
$作{AE\perp BC}于{E}，{DF\perp AE}于{F}，$
$则{CDFE}是矩形，{EF= CD= 10}． $
$在{ \rm{Rt} \triangle AEB}中，∵{\angle AEB= 90^{{\circ} }}，{AB= 17}，{\cos \angle ABC= \frac{3}{5}}，$
$\therefore {BE= AB\cdot \cos \angle ABE= \frac{51}{5}}，$
$\therefore {AE= \sqrt{A{B}^{2} - B{E}^{2}}= \frac{68}{5}}，$
$\therefore {AF= AE-EF= \frac{68}{5}-10= \frac{18}{5}}．$
$\because {\angle ABC+ \angle ADC= 180^{{\circ} }}，{\angle CDF= 90^{{\circ} }}，$
$\therefore {\angle ABC+ \angle ADF= 90^{{\circ} }}，$
$\because {\cos \angle ABC= \frac{3}{5}}，$
$\therefore {\sin \angle ADF= \cos \angle ABC= \frac{3}{5}}．$
$在{ \rm{Rt} \triangle ADF}中，\because \angle AFD={90}^{\circ }，{\sin \angle ADF= \frac{3}{5}}$
$\therefore {AD= \frac{AF}{\sin \angle ADF}= \frac{\frac{18}{5}}{\frac{3}{5}}= 6}$
