证明： (1)$\because AP$、$BP$分别切$\odot O$于点$A$、$B，$
$\therefore \angle PAO = \angle PBO = 90^\circ。$
$\because OP = OP，$$OA = OB，$
$\therefore Rt\triangle PAO \cong Rt\triangle PBO(HL)。$
$\therefore \angle AOP = \angle BOP，$即$\angle AOP = \frac{1}{2}\angle AOB。$
$\because \overset{\frown}{AB} = \overset{\frown}{AB}，$
$\therefore \angle ADB = \frac{1}{2}\angle AOB。$
$\therefore \angle ADB = \angle AOP。$
(2)过点$O$作$OH \perp DE$于点$H，$则$BH = DH。$
在$Rt\triangle OAP$中，$AP = 10，$$\tan \angle AOP = \frac{1}{2}，$
$\therefore OA = 2AP = 20。$
$\because \angle PAO = 90^\circ，$$C$为$OP$的中点，
$\therefore PC = OC = AC。$
$\therefore \angle AOP = \angle OAC。$
由(1)，得
$\angle AOP = \angle ADB = \angle BOP，$
$\therefore \angle OAC = \angle ADB。$
$\therefore DE // OA。$
$\therefore \angle AOP = \angle E = \angle BOP。$
$\therefore BE = OB = 20，$$\tan E = \tan \angle AOP = \frac{1}{2}。$
在$Rt\triangle EHO$中，$\frac{OH}{EH} = \frac{1}{2}。$
设$OH = k，$则$EH = 2k，$
$\therefore BH = 2k - 20。$
在$Rt\triangle BHO$中，$BH^2 + OH^2 = OB^2，$
即$(2k - 20)^2 + k^2 = 20^2，$解得$k = 16$（$k = 0$舍去）。
$\therefore BH = 12。$
$\therefore BD = 2BH = 24。$
$\therefore DE = BD + BE = 44。$
