(1)证明:$\because AB$是$\odot O$的直径，$\therefore \angle ADB = 90^{\circ}，$即$AD\perp BD.$又$\because AB = AC，$$\therefore BD = CD$
(2)解:$\because AB = AC，$$\therefore \angle B = \angle C.$$\because \angle B = \angle E，$$\therefore \angle E = \angle C.$$\therefore CD = DE = 4.$由
(1)，知$AD\perp BC.$$\because CD = 4，$$AC = AB = 5，$$\therefore$在$Rt\triangle ACD$中，由勾股定理，得$AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{5^{2}-4^{2}} = 3$
