Solution: (1) Connect BD. In the right triangle ABC, it is easy to know that AC=5, ∠ BDC = ∠ ADB = 90, …(2 points).

So ∠BDC=∠ABC, and because ∠C=∠C, so △ABC∽Rt△BDC,

So CDBC=BCAC, so CD = bc2ac = 95...(5 points)

(ii) When point E is the midpoint of BC, ED is tangent to ⊙O; It is proved that connecting OD and \de is the center line of Rt△BDC. ∴ed=eb,∴∠ebd=∠edb； ∵ob=od,∴∠obd=∠odb； ∴∠ode=∠odb+∠bde=∠obd+∠ebd=∠abc=90； ∴ED⊥OD, ∴ED and ⊙O are tangent.