So AEB angle =90 degrees,

So AED angle +BEC angle =90 degrees,

Because of the tangent of o and e,

So Angel ·AED = Angel Abel,

Because CE=CB,

So angel ·BEC = angel ·EBC,

So angle ABE+ angle EBC=90 degrees,

Namely: angle ABC=90 degrees,

AB is the diameter of circle o,

So BC is the tangent of circle O.

(2) Solution: Because AD AD, BC are tangents of circle O, and AB is the diameter of circle O,

So AD//BC, angle ADE+ angle BCE= 180 degrees,

Because ad, de and BC are all tangents of circle o,

So OD bisection angle ADE, OC bisection angle BCE,

So angle ADO+ angle BCO=90 degrees,

Because angle ADO+ angle AOD=90 degrees,

So Angel ·BCO = Angel ·AOD,

Same angle BOC= angle AOD,

The triangle AOD is similar to the triangle BOC,

So AD/OB=OA/BC,

Because the diameter AB=2 and the root number is 5,

So the radius OA=OB= root number 5,

Because AD=2,

So 2/ root number 5= root number 5/BC

BC=5/2。