(1) It is proved that the intersection O is OG perpendicular to AD in G, and OH perpendicular to BC in H, connecting OD and OC respectively.

So OGD angle =90 degrees.

Angle OHC=90 degrees

OG, oh oh, respectively, ad AD, BC.

So AG=DG= 1/2AD.

BH=CH= 1/2BC

So triangle OGD and triangle OHC are right triangles.

Because OE divides AEC in two.

So angle OEA= angle OEC= 1/2 angle AEC.

Because OE=OE

So triangle OEG and triangle OEH are congruent (AAS)

So OG = oh

Because OD=OC

So the right triangle OGD congruent right triangle OHC (HL)

So DG=CH

So AD=BC

So arc ABD= arc CDB.

So arc AB= arc CD

So AB=CD

(2) Solution: Because AD is perpendicular to CB.

So the angle AEC=90 degrees

Because angle OEA= angle OEC= 1/2 angle AEC (authentication)

So the angle OEA=45 degrees

Because OGD angle =90 degrees (proved)

So the triangle OGD is a right triangle.

So od 2 = DG 2+og 2.

Because angle OGD+ angle OEA+ angle GOE= 180 degrees.

So the angle θ= 45 degrees

So angle GOE= angle OEA=45 degrees.

So EG=OG

Because the radius of circle O is 5

So OD=5

Because DG=DE+EG

DE= 1

So EG=3

DG=4

Because DG=AG= 1/2AD

So AD=8

So the length of AD is 8.