(1) Angle MCD= Angle CAD+ Angle CDA = Angle CBD+ Angle CBA= Angle ABD.

Angle MCD= Angle BCD= Angle DAB (the equiarc equiangle theorem is used above).

So angle DAB= angle ABD, so AD=BD.

(2)AD=BD, OA = OB = OD, so triangle Addo and triangle BDO are congruent (edge theorem); So Angel Addo = Angel ADO

Extend the intersection AB of DE to F, and the angle FBE= angle EBD (equal arc and equal angle), so ef/de = BF/DB-( 1).

AB=6, so BF=3 AD=BD=3 radical 10, then DF=(90-9) radical.

Answer later. I am going out now. 88