∴∠BAO=30，∠AOB = 90 °, AC = 2ao

∴AO=AB？ cos∠BAO=2√3×cos30 =3

∴AC=6.

(2) Prove: connecting DE, the intersection point D is DF⊥BC, and the vertical foot is point F.

∵ quadrilateral ABCD is a diamond, ∴BD bisects ∠ABC.

⊙D is tangent to the edge AB at point E of ∴DE⊥AB.

∵DF⊥BC

∴DF=DE

∴⊙D is also tangent to the BC edge.