∴ vertical diameter theorem: CP=DP

Point P is the midpoint of OA, that is, AP = OP =1/2oa =1/4ab =1.

CP=DP,OA⊥CD

∴ Quadrilateral ACOD is a diamond (a quadrilateral whose diagonal is bisected vertically is a diamond)

2.∫AB is the diameter, then ∠ ACB = ∠ BPC = 90.

∠ABC=∠CBP

∴△ABC∽△CBP

∴BC/AB=PB/BC

That is, the square of BC =AB×BP=4×3= 12? (OP= 1, OB=2, and then BP=3)

BC=2√3

3.∵ Quadrilateral ACOD is a diamond.

∴AC∥OD, that is, cf∨ed.

AD=AC, then connect BD, arc AC= arc AD, and get ∠ABC=∠ABD=∠OCB.

∠ADO=∠ACO

∫DF is the tangent of a circle, then∠ ADF =∠ Abd =∠ ABC =∠ OCB.

∴∠fdo=∠adf+∠ado=∠ocb+∠aco=∠acb=90

∫CF∫ED，∠FCE=∠ACB=∠FDO=∠FDE=90

That is ∠ FCE = ∠ FDE = 90 degrees.

Quadrilateral CEDF is a rectangle.

So ∠ CEO = 90, which means OE⊥BC.

OC=OB

∴CE= 1/2BC=√3

Pythagorean theorem: OE= 1

∴DE=OD+OE=2+ 1=3

∴S quadrilateral CEDF=CE×DE=√3×3=3√3