(1), since AD passes through the center O and AD is perpendicular to AC, AP vertically divides BC, BD=CD and AB=AC.

Angle ADB= angle CDP=90 degrees, angle ABD= angle CAD, (isosceles triangle with three lines in one), angle DOC+ angle OCD=90 degrees,

Because, PC//AB, so, angle P= angle BAD, angle ECB=2, angle P=2, angle BAD=2, angle CAD,

Connect OC, because OC=OA, so angle OAC= angle OCA, angle DOC= angle OAC= angle OCA=2 angle CAD.

So, angle ECB= angle DOC, so, angle OCE= angle ECB+ angle OCD= angle DOC+ angle OCD=90 degrees,

So EC is perpendicular to CO, and EC is the tangent of circle O.

(2)BD=CD=3, triangle ABD is all equal to triangle ACD, AB=PC=3√3, AD=PD=√(3√3)? +3? =6,

Suppose AP intersects the circle O at f, AF=2OC=2R, and AD*(2R-AD)=DC? ，6*(2R-6)=9，R= 15/4

OD=2R-AD= 15/2-6=3/2, because the angle ECB= angle DOC and the triangle ECF= angle DOC, therefore, DE/CD=DC/OD.

DE=2？ /(3/2)=8/3。