∵AB is the diameter ⊙ O, CD⊥AB,

∴ce=de= 12cd= 12×43=23，

Let OC=x,

BE = 2，

∴OE=x-2，

In Rt△OCE, OC2=OE2+CE2,

∴x2=(x-2)2+(23)2，

Solution: x=4,

∴OA=OC=4,OE=2，

∴AE=6，

At Rt△AED, AD=AE2+DE2=43,

∴AD=CD，

∵AF⊙O tangent,

∴AF⊥AB，

∵CD⊥AB，

∴AF∥CD，

∫CF∨AD，

∴ Quadrilateral FADC is a parallelogram,

AD = CD，

∴ parallelogram FADC is a diamond;

(2) Connection of AC power supply,

The square FADC is a diamond,

∴FA=FC，

∴∠FAC=∠FCA，

AO = CO

∴∠OAC=∠OCA

∴∠FAC+∠OAC=∠FCA+∠OCA

That is ∠ OCF = ∠ OAF = 90.

Namely OC⊥FC,

Point c is on ⊙O,

∴FC is the tangent of⊙ O.