∴∠CAD=∠ABC

∵AB is the diameter ⊙ O, ∴∠ ACB = 90.

∴∠CAD+∠AQC=90

And CE⊥AB, ∴∠ ABC+∠ pcq = 90.

∴∠AQC=∠PCQ

△ ∴ in △PCQ, PC=PQ,

∵CE⊥ Diameter AB, ∴ Arc AC= Arc AE

∴ arc AE= arc CD

∴∠CAD=∠ACE。

△ ∴ in △APC, with PA=PC.

∴PA=PC=PQ

∴P is the external center of△ △ACQ.

(2) Solution: ∵CE⊥ diameter AB in f,

∴, tan∠ABC=, CF=8 in Rt△BCF,

Yes

From Pythagorean theorem, we get

∵AB is the diameter⊙ O,

∴, tan∠ABC= in Rt△ACB,

Get AC=4/3BC = 10.

It is easy to know Rt△ACB∽Rt△QCA, ac 2 = cqxbc.

∴ 。

(3) It is proved that ∵AB is the diameter of ⊙O and ∴∠ ACB = 90.

∴∠DAB+∠ABD=90

And CF⊥AB, ∴∠ abg+∠ g = 90.

∴∠dab=∠g；

∴Rt△AFP∽Rt△GFB，

∴, that is, AFXBF=FPXFG

It is easy to know Rt△ACF∽Rt△CBF.

∴ (or from photography theorem)

∴ FG^2=AFXBF

From (1), PC=PQ, ∴FP+PQ=FP+PC=FC.

∴ 。