∴∠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.
∴ 。