(1) Since AB is the diameter of ⊙O and ∠ ACB = 90, just prove that P is the midpoint of the hypotenuse AQ of Rt△ACQ; According to the vertical diameter theorem, it is easy to know that arc AC= arc AE and c is the midpoint of arc AD, then arc CD= arc AE, that is, ∠PAC=∠PCA. According to the equivalence of equiangular complementary angle, we can also get ∠AQC=∠PCQ, which can prove that AP=PC=PQ, that is, P is the outside of △ACQ.

(2) According to the arc of (1): ∠ABC=∠ACE=∠CAQ, then their tangents are also equal; In Rt△CAF, the length of AC can be obtained by solving the right triangle according to the length of CF and the tangent of ∠ACF, and then in Rt△CAQ, the length of CQ can be obtained according to the tangent of ∠CAQ.

(3) From (1): PQ=CP, the product formula can be changed to: PC? = FP×FG； In Rt△ACB, it is obtained from the projective theorem: PC? =AF×FB, so you only need to prove that AF×FB=FG×FP. The above formula can be converted into a proportional formula, and the triangles where the line segments are located are almost the same, that is, RT △ AFP ∽ RT △ GFB.

(1) Prove: ∵C is the midpoint of AD, ∴ AC? =CD？ ,

∴∠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, ∴ AC? =AE？

∴ AE？ =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/BF=3/4, CF=8 in Rt△BCF,

Bf = 4/3cf = 32/3。

∴, BC=CF in Pythagoras theorem? +BF？ =40/3

∵AB is the diameter⊙ O,

∴, tan∠ABC= AC/BC=3/4, BC=40/3 in Rt△ACB.

Ac = 3/4bc = 10。

It's easy to know Rt△ACB∽Rt△QCA, ∴AC? =CQ×BC

∴ CQ=AC？ /BC= 15/2。

(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，

∴ AFFG=FPBF, that is, AF×BF=FP×FG.

It is easy to know Rt△ACF∽Rt△CBF.

∴FG？ =AF×BF (or derived from projective theorem)

∴FC？ =PF×FG

From (1), PC=PQ, ∴FP+PQ=FP+PC=FC.

∴(FP+PQ)？ =FP×FG