Photography abg

∴ ,

∴∠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,

∴

∴

∴∠CAD=∠ACE，

△ ∴ in △APC, with PA=PC.

∴PA=PC=PQ

P is the external center of △ACQ;

(2)∵CE⊥ Diameter AB is in f,

∴, tan∠ABC=, CF=8 in Rt△BCF,

From Pythagorean theorem, we get

∵AB is the diameter⊙ O,

∴, tan∠ABC= in Rt△ACB,

OK,

It is easy to know Rt△ACB∽Rt△QCA.

∴

∴ ;

(3)∵AB is the diameter ⊙O,

∴∠ACB=90

∴∠DAB+∠ABD=90

CF⊥AB again,

∴∠ABG+∠G=90

∴∠dab=∠g；

∴Rt△AFP∽Rt△GFB，

∴, namely

It is easy to know Rt△ACF∽Rt△CBF.

∴ (or from photography theorem)

∴

From (1), we know that PC=PQ,

∴FP+PQ=FP+PC=FC

∴ 。