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(1)证明:在边AB上截取线段AH,使AH=PC,连接PH,
由正方形ABCD,得∠B=∠BCD=∠D=90°,AB=BC=AD,
∵∠APF=90°,
∴∠APF=∠B,
∵∠APC=∠B+∠BAP=∠APF+∠FPC,
∴∠PAH=∠FPC;
又∵∠BCD=∠DCE=90°,CF平分∠DCE,
∴∠FCE=45°,
∴∠PCF=135°;
又∵AB=BC,AH=PC,
∴BH=BP,即得∠BPH=∠BHP=45°,
∴∠AHP=135°,即得∠AHP=∠PCF;
在△AHP和△PCF中,∠PAH=∠FPC,AH=PC,∠AHP=∠PCF,
∴△AHP≌△PCF,
∴AP=PF.
(2)解:⊙P与⊙G两圆的位置关系是外切.
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延长CB至点M,使BM=DG,连接AM,
由AB=AD,∠ABM=∠D=90°,BM=DG,
得△ADG≌△ABM,即得AG=AM,∠MAB=∠GAD;
∵AP=FP,∠APF=90°,
∴∠PAF=45°,
∵∠BAD=90°,
∴∠BAP+∠DAG=45°,即得∠MAP=∠PAG=45°;
于是,由AM=AG,∠MAP=∠PAG,AP=AP,
得△APM≌△APG,
∴PM=PG,
即得PB+DG=PG,(2分)
∴⊙P与⊙G两圆的位置关系是外切.(1分)
(3)解:由PG∥CF,得∠GPC=∠FCE=45°,(1分)
于是,由∠BCD=90°,得∠GPC=∠PGC=45°,
∴PC=GC.即得DG=BP.(1分)
设BP=x,则DG=x.由AB=2,得PC=GC=2-x,
∵PB+DG=PG,
∴PG=2x.
在Rt△PGC中,∠PCG=90°,得
sin∠GPC==.(1分)
即得
=,
解得
x=2?2,(1分)
∴当
BP=(2?2)时,PG∥CF.(1分)