(1)连结AE,
∵四边形ABCD是正方形,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9f510fb30f2442a7baee4a27d243ad4bd013024b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴∠ACD=∠DAB=45°
∵∠APE=45°,
∴∠APE=∠ACD.
∵∠AFP=∠EFC,
∴△AFP∽△EFC,
∴
=,
∴
=.
∵∠AFE=∠PFC,
∴△AFE∽△PFC,
∴∠AEF=∠FCP=45°
∴△APE是等腰直角三角形,
∴PE=
AP.
(2)∵四边形ABCD是正方形,
∴∠ACD=∠ACB=45°,∠B=∠ADE=∠BAD=∠BCD=90°,AB=BC=CD=AD.
∵△APE是等腰直角三角形,
∴AP=AE.
在Rt△ABP和Rt△ADE中,
,
∴Rt△ABP≌Rt△ADE(HL),
∴BP=DE.
∵∠APE=45°,
∴∠APE=∠ACD.
∵∠AFP=∠EFC,
∴∠PAC=∠CEF
∴△APC∽△EFC,
∴
=.
设BC=CD=x,则CP=x-3,AC=
x,CE=x+3,
∴
=,
解得:x
1=9,x
2=-1(舍去)
∴CB=CD=9,
∴CP=6,CE=12.
∵∠PCG=45°,
∴∠PGC+∠GPC=135°
∵∠PGE=135°,即∠PGC+∠CGE=135°,
∴∠GPC=∠CGE,
∵∠PCG=∠CGE,
∴△PCG∽△GCE,
∴CG
2=CP?CE,
∴CG=6
.
作GK⊥EC于K,
∴∠GKC=∠GKE=90°.
∵∠GCK=45°,
∴∠CGK=45°,
∴CK=KG.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3812b31bb051f81900c5b4c8d9b44aed2f73e754?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
在Rt△CGK中,由勾股定理,得
GK=CG=6,
∴DK=3.
在Rt△GKD,由勾股定理,得
GD=3
.
∵GK=PC=6,且GK∥BC
∴四边形GPCK是平行四边形,
∵∠PCK=90°,
∴四边形GPCK是矩形,
∴PG=CK=6,PG∥ED,
∴∠GPE=∠DEP.
∵∠PNG=∠END,
∴△PNG∽△END,
∴
===2.
∴GN=2ND,
∵GN+ND=GD=3
∴3ND=3
,
∴ND=
∴GN=2
.