Solution:
Connecting GE
∵ quadrilateral ABCD is a square
∴BC=CD,∠BCD=90
∵ Quadrilateral GCEF is a square
∴GC=CE,∠DCE=90
∴∠BCD=∠DCE
∴△BCG≌△DCE(SAS)
∴∠CBG=∠CDE
∠∠CDE+∠CED = 90
∴∠CBG+∠CED=90
∴∠BHE=90
∴BH⊥DE
∫tan∠GBC =√2- 1,BC=CD= 1
∴GC=BC×tan∠GBC=√2- 1=CE
∴dg=dc-gc= 1-(√2- 1)=2-√2
∫In Rt△GCE,GE? =GC? +CE? =(√2- 1)? +(√2- 1)? =6-4√2
Where's DG =(2-√2)? =6-4√2
∴GE? =DG?
∴GE=DG
∴ DH = He (in an isosceles triangle, three lines are on one line)
∴S△GHE= 1/2S△DGE
∴S quadrilateral CEHG
=SRt△GCE+SRt△GHE
= 1/2×CG×CE+ 1/2S△DGE
= 1/2×CG×CE+ 1/2× 1/2×DG×CE
= 1/2×(√2- 1)(√2- 1)+ 1/2× 1/2×(2-√2)×(√2- 1)
=3/2-√2+(3√2/4)- 1
= 1/2-√2/4