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