Point b and point g are symmetrical about the straight line EF,
∴CF=HF,BE=GE.
Let BE=GE=x, then AE = 4-X.
∵ quadrilateral ABCD is a square,
∴∠A=90。
∴AE2+AG2=EG2.
∵B falls at the midpoint g of the edge AD,
∴AG=2，
∴(4-x)2+22=x2.
Solution x = 2.5.
∴BE=2.5.
∵ quadrilateral ABCD is a square,
∴AB∥CD,∠B=90。
Points e and f are on the sides of AB and CD, respectively.
∴ Quadrilateral BCFE is a right-angled trapezoid.
BE = GE = 2.5，AB=4，
∴AE= 1.5.
∴sin∠ 1=35,tan∠ 1=34.
∵∠ 1+∠2=90 ,∠2+∠3=90 ,
∴∠3=∠ 1.
∴sin∠3=sin∠ 1=35，
In Rt△DGP, ∫∠d = 90,
DG=2，sin∠3=DGGP=35，
∴PG= 103，
∴PH=GH-GP=23，
∵∠4=∠3,
∴tan∠4=tan∠3=tan∠ 1=34，
In Rt△HPF, ∫∠h =∠c = 90,
∴FC=HF= 12.
∴S quadrilateral bcfe =12 (fc+be) × BC =12 × (12+2.5 )× 4 = 6.
So the answer is: 6.