∵ABCD is rectangular.
∴AB⊥BC AB⊥AD = BC
∴△ABE is a right triangle
F is the midpoint of AE.
∴AF=BF=BE
∴∠FAB=∠FBA
∴∠DAF=∠CBF(∠DAF=∠DAB+∠BAF，∠CBF=∠CBA+∠FBA)
∴△DAF≌△CBF(AF=BF，∠DAF=∠CBF，AD=BC)
∴∠ADF=∠BCF
∴∠FDC=∠FCD
∴∠FGH=∠FHG
∴fg=fh；
(2) Solution: ∫AC = Ce∠E = 60.
∴△ACE is an equilateral triangle
∴CE=AE=8
∵AB⊥BC
∴BC=BE=？ CE =4
According to Pythagorean theorem AB= 4√3
∴ The area of trapezoidal AECD =? (AD+CE)×CD =？ (4+8) × 4 √ 3 = 24 √ 3.(24 radicals 3)