(1)∵AD∥BC,AB=DC,∴∠DCB=∠ABC; ∵AD=DC,∴∠DCA=∠DAC. ∵AD∥BC,∴∠DAC=∠ACB, 则∠DCB=2∠ACB,所以∠ABC=2∠ACB. ∵AC⊥AB,∴∠ABC+∠ACB=90°, ∴∠ACB=30°,∠ABC=60°;
(2)证明:∵BF=CD,AB=DC,∴BF=AB. ∴∠F=∠BAF. ∵∠ABC=60°,∴∠F=30°. ∴∠ACB=∠F. ∴AC=AF,即:△CAF为等腰三角形;
(3)作AE⊥BC于E. ∵AB=AD=x,∠ABC=60°, ∴AE=x. ∵∠ACB=30°,AC⊥AB, ∴BC=2x. ∴S梯形ABCD=×(x+2x)×x=x2. |