过D作DF∥AB交BC于F,
∵AD∥BC,∴四边形ABFD是平行四边形,∴DF=AB,又AB=CD,
∴DF=CD,∵∠C=60°,∴ΔDFC是等边三角形,
∴∠ABC=60°,
∵AB=AD,∴ABFD是菱形,∠ADB=∠ABD,又∠ADB=∠CBD,
∴∠ABD=∠CBD=1/2∠ABC=30°,
∴AB=2AE=2,
∴SΔDFC=√3/4×2^2=√3,
又S菱形ABFD=2SΔDFC=2√3,
∴S梯形=S菱形ABFD+SΔDFC=3√3.