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延長BE與CD的延長線相交於F.因為AB‖CD,所以∠A=∠1,又∠2=∠3,AE=DE,∴△AEB≌△DEF,∴S△AEB=S△DEF,BE=EF.∴S梯形ABCD=S四邊形EDCB+S△AEB=S四邊形ABCD+S△DEF=S△BFC=2S△BEC=2×2=4,故答案為:4.
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