解:設AD’交BC于O,
方法一:過點B作BE⊥AD’于E,矩形ABCD中,∵AD∥BC,AD=BC,
∠B=∠D=∠BAD=90°,在Rt△ABC中,∵tan∠BAC=
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,
∴∠BAC=60°,∴∠DAC=90°—∠BAC=30°,∵將△ACD沿對角線AC向下翻折,得到△ACD’,∴AD’=AD=BC=
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,∠1=∠DAC=30°,∴∠4=∠BAC—∠1=30°,
又在Rt△ABE中,∠AEB=90°,∴BE=2,∴AE=
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,∴D’E=AD’—AE=
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,∴AE=D’E,即BE垂直平分AD’,∴BD’=AB=4.
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方法二:
矩形ABCD中,∵AD∥BC,AD=BC,∠B=∠D=90°,∴∠ACB=∠DAC,
在Rt△ABC中,∵tan∠BAC=
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,
∴∠BAC=60°,∴∠ACB=90°—∠BAC=30°,
∵將△ACD沿對角線AC向下翻折,得到△ACD’,
∴AD=AD’=BC,∠1=∠DAC=∠ACB=30°,
∴OA=OC,
∴OD’=OB,∴∠2=∠3,
∵∠BOA=∠1+∠ACB=60°, ∠2+∠3=∠BOA,
∴∠2=
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∠BOA=30°,
∵∠4=∠BAC—∠1=30°,∴∠2=∠4,∴BD’=AB=4.解析:
p;【解析】略