题目内容
如图1,在矩形纸片ABCD中,
,其中m≥1,将该矩形沿EF折叠(点E、F分别在边AB、CD上),使点B落在AD边上的点M处,点C落在点N处,MN与CD相交于点P,连接EP.设
,其中0<n≤1.
(1)如图2,当
(即M点与D点重合),
时,则
;
(2)如图3,当
(M为AD的中点),m的值发生变化时,求证:
;
(3)如图1,当
,n的值发生变化时,
的值是否发生变化?说明理由.
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(1)如图2,当
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(2)如图3,当
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(3)如图1,当
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(1)
;(2)证明见解析;(3)
,不发生变化,理由见解析.
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试题分析:(1)由条件可知,当n=1(即M点与D点重合),m=2时,AB=2AD,设AD=a,则AB=2a,由矩形的性质可以得出△ADE≌△NDF,就可以得出AE=NF,DE=DF,在Rt△AED中,由勾股定理就可以表示出AE的值,再求出BE的值就可以得出结论.
(2)延长PM交EA延长线于G,由条件可以得出△PDM≌△GAM,△EMP≌△EMG由全等三角形的性质就可以得出结论.
(3)如图1,连接BM交EF于点Q,过点F作FK⊥AB于点K,交BM于点O,通过证明△ABM∽△KFE,就可以得出
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(1)∵四边形ABCD是矩形,∴AB=CD,AD=BC,∠A=∠B=∠C=∠D=90°.
∵AB=mAD,且n=2,∴AB=2AD.
∵∠ADE+∠EDF=90°,∠EDF+∠NDF=90°,∴∠ADE=∠NDF.
在△ADE和△NDF中,∠A=∠N,AD=ND,∠ADE=∠NDF,
∴△ADE≌△NDF(ASA).∴AE=NF,DE=DF.
∵FN=FC,∴AE=FC.
∵AB=CD,∴AB-AE="CD-CF." ∴BE="DF." ∴BE=DE.
Rt△AED中,由勾股定理,得
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∴BE=2AD-
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∴
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(2)如图3,延长PM交EA延长线于G,∴∠GAM=90°.
∵M为AD的中点,∴AM=DM.
∵四边形ABCD是矩形,∴AB=CD,AD=BC,∠A=∠B=∠C=∠D=90°,AB∥CD.
∴∠GAM=∠PDM.
在△GAM和△PDM中,∠GAM=∠PDM,AM=DM,∠AMG=∠DMP,
∴△GAM≌△PDM(ASA).∴MG=MP.
在△EMP和△EMG中,PM=GM,∠PME=∠GME,ME=ME,
∴△EMP≌△EMG(SAS).∴EG=EP.
∴AG+AE=EP.∴PD+AE=EP,即EP=AE+DP.
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(3)
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如图1,连接BM交EF于点Q,过点F作FK⊥AB于点K,交BM于点O,
∵EM=EB,∠MEF=∠BEF,∴EF⊥MB,即∠FQO=90°.
∵四边形FKBC是矩形,∴KF=BC,FC=KB.
∵∠FKB=90°,∴∠KBO+∠KOB=90°.
∵∠QOF+∠QFO=90°,∠QOF=∠KOB,∴∠KBO=∠OFQ.
∵∠A=∠EKF=90°,∴△ABM∽△KFE.
∴
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∵AB=2AD=2BC,BK=CF,∴
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∴
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