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,矩形及其性质,苏科,版 八,年级下,第,9,章 中心对称图形,平行四边形,9.4.1,D,1,2,3,4,5,B,C,6,7,8,A,答 案 呈 现,温馨,提示,:,点击,进入,讲评,习题链接,4,30,9,10,C,11,如图,四边形,ABCD,的对角线互相平分,要使它变为矩形,需要添加的条件是,(,),A,AB,CD,B,AD,BC,C,AOB,45,D,ABC,90,1,D,【,2021,连云港】,如图,将矩形纸片,ABCD,沿,EF,折叠后,点,D,,,C,分别落在点,D,1,,,C,1,的位置,,ED,1,的延长线交,BC,于点,G,,若,EFG,64,,则,EGB,等于,(,),A,128,B,130,C,132,D,136,A,2,如图,在矩形,ABCD,中,点,E,是,CD,的中点,点,F,是,BC,上一点,且,FC,2,BF,,连接,AE,,,EF,,,AF,.,若,AB,2,,,AD,3,,则,AEF,的大小为,(,),A,30,B,45,C,60,D,不能确定,3,B,【中考,怀化】,如图,在矩形,ABCD,中,,AC,,,BD,相交于点,O,,若,AOB,的面积为,2,,则矩形,ABCD,的面积为,(,),A,4,B,6,C,8,D,10,4,C,【,点拨,】,由,题易知,S,AOD,S,BOC,S,COD,S,AOB,2.,S,矩形,ABCD,4,S,AOB,8.,5,C,【,2021,株洲】,如图,线段,BC,为等腰三角形,ABC,的底边,矩形,ADBE,的对角线,AB,与,DE,交于点,O,,若,OD,2,,则,AC,_,6,4,如图,在矩形,ABCD,中,对角线,AC,,,BD,交于点,O,,点,E,是,BC,上一点,且,AB,BE,,,1,15,,则,2,_,7,30,如图,在矩形,ABCD,中,点,E,在,BC,上,,AE,AD,,,DF,AE,,垂足为,F,.,(1),求证,DF,AB,;,8,证明:在矩形,ABCD,中,,B,90,,,AD,BC,,,AEB,DAF,.,DF,AE,,,DFA,90,.,DFA,B,.,又,AD,EA,,,ADF,EAB,(AAS),DF,AB,.,(2),若,FDC,30,,且,AB,4,,求,AD,的长,解:易知,DAF,FDA,90,,,FDC,FDA,90,,,DAF,FDC,30,.,AD,2,DF,.,又,DF,AB,,,AD,2,AB,24,8.,已知:如图,在矩形,ABCD,中,对角线,AC,与,BD,相交于点,O,,,AE,BD,于点,E,,,BF,AC,于点,F,.,求证:,AE,BF,.,9,证明:,四边形,ABCD,是矩形,,,OA,OB,,,AE,BD,于点,E,,,BF,AC,于点,F,,,AEO,BFO,90,,,又,AOE,BOF,,,AEO,BFO,,,AE,BF,.,已知在矩形,ABCD,中,,BD,是对角线,,AE,BD,于点,E,,,CF,BD,于点,F,.,(1),如图,,求证:,AE,CF,;,10,解:,ABE,,,CDF,,,BCE,,,ADF,.,如图,,在矩形,ABCD,(,AB,BC,),的,BC,边上取一点,E,,使,BE,AB,,作,AEF,90,,交,AD,于点,F,,易证,EA,EF,.,11,(1),如图,,若,EF,与,AD,的延长线交于点,F,,,EA,EF,是否仍然成立?若成立,请说明理由;,解,:,EA,EF,成立,理由:四边形,ABCD,是矩形,,,B,90,,,AD,BC,.,FAE,AEB,.,AB,BE,,,BAE,BEA,45.,FAE,45.,AEF,90,,,AFE,180,90,45,45.,FAE,AFE,.,EA,EF,.,(2),如图,,若四边形,ABCD,是平行四边形,(,AB,BC,),,在,BC,边上取一点,E,,使,BE,AB,,作,AEF,ABE,,交,AD,于点,F,,则,EA,EF,是否成立?若成立,请说明理由,解:,EA,EF,成立理由如下,:,BA,BE,,,BAE,AEB,.,四边形,ABCD,是平行四边形,,,AD,BC,.,ABE,BAD,180.,ABE,BAE,FAE,180.,AEF,ABE,,,AEB,AEF,FEC,180,,,FEC,FAE,.,AD,BC,,,FEC,AFE,,,FAE,AFE,.,EA,EF,.,
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