2018年秋九年级数学上册第23章旋转章末复习随堂检测(新版)新人教版.doc

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1、123 章末复习1在ABCD中,A70,将ABCD绕顶点B顺时针旋转到A 1BC1D1,当C 1D1首次经过顶点C时,旋转角ABA 1_2在等边ABC中,AB10,D是BC的中点,将 ABD绕点A旋转后得到ACE,则线段DE的长度为 _3.如下图,P是矩形ABCD下方一点,将PCD绕P点顺时针旋转60后恰好D点与A点重合,得到PEA,连接EB,问ABE是什么特殊三角形?请说明理由4、如下图,等腰OBD中,ODBD, OBD绕点O逆时针旋转一定角度后得到OAC,此时正好B、D、C在同一直线上,且点D是BC的中点.5.如下图,将ABC绕点B逆时针旋转得到DBE,DE的延长线与AC相交于点F,连接D

2、A、BF, ABC60,BFAF.(1)求证:DABC;(2)猜想线段DF、AF的数量关系,并证明你的猜想23参考答案1、402、 533、【解析】ABE是等边三角形由旋转性质得PAEPDC,所以PAPD,AEDCAB,再由DPA60得P AD是等边三角形,从而得PDCPAEP AB30,所以EAB60,得证【答案】解:ABE是等边三角形,理由如下:由旋转,得PAEPDC,CDA E,PDPA,PDC PAE,DPA60,PAD是等边三角形,PDAPAD60,又CDAB,CDADAB90,PDCPABPAE30,AECDAB,EABPABPAE60,ABE是等边三角形4、(1)由旋转得OACOBD,OCOD又CDBDOD,OCODCD,OCD是等边三角形,COD60,旋转角为60(2)OACOBD,OCD是等边三角形, ACBDCD,OCAODB18060120 , ACDOCAO CD60 ACD是等边三角形,ODOCACAD, 四边形ODAC是菱形5、(1)由旋转得DBEABC60,BDAB, ABD是等边三角形,DAB60,DABABC,DABC;(2)DF2AF,证明 如下:在DF上截取DGAF, 连接BG,则DBGABF,BGBF, DBG ABF,GBFDBE60 , BGF是等边三角形,GFBFAF, DFDGFG2AF

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