上海市2020届初三数学一模提升题汇编第23题(几何证明题)

【2020长宁金山一模】23.（本题满分12分，第（1）小题5分，第（2）小题7分）如图，在ABC中，点D、E分别在边AB、BC上，AE与CD交于点F．若AE平分BAC，AEACAFAB．（1）求证：AECAFD；（2）若CDEG//，交边AC的延长线于点G，求证：BDFCCGCD．（长宁金山）23．（本题满分12分，第（1）小题5分，第（2）小题7分）证明：（1）∵AEACAFAB∴AFAEACAB（1分）∵AE平分BAC∴CAFBAE（1分）∴ABE∽ACF（1分）∴ACFB（1分）又∵BAEBAECCAFACFAFD,∴AECAFD（1分）(2)∵AECAFD，CFEAFD∴AECCFE（1分）∴CEFC（1分）∵CDEG//∴CEGDCBGACF又∵BACF∴GB（2分）∴BCD∽GEC（1分）∴CGBDCECD（1分）∴CGBDFCCD即BDFCCGCD．（1分）第23题图GACBEDF【2020杨浦一模】23．（本题满分12分，每小题各6分）如图，已知在ABC△中，AD是ABC△的中线，DACB，点E在边AD上，CECD.（1）求证：ACBDABAD；（2）求证：22ACAEAD.（杨浦）23．证明：（1）∵CD=CE，∴∠CED=∠CDA.···········································（1分）∴∠AEC=∠BDA.······················································································（1分）又∵∠DAC=∠B，∴△ACE∽△BAD.·························································（1分）∴ACCEABAD=.····························································································（1分）∵AD是ABC△的中线，∴BDCD=.·····················································（1分）∵CD=CE，∴BDCE=.∴ACBDABAD=.·······················································（1分）（2）∵∠DAC=∠B，又∠ACD=∠BCA，∴△ACD∽△BCA.·····································（1分）∴ACCDBCAC=，∴2ACCDCB=?.·································································（1分）∵AD是ABC△的中线，∴2BCCD=，∴222ACCD=.························（1分）∵△ACE∽△BAD，∴CEAEADBD=.··································································（1分）又∵CD=CE=BD，∴2CDADAE=?.······························································（1分）∴22ACADAE=?.·····················································································（1分）第23题图ABCDE【2020徐汇一模】23．（本题满分12分）如图，在ACB中，点D、E、F、G分别在边AB、AC、BC上，ADAB3，AECE2，CGFGBF，DG与EF交于点H．（1）求证：ABHGACFH；（2）联结DF、EG，求证：GEFFDGA．（徐汇）23．证明：（1）∵ADAB3，AECE2，CGFGBF，∴31,31,31,31BCCGBCBFACAEABAD；∴BCBFACAEBCCGABAD,；∴ACDG//，ABEF//；∴CHGF，BHFG；∴HFG∽ABC；∴ABFHACHG；即ABHGACFH．（2）∵ABEF//，ACDG//，∴1FBGFHDGH，1GFCGFHHE；∴FHHEHDGH；∴DFEG//；∴HGEFDG；又HEGHGEFHG，∴HEGFDGFHG；∵HFG∽ABC，∴AFHG；∴GEFFDGA．ABCDEFGH（第23题图）【2020松江一模】23．（本题满分12分,第（1）小题5分，第（2）小题7分）已知：如图，点D、F在△ABC边AC上，点E在边BC上，且DE∥AB，2CDCFCA.（1）求证：EF∥BD；（2）如果ACCFBCCE，求证：2BDDEBA.23．证明：（1）∵DE∥AB∴CDCECACB………（1分）∵2CDCFCA∴CDCFCACD………（1分）∴CECFCBCD………（2分）∴EF∥BD………（1分）(2)∵ACCFBCCE∴CACECBCF∵∠C=∠C∴△CAB∽△CEF………（1分）∴∠CAB=∠CEF………（1分）∵EF∥BD∴∠CBD=∠CEF………（1分）∴∠CBD=∠CAB………（1分）FCBADE(第23题图)FCBADE(第23题图)FCBADE(第23题图)∵DE∥AB，∴∠BDE=∠DBA………（1分）∴△BDE∽△ABD………（1分）∴BDABDEBD∴2BDDEBA………（1分）【2020青浦一模】23．（本题满分12分，第（1）小题6分，第（2）小题6分）已知：如图，在△ABC中，点D在边BC上，AE∥BC，BE与AD、AC分别相交于点F、G，2AFFGFE．（1）求证：△CAD∽△CBG；（2）联结DG，求证：DGAEABAG．23．证明：（1）∵2AFFGFE，∴AFFEFGAF．·························································（1分）又∵∠AFG=∠EFA，∴△FAG∽△FEA．·······················································（1分）∴∠FAG=∠E．·······························································································（1分）∵AE∥BC，∴∠E=∠EBC．···········································································（1分）∴∠EBC=∠FAG．··························································································（1分）又∵∠ACD=∠BCG，∴△CAD∽△CBG．···················································（1分）（2）∵△CAD∽△CBG，∴CACDCBCG．·····························································（1分）又∵∠DCG=∠ACB，∴△CDG∽△CAB．···················································（1分）∴DGCGABCB．·····························································································（1分）∵AE∥BC，∴AEAGCBGC．·········································································（1分）∴AGGCAECB，∴DGAGABAE，·································································（1分）∴DGAEABAG．··············································································（1分）EFGDCBA【2020普陀一模】本题满分12分）23、已知：如图11，四边形ABCD的对角线AC、BD相交于点O，AODBOCSS△△．（1）求证：OACOOBDO；（2）设△OAB的面积为S，kABCD，求证：2(1)ABCDSkS四边形．（普陀）23．证明：（1）过点A作AH⊥BD，垂足为点H.····················································（1分）∵S△AOD=AHDO21,S△AOB=AHOB21,∴OBDOAHOBAHDOSSAOBAOD2121.·····························································（2分）同理，BOCAOBSCOSOA．··········································································（1分）∵AODBOCSS△△，∴DOCOOBOA．···············································································（1分）CDBAO图11（2）∵OACOOBDO，AOBCOD，∴△OCD∽△OAB．·····································································（1分）∴CDDOCOkABBOAO．··································································（1分）22kABCDSSOABOCD．···································································（1分）∵△OAB的面积为S，∴SkSOCD2.·············································（1分）又∵kOBDOSSOABAOD，∴SkSAOD．············································（1分）同理，SkSBOC.······································································（1分）∴AOBBOCCODDOAABCDSSSSS△△△△四边形SkSkSkS2Skk)12(2Sk2)1(．·································································（1分）【2020浦东一模】23．（本题满分12分，其中每小题各6分）如图，