上海市2019年初三中考数学二模汇编-23题几何证明

上海市2019年中考二模数学汇编：23题几何证明闵行23．（本题共2小题，每小题6分，满分12分）如图，已知四边形ABCD是菱形，对角线AC、BD相交于点O，BD=2AC．过点A作AE⊥CD，垂足为点E，AE与BD相交于点F．过点C作CG⊥AC，与AE的延长线相交于点G．求证：（1）△ACG≌△DOA；（2）2DFBDDEAG．宝山23．（本题满分12分，第(1)、第(2)小题满分各6分）如图，在矩形ABCD中，E是AB边的中点，沿EC对折矩形ABCD，使B点落在点P处，折痕为EC，联结AP并延长AP交CD于F点，（1）求证：四边形AECF为平行四边形；（2）如果PA=PC，联结BP，求证：△APB≅△EPC．ABCDOEGF（第23题图）ABCDOEHF第23题图崇明23．（本题满分12分，每小题满分各6分）如图7，在直角梯形ABCD中，90ABC，ADBC∥，对角线AC、BD相交于点O．过点D作DEBC，交AC于点F．（1）联结OE，若BEAOECOF，求证：OECD∥；（2）若ADCD且BDCD，求证：AFDFACOB．奉贤23．（本题满分12分，每小题满分各6分）已知：如图8，正方形ABCD，点E在边AD上，AF⊥BE，垂足为点F，点G在线段BF上，BG=AF．（1）求证：CG⊥BE；（2）如果点E是AD的中点，联结CF，求证：CF=CB．金山22.已知：如图，菱形ABCD的对角线AC与BD相交于点O，若DBCCAD.（1）求证：ABCD是正方形.（2）E是OB上一点，CEDH，垂足为H，DH与OC相交于点F，求证：OFOE.ABCDOEF图7ABCDFGE图8普陀23．（本题满分12分）已知：如图10，在四边形ABCD中，ADBC，点E在AD的延长线上，ACEBCD，ECEDEA2．（1）求证：四边形ABCD为梯形；（2）如果ECABEAAC，求证：ABEDBC2．杨浦23.已知：在ABC中，AB=BC，∠ABC=90°，点D、E分别是边AB、BC的中点，点F、G是边AC的三等分点，DF、EG的延长线相交于点H，联结HA、HC.求证：（1）四边形FBGH是菱形；（2）四边形ABCH是正方形.长宁23.（本题满分12分，第（1）小题5分，第（2）小题7分）图10ABCDE如图5，平行四边形ABCD的对角线BDAC、交于点O，点E在边CB的延长线上，且90EAC，ECEBAE2．（1）求证：四边形ABCD是矩形；（2）延长AEDB、交于点F，若ACAF，求证：BFAE．黄浦嘉定23.静安图5ABCDEFO松江徐汇答案闵行23．证明：（1）在菱形ABCD中，AD=CD，AC⊥BD，OB=OD．∴∠DAC=∠DCA，∠AOD=90°．……………………………（1分）∵AE⊥CD，CG⊥AC，∴∠DCA+∠GCE=90°，∠G+∠GCE=90°．∴∠G=∠DCA．…………………………………………………（1分）∴∠G=∠DAC．…………………………………………………（1分）∵BD=2AC，BD=2OD，∴AC=OD．……………………（1分）在△ACG和△DOA中，∵∠ACG=∠AOD，∠G=∠DAC，AC=OD，∴△ACG≌△DOA．……………………………………………（2分）（2）∵AE⊥CD，BD⊥AC，∴∠DOC=∠DEF=90°．…………（1分）又∵∠CDO=∠FDE，∴△CDO∽△FDE．…………………（1分）∴CDODDFDE．即得ODDFDECD．……………………（2分）∵△ACG≌△DOA，∴AG=AD=CD．……………………（1分）又∵12ODBD，∴2DFBDDEAG．…………………（1分）宝山23.（1）证明：由折叠得到EC垂直平分BP，………………1分设EC与BP交于Q，∴BQ=EQ………………1分∵E为AB的中点，∴AE=EB，………………1分∴EQ为△ABP的中位线，∴AF∥EC，………………2分∵AE∥FC，∴四边形AECF为平行四边形；………………1分（2）∵AF∥EC，∴∠APB=∠EQB=90°………………1分由翻折性质∠EPC=∠EBC=90°，∠PEC=∠BEC………………1分∵E为直角△APB斜边AB的中点，且AP=EP，∴△AEP为等边三角形,∠BAP=∠AEP=60°，………………1+1分60260180CEBCEP………………1分在△ABP和△EPC中，∠BAP=∠CEP，∠APB=∠EPC，AP=EP∴△ABP≌△EPC（AAS），………………1分崇明23．（本题满分12分，每小题满分各6分）证明(1)∵90ABD,BCDE∴//ABDE………………………………………………………………（1分）∴AOBOOFOD………………………………………………………………（2分）∵BEAOECOF∴AOBEOFEC………………………………………………………………（2分）∴//OECD…………………………………………………………………（1分）(2)∵BCAD//，//ABDE，∴四边形ABED为平行四边形又∵90ABD∴四边形ABED为矩形……………………………………………………（1分）∴ADBE，90ADE又∵CDBD∴90BDCBDECDE90BDEADBADE∴CDEADB…………………………………………………………（1分）ADCD∴DCADAC∴ASACDFADO..…………………………………………………（1分）∴ODDFDEAB//∴AFBEADACBCBC…………………………………………………………（1分）∵BCAD//∴BODFBOODBCAD…………………………………………………………（1分）∴AFDFACOB…………………………………………………………………（1分）奉贤22．证明：（1）∵四边形ABCD是正方形，∴ABBC．90ABC??．·············（1分）∵AF⊥BE，∴90FABFBA．∵90FBACBG，∴FABCBG．··········································（1分）又∵AFBG，∴△AFB△BGC．····························································（2分）∴AFBBGC．·····························································································（1分）∵90AFB，∴90BGC，即CG⊥BE．···········································（1分）（2）∵ABFEBA，90AFBBAE，∴△AEB∽△FAB．∴AEAFABBF．·································································（3分）∵点E是AD的中点，ADAB，∴12AEAB．∴12AFBF．··························（1分）∵AFBG，∴12BGBF，即FGBG．··························································（1分）∵CG⊥BE，∴CFCB．····················································································（1分）金山23.（1）证明：∵四边形ABCD是菱形，∴BCAD//，DACBAD2,DBCABC2;（2分）∴180ABCDAB;（1分）∵DBCCAD;∴ABCBAD,（1分）∴1802BAD;∴90BAD;（1分）∴四边形ABCD是正方形.（1分）（2）证明：∵四边形ABCD是正方形；∴BDAC，BDAC，ACCO21,BODO21；（1分）∴90DOCCOB，DOCO；（1分）∵CEDH，垂足为H；∴90DHE，90DEHEDH；（1分）又∵90DEHECO；∴EDHECO；（1分）∴ECO≌FDO；（1分）∴OFOE.（1分）普陀23．证明：（1）∵ACEBCD，∴DCEBCA．······················································（1分）∵ECEDEA2，∴EDECECEA．·······································································（1分）又∵E是公共角，∴△EDC∽△ECA．·····························································（1分）∴DCECAE．·································································································（1分）∴BCACAE．∴AD∥BC．·············································································································（1分）∵ADBC，∴AB与CD不平行．∴四边形ABCD是梯形．···························································································（1分）（2）∵△EDC∽△ECA．∴ECCDEAAC．∵ECABEAAC，∴ABDC．··············································································（1分）∴四边形ABCD是等腰梯形．···············································································（1分）∴BDCB．···································································································（1分）∵AD∥BC．∴EDCDCB．∴EDCB．∵ECDACB，∴△EDC∽△ABC．·····················································（1分）∴EDDCABBC．········································································································（1分）∴ABEDBC2．·····························································································（1分）杨浦23.（1）证明略（2）证明略长宁23．（本题满分12分，第（1）小题5分，第（2）小题7分）证明：（1）∵ECEBAE2∴AEEBECAE又∵CEAAEB∴AEB