2019年上海各区初三二模数学试卷23题专题汇编(教师版)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 ∠=︒,AD BC ∥,对角线AC 、BD 相交于点O . 过点D 作DE BC ⊥,交AC 于点F . (1)联结OE ,若BE AOEC OF=,求证:OE CD ∥; (2)若AD CD =且BD CD ⊥,求证:AF DFAC OB=. 23.(本题满分12分,每小题满分各6分) 证明(1)∵90ABD ∠=︒,BC DE ⊥∴//AB DE ………………………………………………………………(1分) ∴AO BOOF OD=………………………………………………………………(2分) ∵BE AOEC OF =∴AO BEOF EC=……… ………………………………………………………(2分) ∴//OE CD …………………………………………………………………(1分) (2)∵BC AD //,//AB DE ,∴四边形ABED 为平行四边形 又∵90ABD ∠=︒∴四边形ABED 为矩形 ……………………………………………………(1分) ∴AD BE =,90ADE ∠=︒ 又∵CD BD ⊥∴90BDC BDE CDE ∠=∠+∠=︒︒=∠+∠=∠90BDE ADB ADE∴CDE ADB ∠=∠ …………………………………………………………(1分)AD CD =∴DCA DAC ∠=∠∴()A S A CDF ADO ..∆≅∆…………………………………………………(1分) ∴OD DF =DE AB //ABCDOE F图7∴AF BE ADAC BC BC==…………………………………………………………(1分) ∵BC AD //∴BODFBO OD BC AD ==…………………………………………………………(1分) ∴AF DFAC OB=…………………………………………………………………(1分) 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 .23.证明:(1)∵四边形ABCD 是正方形,∴AB BC =.90ABC. ··········· (1分) ∵AF ⊥BE ,∴90FAB FBA ∠+∠=︒.∵90FBA CBG ∠+∠=︒,∴FAB CBG ∠=∠. ································ (1分) 又∵AF BG =,∴△AFB ≅△BGC . ············································· (2分)∴AFB BGC ∠=∠. ······································································· (1分) ∵90AFB ∠=︒,∴90BGC ∠=︒,即CG ⊥BE . ······························· (1分) (2)∵ABF EBA ∠=∠,90AFB BAE ∠=∠=︒,∴△AEB ∽△FAB .∴AE AFAB BF=. ················································· (3分) ∵点E 是AD 的中点,AD AB =,∴12AE AB =.∴12AF BF =.··················· (1分) ∵AF BG =,∴12BG BF =,即FG BG =.············································ (1分) ∵CG ⊥BE ,∴CF CB =. ······························································· (1分)ABCD FG E 图8闵行(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BD AC 、相交于点O ,AC BD 2=,过点A 作CD AE ⊥,垂足为点E ,AE 与BD 相交于点F ,过点C 作AC CG ⊥,与AE 的延长线相交于点G . 求证:(1)DOA ACG ∆∆≌;(2)AG DE BD DF ⋅=⋅223.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)∴ CD OD DF DE=.即得 OD DF DE CD ⋅=⋅. ……………………(2分) ∵ △ACG ≌△DOA ,∴ AG = AD = CD . ……………………(1分)又∵ 12OD BD =,∴ 2DF BD DE AG ⋅=⋅.…………………(1分)嘉定23.(本题满分12分,第(1)小题6分、第(2)小题6分)如图6,在矩形ABCD 中,点E 是边AB 的中点,△EBC 沿直线EC 翻折,使B 点落在矩形ABCD 内部的点P 处,联结AP 并延长AP 交CD 于点F ,联结BP 交CE 于点Q . (1)求证:四边形AECF 是平行四边形; (2)如果PE PA =,求证:△APB ≌△EPC .23.(1)证明:由翻折得:EC 垂直平分BP ………………1分∴EQ BQ = ………………1分 ∵点E 为AB 的中点,∴EB AE = ………………1分 ∴EQ 是△ABP 的中位线,∴EC ∥AF ,……………1分 ∵四边形ABCD 是矩形∴AE ∥FC ………………1分 ∴四边形AECF 是平行四边形. ………………1分(2)∵AE ∥FC ,∴EQB APB ∠=∠ ………………1分由翻折得: ︒=∠90EQB ,︒=∠90EPC∴︒=∠=∠90EPC APB ………………1分 由翻折得:EB PE =,BEC PEC ∠=∠∵PE PA =,EB AE = ∴AE PE PA ==∴△AEP 是等边三角形,∴︒=∠=∠60AEP PAB …………1分 ∵︒=∠+∠+∠180BEC PEC AEP∴︒=∠60PEC ………………1分AB DCF PEQ图6∴PEC PAB ∠=∠ ………………1分 ∵PE PA =,∴△APB ≌△EPC ………………1分 黄埔23.(本题满分12分)如图6,已知四边形ABCD ,AD ∥BC ,对角线AC 、BD 交于点O ,DO =BO ,过点C 作CE ∥AC ,交BD 的延长线于点E ,交AD 的延长线于点F ,且满足DCE ACB ∠=∠. (1)求证:四边形ABCD 是矩形; (2)求证:DE ADEF CD=.23. 证明:(1)∵AD ∥BC ,∴AD DOBC BO=, ∵DO =BO ,∴AD BC =,--------------------(2分)∴四边形ABCD 是平行四边形. ------------------------------------------------------------------------(1分) ∵CE ⊥AC ,∴90ACD DCE ∠+∠=︒,∵DCE ACB ∠=∠,∴90ACB ACD ∠+∠=︒,即90BCD ∠=︒,------------------------(2分) ∴四边形ABCD 是矩形. --------------------------------------------------------------------------------------(1分)(2)∵四边形ABCD 是矩形,∴AC BD =,90ADC ∠=︒---------------------------------------(2分)∵AD ∥BC ,∴DE EFBD FC=.--------------------------------------------------------------------------------(1分) ∴DE EFAC FC =,------------------------------------------------------------------------------------------------(1分) ∴DE AC EF FC=,∵90ADC ACF ∠=∠=︒, ∴cot AC ADDAC FC CD∠==,----------------------------------------------------------------------------------(1分) ∴DE AD EF CD =.--------------------------------------------------------------------------------------------------(1分)ABC DEF图6OA B CDO E H F 第23题图金山22. 已知:如图,菱形ABCD 的对角线AC 与BD 相交于点O ,若DBC CAD ∠=∠.(1)求证:ABCD 是正方形.(2)E 是OB 上一点,CE DH ⊥,垂足为H ,DH 与OC 相交于点F ,求证:OF OE =.23.(1)证明:∥四边形ABCD 是菱形,∥BC AD //,DAC BAD ∠=∠2,DBC ABC ∠=∠2; (2分) ∥180=∠+∠ABC DAB ; (1分) ∥DBC CAD ∠=∠;∥ABC BAD ∠=∠, (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;∥BD AC ⊥,BD AC =,AC CO 21=,BO DO 21=; (1分) ∥90=∠=∠DOC COB ,DO CO =; (1分) ∥CE DH ⊥,垂足为H ;∥90=∠DHE ,90=∠+∠DEH EDH ; (1分) 又∥90=∠+∠DEH ECO ;∥EDH ECO ∠=∠; (1分) ∥ECO ∆≌FDO ∆; (1分) ∥OF OE =. (1分)普陀23.(本题满分12分)已知:如图10,在四边形ABCD 中,AD BC <,点E 在AD 的延长线上, ACE BCD ∠=∠,EC ED EA =⋅2. (1)求证:四边形ABCD 为梯形; (2)如果EC ABEA AC=,求证:AB ED BC =⋅2.23.证明:(1)∵ ACE BCD ∠=∠,∴DCE BCA ∠=∠. ········································· (1分)∵EC ED EA =⋅2,∴ED ECEC EA=. ······················································ (1分) 又∵E ∠是公共角,∴△EDC ∽△ECA . ·············································· (1分) ∴DCE CAE ∠=∠. ········································································· (1分) ∴BCA CAE ∠=∠.∴AD ∥BC . ·················································································· (1分) ∵AD BC <,∴AB 与CD 不平行.∴四边形ABCD 是梯形. ····································································· (1分) (2)∵△EDC ∽△ECA .∴EC CDEA AC =. ∵EC AB EA AC=,∴AB DC =.··························································· (1分) ∴四边形ABCD 是等腰梯形. ···························································· (1分) ∴B DCB ∠=∠. ··········································································· (1分) ∵AD ∥BC .∴EDC DCB ∠=∠.图10A BCD E∴EDC B ∠=∠.∵ECD ACB ∠=∠,∴△EDC ∽△ABC . ········································ (1分) ∴ED DCAB BC=. ··············································································· (1分) ∴AB ED BC =⋅2. ······································································ (1分) 徐汇22. (本题满分(12分),第(1)题满分6分,第(2)小题满分6分) 如图,已知梯形ABCD 中,E AC AB BC AD ,,=∥是边BC 上的点,且CAD AED ∠=∠,DE 交AC 于点F(1) 求证:DAF ABE ∽△△(2) 当EC AE FC AC ⋅=⋅时,求证:BE AD = 23. :(1)BC AD // ACB CAD ∠=∠∴ AC AB = ACB B ∠=∠∴ 又CAD AED ∠=∠CAD AED ACB B ∠=∠=∠=∠∴ 又CED AED BAE B ∠+∠=∠+∠ CED BAE ∠=∠∴又BC AD // CED ADF ∠=∠∴ ADF BAE ∠=∠∴ CAD ABE ∠=∠ ABE ∆∴相似于DAF ∆(2)由(1)知ABE ∆∴相似于DAF ∆AF BE AD AB =∴AFADBE AB =∴ BC AD // FC AF EC AD =∴FCECAF AD =∴ FC ECBE AB =∴ 由(1)知:CED BAE CED B ∠=∠∠=∠,ABE ∆∴相似于ECF ∆ FC BE EC AB =∴ FCEC BE AB =∴ EC AE FC AC ⋅=⋅ FCECAE AC =∴AEAC BE AB =∴ 又AC AB = AE BE =∴ BAE B ∠=∠∴又AED B ∠=∠ AED BAE ∠=∠∴DE AB //∴ 又BC AD //∴四边形ABED 是平行四边形 BE AD =∴杨浦1、 (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,AB=BC ,∠ABC=90°,点D 、E 分别是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H ,联结HA 、HC 求证:(1)四边形FBGH 是菱形 (2)四边形ABCH 是正方形23.证明(1):∵点F 、G 是边AC 的三等分点,∴F 、G 分别是AG 、CF 的中点, ∵点D 是AB 的中点,∴DF //BG ,即FH //BG . ........................ (2分)同理: GH // BF . ........................................................................... (1分) ∴四边形FBGH 是平行四边形. .................................................. (1分) ∵AB =BC ,∴∠BAC =∠ACB .∵点F 、G 是边AC 的三等分点,∴AF =CG .∴△ABF ≌△CBG . ∴BF =BG. .................................................... (1分) ∴平行四边形FBGH 是菱形. ....................................................... (1分)证明(2)联结BH ,交FG 于点O ,∵四边形FBGH 是平行四边形,∴OB =OH ,OF =OG . ............ (2分) ∵AF =CG ,∴OA =OC . ................................................................. (1分) ∴四边形ABCH 是平行四边形. .................................................. (1分) ∵∠ABC =90°,∴平行四边形ABCH 是矩形. .......................... (1分)∵AB =BC ,∴矩形ABCH 是正方形. (1分)长宁23.(本题满分12分,第(1)小题5分,第(2)小题7分)如图5,平行四边形ABCD 的对角线BD AC 、交于点O ,点E 在边CB 的延长线上,且︒=∠90EAC ,EC EB AE ⋅=2. (1)求证:四边形ABCD 是矩形;(2)延长AE DB 、交于点F ,若AC AF =,求证:BF AE =.23.(本题满分12分,第(1)小题5分,第(2)小题7分)证明:(1)∵EC EB AE ⋅=2 ∴AEEB EC AE =又 ∵CEA AEB ∠=∠ ∴AEB ∆∽CEA ∆ (2分) ∴EAC EBA ∠=∠∵︒=∠90EAC ∴︒=∠90EBA (1分) 又 ∵︒=∠+∠180CBA EBA ∴︒=∠90CBA (1分) ∵四边形ABCD 是平行四边形∴四边形ABCD 是矩形 (1分)(2)∵ AEB ∆∽CEA ∆ ∴ AC AB AE BE = 即 ACAE AB BE = , ECA EAB ∠=∠ (2分)∵四边形ABCD 是矩形 ∴BD AC =又 ∵BD OB 21=, AC OC 21= ∴OC OB = ∴ECA OBC ∠=∠ 又 ∵OBC EBF ∠=∠ ECA EBA ∠=∠ ∴EAB EBF ∠=∠又∵F F ∠=∠ ∴EBF ∆∽BAF ∆ (3分)∴AB BE AF BF = ∴ACAEAF BF =(1分) ∵AC AF = ∴AE BF = (1分)图5AB CDE FO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P A=PC ,联结BP ,求证:∥APB ≅∥EPC .第23题图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 ,∥∥A PB =∥EQB =90° ………………1分由翻折性质∥E PC =∥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︒=︒-︒=∠=∠60260180CEB CEP ………………1分 在∥ABP 和∥EPC 中, ∥BAP =∥CEP ,∥APB=∥E PC ,AP =EP ∥∥ABP ∥∥EPC (AAS ), ………………1分松江23.(本题满分12分,每小题各6分)如图,已知□ABCD 中,AB=AC ,CO ⊥AD ,垂足为点O ,延长CO 、BA 交于点E ,联结DE . (1)求证:四边形ACDE 是菱形;(2)联结OB ,交AC 于点F ,如果OF=OC ,求证:22AB BF BO =⋅.23.证明:(1)∵四边形ABCD 是平行四边形∴AB ∥DC ,AB=DC ………………………………………………………………(1分) ∵AB=AC ,∴AC=DC ……………………………………………………………(1分) ∵CO ⊥AD ,∴AO=DO …………………………………………………………(1分) ∵EO AOCO DO=,∴EO=CO ………………………………………………………(1分) ∴四边形ACDE 是平行四边形……………………………………………………(1分) ∵AC=DC ,∴四边形ACDE 是菱形……………………………………………(1分) (2)∵ OF=OC ,∴∠OFC=∠OCF ……………………………………………(1分) ∵AE=AC ,∴∠OCF=∠BEO∵∠OFC=∠BF A ,∴∠BF A=∠BEO …………………………………………(1分) ∵∠ABF=∠OBE …………………………………………………………………(1分) ∴△BF A ∽△BEO ,∴AB BFBO BE=………………………………………………(1分) ∴AB ·BE=BF ·BO ,∵AE=AC=AB ,∴BE=2AB ………………………………(1分) ∴22AB BF BO =⋅………………………………………………………………(1分)(第23题图)OECBA静安22.(本题满分10分,第(1)小题满分5分,第(2)小题满分5分)已知:如图5,在矩形ABCD 中,过AC 的中点M 作EF ⊥AC , 分别交AD 、BC 于点E 、F . (1)求证:四边形AECF 是菱形; (2)如果2CD BF BC =⋅,求∠BAF 的度数.22.(本题满分10分,第(1)小题5分,第(2)小题5分) 证明:(1)∵四边形ABCD 为矩形,∴AD //BC , ∴∠1=∠2...........................................(1分)∵点M 为AC 的中点,∴AM =CM .在△AME 与△CMF 中,12AM CM AME CMF ∠=∠⎧⎪=⎨⎪∠=∠⎩..............................................(1分) ∴△AME ≌△CMF ...........................................(1分) ∴AE =CF .∴四边形AECF 为平行四边形. ·································································· (1分) 又∵EF ⊥AC ,∴平行四边形AECF 为菱形. ····································································· (1分) (2)∵2CD BF BC =⋅,∴CD BC BF CD =.又∵四边形ABCD 为矩形,∴AB =CD ,∴AB BC BF AB =. ··········································································· (1分)又∵∠ABF =∠CBA ,∴△ABF ∽△CBA . ·················································································· (1分) ∴∠2=∠3. ···························································································· (1分) ∵四边形AECF 为菱形,∴∠1=∠4,即∠1=∠3=∠4. ····································································· (1分) ∵四边形ABCD 为矩形, ∴∠BAD =∠1+∠3+∠4=90°,∴即∠1=30°. ······················································································· (1分)图5CFEDA BM图5CF EDA B M 124323.(本题满分12分,第(1)小题满分8分,第(2)小题满分4分)已知:如图6,△ABC 内接于⊙O ,AB ﹦AC ,点E 为弦AB 的中点,AO 的延长线交BC 于点D ,联结ED .过点B 作BF ⊥DE 交AC 于点F .(1)求证:∠BAD ﹦∠CBF ; (2)如果OD ﹦DB .求证:AF =BF .证明:(1)∵AB ﹦AC , ∴AB AC =. ........................(1分)∵直线AD 经过圆心O , ..................................................(1分) ∴AD ⊥BC ,BD=CD . ....................................................(1分) ∵点E 为弦AB 的中点, ∴DE 是△ABC 的中位线. ∴DE ∥AC . ......................................................................(1分) ∵BF ⊥DE ,∴∠1=90°, ∴∠2=90°.......................................................................(1分) ∴∠CBF +∠ACB ﹦90°.∵AB ﹦AC ,∴∠ABC ﹦∠ACB , .....................................(1分)∴∠CBF +∠ABC ﹦90°..................................................(1分)又∵AD ⊥BC ,∴∠BAD +∠ABC ﹦90°,∴∠BAD ﹦∠CBF ..............................................................(1分)(2)联结OB .∵AD ⊥BC ,OD ﹦DB ,∴△ODB 是等腰直角三角形........................................................................................................(1分)∴∠BOD ﹦45°. ∵OB=OA ,∴∠OBA ﹦∠OAB .∵∠BOD ﹦∠OBA +∠OAB ,∴∠BAO=12∠BOD=22.5°. .....................................................................................................(1分)∵AB=AC ,且AD ⊥BC , ∴∠BAC=2∠BAO=45°. ∵∠2=90°,即BF ⊥AC ,∴在△ABF 中,∠ABF =180904545--=,................................................................................(1分)图6BCDEF OA· 图6 B C DE F O A·12OE第23题图 C A B D F∴∠ABF =∠BAC ,∴AF =BF ..........................................................................................................................................(1分) 虹口23.(本题满分12分,第(1)小题6分,第(2)小题6分)如图,在□ABCD 中,AC 与BD 相交于点O ,过点B 作BE ∥AC ,联结OE 交BC 于点F ,点F 为BC 的中点.(1)求证:四边形AOEB 是平行四边形;(2)如果∠OBC =∠E ,求证:=BO OC AB FC ⋅⋅.23.(1)证明:∵BE ∥AC ∴OC CFBE BF=∵点F 为BC 的中点 ∴CF=BF ∴OC=BE ∵四边形ABCD 是平行四边形 ∴AO=CO ∴AO=BE∵BE ∥AC ∴四边形AOEB 是平行四边形(2)证明:∵四边形AOEB 是平行四边形 ∴∠BAO =∠E ∵∠OBC =∠E ∴∠BAO =∠OBC∵∠ACB =∠BCO ∴△COB ∽△CBA ∴BO BC AB AC =∵四边形ABCD 是平行四边形 ∴AC =2OC ∵点F 为BC 的中点 ∴BC =2FC ∴BO FC AB OC= 即=BO OC AB FC⋅⋅青浦23.(本题满分12分,第(1)、(2)小题,每小题6分)已知:如图9,在菱形ABCD 中,AB =AC ,点E 、F 分别在边AB 、BC 上,且AE =BF ,CE 与AF 相交于点G . (1)求证:∠FGC =∠B ;(2)延长CE 与DA 的延长线交于点H ,求证:.23.证明:(1)∵四边形ABCD 是菱形, ∴AB =BC . ··········································································· (1分)∵AB =AC ,∴AB =BC =AC ,∴∠B =∠BAC =60°. ··························· (1分) 在△EAC 与△FBA 中,∵EA =FB ,∠EAC =∠FBA ,AC =BA , ∴△EAC ≌△FBA , ································································ (1分) ∴∠ACE =∠BAF ,·································································· (1分) ∵∠BAF+∠F AC =60°,∴∠ACE +∠F AC =60°,∴∠FGC =60°, ······· (1分) ∴∠FGC =∠B . ····································································· (1分) (2)∵四边形ABCD 是菱形,∴∠B =∠D ,AB =DC ,AB //DC , ················································ (1分) ∴∠BEC =∠HCD , ································································· (1分) ∴△BEC ∽△DCH , ······························································· (1分)∴=BE ECDC CH, ····································································· (1分) ∴⋅=⋅BE CH EC DC .∵AB =AC ,∴CD =AC , ··························································· (1分) ∵△EAC ≌△FBA , ∴EC =F A ,∴⋅=⋅BE CH AF AC . ························································· (1分)BE CH AF AC ⋅=⋅GF EDA BC图9。