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一、选择题1．（市海淀区八年级期末）某小区有一块边长为a的正方形场地，规划修建两条宽为b的绿化带．方案一如图甲所示，绿化带面积为S甲；方案二如图乙所示，绿化带面积为S乙．设0kSabS甲乙，下列选项

中正确的是bbbbaaaabbbbbbbbaaaabbbb甲乙A．012kB．112kC．312kD．232k答案：B二、填空题2.（市师达中学八年级第一学期第二次月考）3.（西城区二模）如图，在矩形

ABCD中，顺次连接矩形四边的中点得到四边形EFGH.若AB=8，AD=6，则四边形EFGH的周长等于．答案：204.（西城区二模）我们知道：四边形具有不稳定性.如图，在平面直角坐标系xOy中，矩形ABCD的边AB在x轴上，(3,0)A，(4,0)B，边AD长为5.现固定边AB，“

推”矩形使点D落在y轴的正半轴上（落点记为D），相应地，点C的对应点C的坐标为.5、（平谷区第一学期期末）12．已知菱形ABCD中，∠B=60°，AB=2，则菱形ABCD的面积是．答案：23三、解答题6．（

石景山区初三毕业考试）问题:将菱形的面积五等分．小红发现只要将菱形周长五等分，再将各分点与菱形的对角线交点连接即可解决问题．如图，点O是菱形ABCD的对角线交点，5AB，下面是小红将菱形ABCD面积五等分的操作与证明思路，请补充完整.（1）在AB边上取点E，使4AE，连接OA，OE；（

2）在BC边上取点F，使BF，连接OF；（3）在CD边上取点G，使CG，连接OG；（4）在DA边上取点H，使DH，连接OH．由于AE＋＋＋.可证S△AOE==EOFBFOGCGOHDSSS四边形四边形四边形S△HOA.解：3，2，1；„„„„„„2分

EB、BF；FC、CG；GD、DH；HA.„„„„„„4分7、（市师达中学八年级第一学期第二次月考）8.（西城区二模）如图，在Rt△ABC中，90ACB，CD⊥AB于点D，BE⊥AB于点B，BE=CD，连接CE，DE．（1）

求证：四边形CDBE为矩形；（2）若AC=2，1tan2ACD，求DE的长．OHGFEDCBA图1（1）证明：如图2.∵CD⊥AB于点D，BE⊥AB于点B，∴90CDADBE．∴CD∥BE．„„„„„„„„„„„„„1分又∵

BE=CD，∴四边形CDBE为平行四边形．„„„„„2分又∵90DBE，∴四边形CDBE为矩形．„„„„„„„„„„„„„„„„„„3分（2）解：∵四边形CDBE为矩形，∴DE=BC．„„„„„„„„„„„„„„„„„„„„„„„„„4分∵在Rt△ABC中

，，CD⊥AB，可得．∵，∴．∵在Rt△ABC中，，AC=2，，∴．∴DE=BC=4．„„„„„„„„„„„„„„„„„„„„„„„5分9．（石景山区初三毕业考试）如图，在四边形ABCD中，90ABCD°，210BCCD，CEAD于点E．（1）求

证：AECE；（2）若tan3D，求AB的长．（1）证明：（法一）过点B作BH⊥CE于H，如图1．∵CE⊥AD，∴∠BHC＝∠CED＝90°，190D．∵∠BCD＝90°，∴1290，∴2

D．又BC＝CD∴BHC△≌CED△．∴BHCE．∵BH⊥CE，CE⊥AD，∠A＝90°，BACED12BACHDE图2∴四边形ABHE是矩形，∴AEBH．∴AECE．„„„„„„3分（法二）过点C作CH⊥AB交AB的延长线于H．图略，证明略．（2）解：∵四边形ABHE是矩

形，∴ABHE．∵在RtCED△中，tan3CEDDE,设,3DExCEx，∴10210CDx．∴2x．∴2DE,6CE．„„„„„„4分∵2CHDE．∴624ABHE．10．（燕山地区一模）如图

，在△ABC中，D,E分别是AB,AC的中点，BE=2DE,延长DE到点F，使得EF=BE,连接CF．（1）求证：四边形BCFE是菱形；（2）若∠BCF=120°，CE=4，求菱形BCFE的面积．（1）证明：∵点D,E,是AB,AC中点∴DE∥BC,DE=

12BC……………………….1′又BE=2DE,即DE=12BE∴BC=BE又EF=BE∴EF∥BC,EF=BC∴四边形BCFE是平行四边形……………………….2′又EF=BE∴四边形BCFE是菱形……………………….3′（2）∵四边形BCFE是菱形∴BC=BE又∠BCF=120°∴∠BCE=

60°∴△BCE是等边三角形∴连结BF交EC于点O．∴BF⊥EC在Rt△BOC中，BO=32242222OCBC……………………….4′ABCDEF322322121OCBOSBOC∴∴……………………….5′11．（延庆区初

三统一练习）如图，Rt△ABC中，∠ABC=90°，点D，F分别是AC，AB的中点，CE∥DB，BE∥DC．（1）求证：四边形DBEC是菱形；（2）若AD=3，DF=1，求四边形DBEC面积.解：（1）在Rt

△ABC中，∵CE//DC，BE//DC∴四边形DBEC是平行四边形∵D是AC的中点，∠ABC=90°∴BD=DC……1分∴四边形DBEC是菱形……2分（2）∵F是AB的中点∴BC=2DF=2,∠AFD=∠ABC=90°在Rt△AFD中,……3分∴……4分……5分12．（

西城区九年级统一测试）如图，在ABD△中，ABDADB，分别以点B，D为圆心，AB长为半径在BD的右侧作弧，两弧交于点C，分别连接BC，DC，AC，记AC与BD的交点为O．（1）补全图形，求AOB的度数并说明理由;（2）若5AB，3cos5ABD

，求BD的长．FEDCBA38324BCFES菱形BDA解：（1）补全的图形如图2所示．„„„„„„„„„„„„„„„„„„„„„„„1分∠AOB=90．证明：由题意可知BC=AB，DC=AB．∵在△ABD中，=ABD

ADB，∴AB=AD．∴BC=DC=AD=AB．∴四边形ABCD为菱形．„„„„„„„„2分∴AC⊥BD．∴∠AOB=90．„„„„„„„„„„„3分（2）解：∵四边形ABCD为菱形，∴OB=OD．„„„„„„„„„„„„„„„„„„„„„„„„„„4分在Rt

△ABO中，90AOB，AB=5，3cos5ABD，∴cos3OBABABD．∴2=6BDOB．„„„„„„„„„„„„„„„„„„„„„„„5分13.（门头沟区初三综合练习）在矩形ABCD中，连接AC,AC的垂直平分线交AC于点O,分别交AD、B

C于点E、F，连接CE和AF．（1）求证：四边形AECF为菱形；（2）若AB=4，BC=8，求菱形AECF的周长．（1）证明：∵EF是AC的垂直平分线，∴AO=OC，∠AOE=∠COF=90°，„„„„„„„„1分∵四边形ABCD是矩形，∴AD∥BC，∴∠EAO=∠FCO

，在△AEO和△CFO中，∵∠EAO=∠FCO，AO=CO，∠AOE=∠COF，∴△AEO≌△CFO（ASA），FEOABCDFEOABCD图2∴OE=OF．„„„„„2分又∵OA=OC，∴四边形AECF是平行四边形，又∵EF⊥AC，∴平行四边形AECF是菱

形；„„„„„3分（2）设AF=x，∵EF是AC的垂直平分线，∴AF=CF=x，BF=8﹣x，„„„„„„„„„„„„„„„4分在Rt△ABF中，由勾股定理得：AB2+BF2=AF2，42+（8﹣x）2=x2，解得x=5，∴AF=5，∴菱形AECF的周长为20．„„„„„„„5分14

.（通州区一模）答案：15．（平谷区中考统一练习）如图，在□ABCD中，BF平分∠ABC交AD于点F，AE⊥BF于点O，交BC于点E，连接EF．（1）求证：四边形ABEF是菱形；（2）连接CF，若∠ABC

=60°，AB=4，AF=2DF，求CF的长．（1）证明：∵BF平分∠ABC，∴∠ABF=∠CBF．····································································1∵□ABCD，∴AD∥BC．∴∠AFB=∠CBF．∴∠

ABF=∠AFB．∴AB=AF．∵AE⊥BF，∴∠ABF+∠BAO=∠CBF+∠BEO=90°．∴∠BAO=∠BEO．∴AB=BE．∴AF=BE．∴四边形ABEF是平行四边形．∴□ABEF是菱形．··················

················································2（2）解：∵AD=BC，AF=BE，∴DF=CE．∴BE=2CE．∵AB=4，∴BE=4．∴CE=2．过点A作AG

⊥BC于点G．························································3∵∠ABC=60°，AB=BE，∴△ABE是等边三角形．∴BG=GE=2．∴AF=CG=4．····························

·············································4∴四边形AGCF是平行四边形．∴□AGCF是矩形．∴AG=CF．在△ABG中，∠ABC=60°，AB=4，O

ECBDAFGOECBDAF∴AG=23．∴CF=23．··········································································516．（顺义区初三练

习）如图，四边形ABCD中，AD∥BC，∠A=90°，BD=BC，点E为CD的中点，射线BE交AD的延长线于点F，连接CF．（1）求证：四边形BCFD是菱形；（2）若AD=1，BC=2，求BF的长．（1）证明：∵BD=BC，点E是CD的中点，∴∠1=∠2．„„„„„„„„„„„„„„„„

„„„„1分∵AD∥BC，∴∠2=∠3．∴∠1=∠3．„„„„„„„„„„„2分∴BD=DF．∵BD=BC，∴DF=BC．又∵DF∥BC，∴四边形BCFD是平行四边形．∵BD=BC，∴□BCFD是菱形．

„„„„„„„„„„„„„„„„„„„„3分（2）解：∵∠A=90，AD=1，BD=BC=2，∴223ABBDAD．∵四边形BCFD是菱形，∴DF=BC=2．„„„„„„„„„„„„„„„„„„„„„„4分∴AF=AD+DF=3．∴22

3923BFABAF．„„„„„„„„„„„„5分217．（顺义区初三练习）如图，矩形ABCD中，点E是CD延长线上一点，FEABCD321FEABCDDBECAF第21题图且DE=DC，求证：∠E=∠BAC.EABCD证明：∵四边形ABCD是矩

形，∴∠ADC=90，AB∥CD．„„„„„„„„„„„„„„„„„„„1分∵DE=DC，∴AE=AC．„„„„„„„„„„„„„„„„„„„„„„„„„2分∴∠E=∠ACE．„„„„„„„„„„„„„„„„„„„„„„„„3分∵AB∥CD，∴∠BAC=∠A

CE．„„„„„„„„„„„„„„„„„„„„„„„4分∴∠E=∠BAC．„„„„„„„„„„„„„„„„„„„„„„„5分18．（海淀区第二学期练习）如图，□ABCD的对角线,ACBD相交于点O，且AE∥BD，BE∥AC，OE=CD.（1）

求证：四边形ABCD是菱形；（2）若AD=2，则当四边形ABCD的形状是_______________时，四边形AOBE的面积取得最大值是_________________.（1）证明：∵AEBD∥，BEAC∥，∴四边形AEBO是平行四边形.„

„„„„„1分∵四边形ABCD是平行四边形，∴DCAB.∵OECD,∴OEAB.∴平行四边形AEBO是矩形.„„2分∴90BOA.∴ACBD.∴平行四边形ABCD是菱形.„„3分(2)正方形；„„4分2.„5分19.（怀柔区一模）直角三角形ABC中，∠B

AC=90°，D是斜边BC上一点，且AB=AD，过点C作CE⊥AD，交AD的延长线于点E，交AB延长线于点F.CBEOADyx–1–2–3–4–512345–1–2–3–4–512345O(1)求证：∠ACB=∠DCE；(2)

若∠BAD=45°，2+2AF，过点B作BG⊥FC于点G，连接DG．依题意补全图形，并求四边形ABGD的面积．解：(1)∵AB=AD，∴∠ABD=∠ADB，………………………………1分∵∠ADB=∠CDE，∴∠ABD=∠CDE.∵∠BAC=90°，∴∠ABD+∠ACB=

90°.∵CE⊥AE，∴∠DCE+∠CDE=90°.∴∠ACB=∠DCE.…………………………………2分（2）补全图形,如图所示:…………………………3分∵∠BAD=45°,∠BAC=90°,∴∠BAE=∠CAE=45°,∠F=∠ACF=45°,∵AE⊥CF,BG⊥CF,∴AD∥BG.∵

BG⊥CF,∠BAC=90°，且∠ACB=∠DCE,∴AB=BG.∵AB=AD，∴BG=AD.∴四边形ABGD是平行四边形.∵AB=AD∴平行四边形ABGD是菱形.…………………………………………………………………4分设AB=BG=GD=AD=x，∴BF=2BG=2x.∴AB+BF=x+

2x=2+2.∴x=2，过点B作BH⊥AD于H.∴BH=22AB=1.∴S四边形ABDG=AD×BH=2.……………………………………………………………………5分20．（市朝阳区一模）如图，在菱形ABCD中，AC和BD相交于点O，过点O的线段EF与一组对边AB，CD分别

相交于点E，F．(1)求证：AE=CF；（2）若AB=2，点E是AB中点，求EF的长．DHGBECAFDGBECAF解（1）证明：∵四边形ABCD是菱形，∴AO=CO，AB∥CD.„„„„„„„„„„„„„„„„„„„1分∴∠EAO=∠FCO，∠AEO=∠CFO.∴△AOE≌

△COF.„„„„„„„„„„„„„„„„„„„2分∴AE=CF.„„„„„„„„„„„„„„„„„„„„„„„„3分（2）解：∵E是AB中点，∴BE=AE=CF．∵BE∥CF，∴四边形BEFC是平行四边形.„„„„„„„„„„„„„„„4分∵AB=2，∴EF=BC=AB=2．„„

„„„„„„„„„„„„„„„„„„5分21.（市大兴区检测）如图，矩形ABCD的对角线AC、BD交于点O，且DE=OC，CE=OD．（1）求证：四边形OCED是菱形；（2）若∠BAC＝30°，AC＝4，求菱形OCED的面积．（1）证明：∵DE=OC，CE=OD，∴四边形OCED是平行四边形……

…………………………1分∵矩形ABCD，∴AC=BD，OC=12AC，OD=12BD.∴OC=OD.∴平行四边形OCED是菱形………………………………2分（2）解：在矩形ABCD中，∠ABC=90°，∠BAC=30°，AC＝4，∴BC=2.∴AB=DC=23.…………

………………………………………3分连接OE，交CD于点F.∵四边形OCED为菱形，∴F为CD中点.∵O为BD中点，∴OF=12BC=1.∴OE＝2OF＝2…………………………………………………4分∴S菱形O

CED＝12OE·CD=12×2×23＝23…………………………………………………5分22．（丰台区一模）已知：如图，菱形ABCD，分别延长AB，CB到点F，E，使得BF=BA，BE=BC，连接AE，EF，FC，CA．

（1）求证：四边形AEFC为矩形；（2）连接DE交AB于点O，如果DE⊥AB，AB=4，求DE的长．（1）证明：∵BF=BA，BE=BC，∴四边形AEFC为平行四边形.………………………1分∵四边形ABCD为菱

形，∴BA=BC.∴BE=BF.∴BA+BF=BC+BE，即AF=EC.∴四边形AEFC为矩形.………………………2分（2）解：连接DB.由（1）知，AD∥EB，且AD=EB.∴四边形AEBD为平行四边形∵DE⊥AB，∴四边形AEBD为菱形.∴AEEB

，AB2AG，ED2EG.………………………4分∵矩形ABCD中，EBAB，AB=4，∴AG2，AE4.∴Rt△AEG中，EG=23.∴ED=43.………………………5分（其他证法相应给分）23.（昌平区二模）

如图，已知△ACB中，∠ACB=90°，CE是△ACB的中线，分别过点A、点C作CE和AB的平行线，交于点D．（1）求证：四边形ADCE是菱形；（2）若CE=4，且∠DAE=60°，求△ACB的面积．ABCEDFEFDCBAG答案．(1)证明：∵AD//CE，CD//AE∴四边形AECD为平

行四边形„„„„„„„„„1分∵∠ACB=90°，CE是△ACB的中线∴CE=AE„„„„„„„„„„„„„2分∴四边形ADCE是菱形（2）解：∵CE=4，AE=CE=EB∴AB=8，AE=4∵四边形ADCE是菱形,∠DAE=60°∴∠CAE=

30°„„„„„„„„„„„„„3分∵在Rt△ABC中，∠ACB=90°，∠CAB=30°，AB=83cos2ACCABAB，142CBAB∴AC=43„„„„„„„„„„„„„4分∴1832ABCSACBC„„„„„„„„„„„„„„„„„„„5分24.（东城区二模）如

图，在菱形ABCD中，BAD，点E在对角线BD上.将线段CE绕点C顺时针旋转，得到CF，连接DF.（1）求证：BE=DF；（2）连接AC，若EB=EC，求证：ACCF.答案21.(1)证明：∵四边形ABCD是菱形，∴=BCDC，BADBCD∠∠.∵ECF∠，∴BCDEC

F∠.∴=BCEDCF.∵线段CF由线段CE绕点C顺时针旋转得到，∴=CECF.在BEC△和DFC△中，BCDCBCEDCFCECF，，，DECBA∴BEC△≌SASDFC△.∴=.BEDF---

-------------------------------------------------------------------2分(2)解：∵四边形ABCD是菱形，∴ACBACD∠，ACBD.∴+90ACBEBC∠.∵=E

BEC，∴=EBCBCE.由（1）可知，∵=EBCDCF，∴+90DCFACDEBCACB∠.∴90ACF∠.∴ACCF.---------------------------------------------------------------

------5分25、（房山区二模）已知：如图，四边形ABCD中，AD∥BC，AD=CD，E是对角线BD上一点，且EA=EC．（1）求证：四边形ABCD是菱形；（2）如果∠BDC=30°，DE=2，EC=3，求CD的

长．解：（1）∵AD=CD,EA=EC,DE=DE∴△ADE≌△CDE∴∠ADE=∠CDE∵AD∥BC∴∠ADB=∠DBC∴∠DBC=∠BDC∴BC=CD∴AD=BC又∵AD∥BC∴四边形ABCD是平行四边形„„„„„

„„„„„„„„„„„„„„2′∵AD=CD∴四边形ABCD是菱形„„„„„„„„„„„„„„„„„„„„„„3′（2）作EF⊥CD于F∵∠BDC=30°，DE=2∴EF=1,DF=3„„„„„„„„„„„„„„„„„„„„„„„„„„4′∵CE=3ADCBE∴CF=22∴CD=22+3„

„„„„„„„„„„„„„„„„„„„„„„„„5′26.（丰台区二模）如图，BD是△ABC的角平分线，过点D作DE∥BC交AB于点E，DF∥AB交BC于点F．（1）求证：四边形BEDF为菱形；（2）如果∠A=90°，∠C=30°，BD=12，求菱形BEDF的面积．答案

．（1）证明：∵DE∥BC，DF∥AB，∴四边形BEDF为平行四边形………………1分∴∠1=∠3.∵BD是△ABC的角平分线，∴∠1=∠2.∴∠2=∠3.∴BF=DF.∴四边形BEDF为菱形.………………………

2分（2）解：过点D作DG⊥BC于点G，则∠BGD=90°.∵∠A=90°，∠C=30°，∴∠ABC=60°.由（1）知，BF=DF，∠2=30°，DF∥AB，∴∠DFG=∠ABC=60°.∵BD=12，∴在Rt△BDG中，DG=6.∴在Rt△FDG中，DF

=43.………………………4分∴BF=DF=43.∴S菱形BEDF243BFDG.………………………5分（其他证法相应给分）27.（海淀区二模）如图，在四边形ABCD中，ABCD，BD交AC于G，E是BD的中点，连接AE并延长，交CD

于点F，F恰好是CD的中点.（1）求BGGD的值；（2）若CEEB，求证：四边形ABCF是矩形.答案．（1）解：∵AB∥CD，∴∠ABE=∠EDC.∵∠BEA=∠DEF，∴△ABE∽△FDE.∴ABBEDFDE.∵E是BD的中点，FDECBAG321ABCEDFEGFA

BCD∴BE=DE.∴AB=DF.∵F是CD的中点，∴CF=FD.∴CD=2AB.∵∠ABE=∠EDC，∠AGB=∠CGD，∴△ABG∽△CDG.∴12BGABGDCD.（2）证明：∵AB∥CF，AB=CF，∴四边形ABC

F是平行四边形.∵CE=BE，BE=DE，∴CE=ED.∵CF=FD，∴EF垂直平分CD.∴∠CFA=90°.∴四边形ABCF是矩形.