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八年级数学参考答案第1页（共6页）2020-2021学年第一学期八年级期中测试-数学试题卷参考答案及评分建议一、选择题：本题共10小题，每小题4分，共40分．12345678910BCACDBADCD二、填空题：

本题共6小题，每小题4分，共24分．11．九12．20°13．1514．1615．=16．8三、解答题：本题共9小题，共86分．17．（本小题满分8分）证明：∵∠BAD=∠CAE，∴∠BAD+∠DAC=∠CAE+∠DAC，即∠BAC=∠DAE．························

·····················································2分在△ABC和△ADE中，BACDAEABADBD，·························································

····················5分∴△ABC≌△ADE(ASA)．···································································

·6分∴BC=DE．·······················································································8分18．（本小题满分8分）解：（1）如图，△A′B′

C′即为所求．····························································5分（2）A′(2，3)B′(3，1)C′(－1，－2)·······································

·················8分19．（本小题满分8分）证明：∵AD⊥BC，∴∠BDF=∠ADC=90°．·······································································2分∵∠ABC=45°，∴

∠BAD=∠ABC=45°，∴BD=AD．·······················································································4分∵AD⊥BC，BE⊥AC，∴∠C+∠DAC=90°

，∠C+∠CBE=90°，八年级数学参考答案第2页（共6页）∴∠CBE=∠DAC．···································································

··········6分在△BDF和△ADC中，CBEDACBDADBDFADC，············································································7分∴△B

DF≌△ADC(ASA)．····································································8分20．（本小题满分8分）证明：∵AD是∠BAC的平分线

，DE⊥AB，DF⊥AC，∴DE=DF，∠AED=∠AFD=90°，∠EAD=∠FAD，∴点D在EF的垂直平分线上．····················································

·········2分在△AED和△AFD中，AEDAFDEADFADADAD，∴△AED≌△AFD(AAS)，····················································

················4分∴AE=AF，∴点A在EF的垂直平分线上，·····························································6分∴AD所在直线是EF的垂直平分线，即AD垂直平分EF．·····

······································································8分21．（本小题满分8分）解：（1）如图，DE所在直线是AC的垂直平分线．······

··················································3分（2）如上图，连接AD．∵AB=AC，∠BAC=120°，∴∠C=∠B=30°．············

································································4分由（1）知，DE所在直线是AC的垂直平分线，∴AD=CD，················································

····································5分∴∠DAC=∠C=30°，∴AD=CD=2DE，∠BAD=120°－30°=90°，············································6分

∴△ABD是直角三角形，且∠B=30°，∴BD=2AD=4DE．∵DE=2cm，∴CD=4cm，BD=8cm，∴BC=BD+CD=12cm，即BC的长为12cm．···············································

························8分八年级数学参考答案第3页（共6页）22．（本小题满分10分）解：（1）∵l1垂直平分AB，∴DB=DA，····························································

························1分同理EA=EC．·················································································2分∵

△ADE的周长为12cm，∴DA+DE+EA=12cm，∴BC=BD+DE+EC=DA+DE+EA=12cm，即BC的长为12cm．···················································

····················4分（2）如图，连接OA，OB，OC．∵l1垂直平分AB，∴OB=OA，····················································································5分同理OA=OC

．∵OA的长为8cm，∴OA=OB=OC=8cm．······································································7分由（1）可

知，BC=12cm，∴△OBC的周长为：OB+OC+BC=8+8+12=28(cm)．······························10分23．（本小题满分10分）证明：（1）∵AD，CE分别是∠BAC，

∠BCA的平分线，∴12FACBAC，12FCABCA．·········································2分∵∠BAC+∠BCA=180°－∠B，∴111()(180)902

22FACFCABACBCABB．···································································································4分∵∠EFA=∠FAC

+∠FCA，∴1902EFAB．································································5分（2）如图，过点F作FG⊥BC于点G，作FH⊥AB于点H，作FM⊥AC于点M

．·································································································6分八年级数学参考答案第4页（共6页）∵AD，CE分别是∠BAC，∠BCA的平分线，∴FG=FM=

FH．由（1）可知，1902EFAB，又∵∠B=60°，∴∠EFA=60°，∴∠AFC=∠EFD=120°，即∠EFH+∠DFH=120°．∵∠DFG+∠DFH=360°－90°×2－60°=120°，∴∠EF

H=∠DFG．·······································································8分在△EFH和△DFG中，90EHFDGFFHFGEFHDFG，∴△EFH≌△D

FG(ASA)，∴EF=DF．···············································································10分24．（本小题满分12分）（1）证明：∵△ABC

是等边三角形，∴∠ABQ=∠CAP=60°，AB=CA．∵点P，Q的运动速度相同，∴AP=BQ．················································································2分在△ABQ与△CAP中，A

BCAABQCAPAPBQ，∴△ABQ≌△CAP(SAS)．·····························································4分（2）解：当点P，Q分别在AB，BC边上运动时，∠QMC的大小不变．

···············5分由（1）可知，△ABQ≌△CAP，∴∠BAQ=∠ACP．·······························································

··········6分∵∠QMC是△ACM的外角，∴∠QMC=∠ACP+∠MAC=∠BAQ+∠MAC=∠BAC．∵△ABC是等边三角形，∴∠BAC=60°，∴∠QMC=60°，即当点P，Q分别在AB，BC边上运动时，∠QMC的度数为60°．·········8分（3）解：当

点P，Q分别在AB，BC的延长线上运动时，∠QMC的大小不变．易证△ABQ≌△CAP，∴∠BAQ=∠ACP．·······································································10分∵∠QMC是△APM的外角，∴∠

QMC=∠BAQ+∠APM=∠ACP+∠APM=180°－∠PAC．∵△ABC是等边三角形，∴∠PAC=60°，∴∠QMC=180°－60°=120°，八年级数学参考答案第5页（共6页）即当点P，Q分别在AB，BC的延长线上运动时，

∠QMC的度数为120°．································································································12分25

．（本小题满分14分）（1）证明：∵AB=AC，D是边BC的中点，∴AD所在直线是BC的垂直平分线，··············································2分∴FB=FC．·········

·······································································3分（2）证明：∵△ACE是等边三角形，∴∠EAC=∠ACE=60°，AC=AE．·····················

·······························4分∵AB=AC，∴AB=AE，∴∠ABF=∠FEA．由（1）可知，FB=FC，又∵AF=AF，AB=AC，∴△ABF≌△ACF(SSS)，·········

·····················································6分∴∠ABF=∠FCA，∴∠FEA=∠FCA．·····························································

·········8分（3）解：如图，延长AD至点P处，使DP=AD，连接CP．∵AB=AC，D是边BC的中点，∴∠ADC=∠PDC=90°．∵△ACE是等边三角形，∴AC=CE，∠EAC=60°．∵AD=DP，∠ADC=∠PDC，CD=CD，∴△ADC≌

△PDC(SAS)，∴AC=CP=CE，∠ACD=∠PCD．····················································10分由（2）可知，∠FEA=∠FCA，∵∠AM

C=∠FME，∴∠EFC=∠EAC=60°．由（1）可知，BF=CF，∴∠BFD=∠CFD=(180°－60°)÷2=60°，∴∠FCD=90°－60°=30°．·················································

·············11分∵∠FCA=∠FCD－∠ACD，∴∠FCA=30°－∠ACD．∵∠ECF=∠ECA－∠FCA，∴∠ECF=∠ECA－(30°－∠ACD)=∠ECA－30°+∠ACD=30°+∠ACD．∵∠FCP=∠FCD+∠PCD，八年级数学参考答案第6页（

共6页）∴∠FCP=30°+∠ACD，∴∠ECF=∠FCP．∵FC=FC，CE=CP，∴△ECF≌△PCF(SAS)，∴FE=FP，··································································

···············12分∴FE=FD+AD，···········································································13分∴FD=FE－

AD=8－2=6，∴AF=FD－AD=4．······································································14分