江苏省徐州市2022-2023学年八年级上学期期中检测数学试卷（含答案）

这是一份江苏省徐州市2022-2023学年八年级上学期期中检测数学试卷（含答案），共7页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。

2022-2023学年度第一学期八年级数学学科期中抽测 评分标准一、选择题（本大题有8小题，每小题3分，共24分）题号12345678选项BABCCDAD二、填空题（每题共8小题，每题4分，共32分．说明：第16题只看对1个2分，对2个3分，对3个4分） 9．   30        10．     100         11．   10        12．∠B=∠C（答案不唯一）          13．  15        14．     20        15．   90      16．  70°或40°或55°    三、解答题（本大题有9小题，共84分）
17．（每空1分，共8分）已知，∠ABO=∠CDO，∠ABO=∠CDO，∠AOB=∠COD，对顶角相等，AAS，全等三角形的对应边相等               18．∵AD平分∠BAC，·····························1分DE⊥AB，DF⊥AC，垂足为E，F·····························2分∴DE=DF，······················3分∠BED=∠CFD=90°，·············4分在Rt△BED和Rt△CFD中，∠BED=∠CFD=90°，·····························6分∴Rt△BED≌Rt△CFD，············7分∴BE=CF，··················8分19．（1）如图所示··············4分（每个点1分）    （2）如图所示················7分（3）16···················10分          20． （1）证明：∵AE∥CF，∴∠AEF=∠CFE，················1分∴∠AEB=∠CFD，················2分∵DE=BF，∵DF=BE，······················4分在△ABE和△CFD中，·····························7分∴△ABE≌△CFD，···············8分（2）AB=CD，AB∥CD·············10分21．证明：在等边△ABC中，AB=BC，∠ABC=∠ACB=60°， 2分∵BD是边AC上的高，∴∠ABD=∠CBD=30°,···············4分∵CE=CD，∴∠CDE=∠CED，··················6分∵∠ACB是△CDE的外角，∴∠CED=30°，··················8分∴∠DBC=∠CED，··················9分∴BD＝DE． 10分22．（1）在△ABC中，∵，，······························································3分∴，································································4分故△ABC是直角三角形，且∠ACB=90°，····································5分∵点D是AB的中点，∴CD=BD=,···························································7分∴∠B=∠BCD=50°，···················································8分∴∠DCA=∠ACB－∠BCD=90°－50°=40°．·································9分（2）······························································12分
23．（1）如图·································6分（2）解：∵                                ······································8分           又∵·····································10分             ∴······································12分   24．在△ABC中，∵AB=AC，∠BAC=90°，∴∠B=∠C=45°，  1分又∵点D是BC的中点，∴AD=BD=，且AD⊥BC，∠BAD=∠CAD= 2分∴∠ADN+∠BDN=90°，又∵△DEF是直角三角尺，∴∠EDF =90°，即∠ADN+∠ADM=90°,∴∠BDN=∠ADM  3分在△BDN和△ADM中∴△BDN≌△ADM， ···············5分∴DN=DM；··························································6分
（2）∵△BDN≌△ADM∴BN= AM，∠BND=∠AMD，DN=DM········································7分∴∠BNF=∠AME，且由于△DEF是含45°直角三角尺，∴DF=DE，∴DF－DN=DE－DM即FN=EM····························································8分在△BNF和△AME中∴△BNF≌△AME，∴AE=BF；···························································9分
（3）作图正确（如图所示）·············································10分猜想：AE⊥BF，理由如下：··············································11分∵△BNF≌△AME，∴∠BFN=∠AEM， ∵∠FDE=90°，∴∠AEM+∠APD=90°又∵∠APD=∠FPQ，···················································12分∴∠FPQ+∠BFN=90°，∴∠FQP =90°，······················································13分∴AE⊥BF．··························································14分