初中数学新教师培训第三次考核试卷教师版

1、新教师培训第四次试卷时间：60 分钟满分：100 分（10 分）1.下面是“以已知线段为直径作圆”的尺规作图过程.请回答：该作图的依据是：垂直平分线的判定；垂直平分线的圆的定义（10 分）2.如图，CD 为O 的直径，点 B 在O 上，连接 BC、BD，过点 B 的切线 AE 与CD 的延长线交于点 A， ÐAEO = ÐC ，OE 交 BC 于点 F.（1）求证：OEBD；（2）当O 的半径为 5， sin ÐDBA = 2 时，求 EF 的长.5已知：线段 AB.A求作：以 AB 为直径的O. 作法：如图，1（1） 分别以 A，B 为圆心，大于 AB 的长为半

2、径2作弧，两弧相交于点 C，D；（2） 作直线 CD 交 AB 于点 O；EBFCADO(1) 证明：连接 OBCD 为O 的直径 ， ÐCBD = ÐCBO +ÐOBD = 90° .AE 是O 的切线， ÐABO = ÐABD +ÐOBD = 90° .1 分 ÐABD = ÐCBO .EBFCADOOB、OC 是O 的半径，OB=OC. ÐC = ÐCBO . ÐC = ÐABD . ÐE = ÐC ， ÐE = 

3、08;ABD . OEBD2 分2(2)解：由（1）可得 sinC= DBA=，5= BD = 2CD5在 Rt OBE 中, sinC,OC=5BD = .43 分 ÐCBD = ÐEBO = 90° ， ÐE = ÐC ，CBDEBO.BD = CD. BOEOEO = 252.4 分OEBD，CO=OD，CF=FB.OF = 1 BD = 2.2EF = OE - OF = 21.2 5 分（40 分）3.在ABC 中，AB=AC，A=60°，点 D 是 BC 边的中点，作射线 DE，与边 AB交于点 E，射线 DE 绕点 D

4、顺时针旋转 120°，与直线 AC 交于点 F（1）依题意将图 1 补全；（2）通过观察、实验提出猜想：在点 E 运动的过程中，始终有 DE=DF把这个猜想与同学们进行交流，通过讨论，形成了证明该猜想的几种想法：想法 1：由点 D 是 BC 边的中点，通过构造一边的平行线，利用全等三角形，可证 DE=DF；想法 2：利用等边三角形的对称性，作点 E 关于线段 AD 的对称点 P，由BAC 与EDF 互补，可得AED 与AFD 互补，由等角对等边，可证 DE=DF；想法 3：由等腰三角形三线合一，可得 AD 是BAC 的角平分线，由角平分线定理，构造点D 到 AB，AC 的高，利用全等

5、三角形，可证 DE=DF.请你参考上面的想法，帮助证明 DE=DF（选法即可）；（3）在点 E 运动的过程中，直接写出 BE，CF，AB 之间的数量关系AAEEBB图 1备用图解：（1）如图 1， ·································

6、83;·················································

7、83;············1AEFCBD图 1（2）AAAEMPFEENFCCBBD图 3BD图 4图 2想法 1 证明：如图 2，过 D 作 DGAB，交 AC 于 G， ·························&

8、#183;······················2点 D 是 BC 边的中点，1DG=AB2CDG 是等边三角形EDB+EDG=120°FDG+EDG=120°，EDB =FDG ··············

9、83;·················································

10、83;····························3BD=DG，B=FGD=60°，BDEGDF ················

11、3;·················································

12、3;·························4DE=DF ·······················

13、83;·················································

14、83;······························5想法 2 证明：如图 3，连接 AD，点 D 是 BC 边的中点，AD 是ABC 的对称轴作点 E 关于线段 AD 的对称点 P，点 P 在边 AC 上， ······&#

15、183;············································2ADEADPDE=DP，AED=APDBAC+EDF=180&

16、#176;，AED+AFD=180°APD+DPF=180°，AFD=DPF ··········································

17、;··················································

18、;··3DP=DF···············································

19、;··················································

20、;·······4DE=DF ·········································

21、3;·················································

22、3;············5想法 3 证明：如图 4，连接 AD，过 D 作 DMAB 于 M，DNAB 于 N， ························2点 D 是 BC 边的中点，AD 平分BACDMAB 于 M，DNAB

23、于 N，DM=DN················································

24、83;·················································

25、83;····3A=60°，MDE+EDN=120°FDN+EDN=120°，MDE=FDNRtMDERtNDF ··································

26、3;·················································

27、3;··4DE=DF ··············································

28、83;·················································

29、83;·······51（3）当点 F 在 AC 边上时， BE + CF =AB ； ································2·······················6当点 F 在 AC 延