初中數(shù)學新教師培訓第三次考核試卷教師版

1、新教師培訓第四次試卷時間：60 分鐘滿分：100 分（10 分）1.下面是“以已知線段為直徑作圓”的尺規(guī)作圖過程.請回答：該作圖的依據(jù)是：垂直平分線的判定；垂直平分線的圓的定義（10 分）2.如圖，CD 為O 的直徑，點 B 在O 上，連接 BC、BD，過點 B 的切線 AE 與CD 的延長線交于點 A， ÐAEO = ÐC ，OE 交 BC 于點 F.（1）求證：OEBD；（2）當O 的半徑為 5， sin ÐDBA = 2 時，求 EF 的長.5已知：線段 AB.A求作：以 AB 為直徑的O. 作法：如圖，1（1） 分別以 A，B 為圓心，大于 AB 的長為半

2、徑2作弧，兩弧相交于點 C，D；（2） 作直線 CD 交 AB 于點 O；EBFCADO(1) 證明：連接 OBCD 為O 的直徑 ， ÐCBD = ÐCBO +ÐOBD = 90° .AE 是O 的切線， ÐABO = ÐABD +ÐOBD = 90° .1 分 ÐABD = ÐCBO .EBFCADOOB、OC 是O 的半徑，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°，點 D 是 BC 邊的中點，作射線 DE，與邊 AB交于點 E，射線 DE 繞點 D

4、順時針旋轉(zhuǎn) 120°，與直線 AC 交于點 F（1）依題意將圖 1 補全；（2）通過觀察、實驗提出猜想：在點 E 運動的過程中，始終有 DE=DF把這個猜想與同學們進行交流，通過討論，形成了證明該猜想的幾種想法：想法 1：由點 D 是 BC 邊的中點，通過構(gòu)造一邊的平行線，利用全等三角形，可證 DE=DF；想法 2：利用等邊三角形的對稱性，作點 E 關(guān)于線段 AD 的對稱點 P，由BAC 與EDF 互補，可得AED 與AFD 互補，由等角對等邊，可證 DE=DF；想法 3：由等腰三角形三線合一，可得 AD 是BAC 的角平分線，由角平分線定理，構(gòu)造點D 到 AB，AC 的高，利用全等

5、三角形，可證 DE=DF.請你參考上面的想法，幫助證明 DE=DF（選法即可）；（3）在點 E 運動的過程中，直接寫出 BE，CF，AB 之間的數(shù)量關(guān)系A(chǔ)AEEBB圖 1備用圖解：（1）如圖 1， ·································

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

7、83;············1AEFCBD圖 1（2）AAAEMPFEENFCCBBD圖 3BD圖 4圖 2想法 1 證明：如圖 2，過 D 作 DGAB，交 AC 于 G， ·························&

8、#183;······················2點 D 是 BC 邊的中點，1DG=AB2CDG 是等邊三角形EDB+EDG=120°FDG+EDG=120°，EDB =FDG ··············

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

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

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14、83;······························5想法 2 證明：如圖 3，連接 AD，點 D 是 BC 邊的中點，AD 是ABC 的對稱軸作點 E 關(guān)于線段 AD 的對稱點 P，點 P 在邊 AC 上， ······&#

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

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

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

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

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22、3;············5想法 3 證明：如圖 4，連接 AD，過 D 作 DMAB 于 M，DNAB 于 N， ························2點 D 是 BC 邊的中點，AD 平分BACDMAB 于 M，DNAB

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

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

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

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29、83;·······51（3）當點 F 在 AC 邊上時， BE + CF =AB ； ································2·······················6當點 F 在 AC 延