2016-2017年河北省承德市承德县初二（上）期末数学试卷含参考答案

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∴BM=CN．

（2）结论BM=CN仍成立．理由如下： ∵△ABC、△AMN是等边三角形， ∴AB=AC，AM=AN，∠BAC=∠MAN=60°， ∴∠BAM=∠CAN， ∵在△BAM和△CAN中，

，

∴△BAM≌△CAN（SAS）， ∴BM=CN．

26．（14分）如图，在△ABC中，AB=AC，D是BA延长线上的一点，点E是AC的中点．

（1）操作：利用尺规按下列要求作图，并在图中标明相应字母（保留作图痕迹，不写作法）．

①作∠DAC的平分线AM．②连接BE并延长交AM于点F．

（2）猜想：试猜想AF与BC有怎样的位置关系和数量关系，并说明理由．（3）探究：当AF与EC有怎样的数量关系时，△ABC是等边三角形，并说明理由．

【解答】解：（1）如图所示：

第17页（共19页）

（2）∵AB=AC， ∴∠ABC=∠ACB， ∵AM平分∠DAC， ∴∠DAF=∠FAC=∠DAC， ∵∠ABC+∠ACB=∠DAC， ∴∠ABC=∠ACB=∠DAC， ∴∠DAF=∠ABC， ∴AF∥BC， ∴∠AFE=∠CBE， ∵E是AC的中点， ∴AE=EC，

在△AEF和△CEB中，

，

∴△AEF≌△CEB（AAS）， ∴AF=BC．

（3）当AF=2EC时，△ABC是等边三角形．理由：∵AF=2EC，EC=AE， ∴AF=AC，

∵AF=BC，AB=AC， ∴AB=AC=BC，

∴△ABC是等边三角形．

第18页（共19页）

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