如图,Rt△ABC中,∠ABC=90°,以AB为直径的⊙O交AC于点D,E是BC的中点,连接DE、OE. (1)判断DE与⊙
如图,Rt△ABC中,∠ABC=90°,以AB为直径的⊙O交AC于点D,E是BC的中点,连接DE、OE.(1)判断DE与⊙O的位置关系并说明理由;(2)求证:(3)若ta...
如图,Rt△ABC中,∠ABC=90°,以AB为直径的⊙O交AC于点D,E是BC的中点,连接DE、OE. (1)判断DE与⊙O的位置关系并说明理由; (2)求证: (3)若tanC= ,DE=2,求AD的长.
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情轻au7GD
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(1)证明∠EDO=∠EBO=90°,所以DE与⊙O相切 (2)通过证明AC="2OE" ,BC 2 =CD·AC得BC 2 =2CD·OE (3) |
试题分析:(1) DE与⊙O相切 理由如下:连接OD,BD, ∵AB是直径,∴∠ADB=∠BDC=90° ∵E是BC的中点,∴DE=BE=CE,∴∠EDB=∠EBD, ∵OD=OB,∴∠OBD=∠ODB. ∴∠EDO=∠EBO=90° ∴DE与⊙O相切 (2)证明:由题意得OE是的 ![](https://iknow-pic.cdn.bcebos.com/ae51f3deb48f8c54be98969d39292df5e0fe7f3c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ABC的中位线,∴AC=2OE ∵∠ABC=∠BDC=90 0 ,∠C=∠C ,∴ ![](https://iknow-pic.cdn.bcebos.com/ae51f3deb48f8c54be98969d39292df5e0fe7f3c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ABC∽ ![](https://iknow-pic.cdn.bcebos.com/ae51f3deb48f8c54be98969d39292df5e0fe7f3c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) BDC ∴ ![](https://iknow-pic.cdn.bcebos.com/a8014c086e061d95bddf648a78f40ad163d9cae3?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,∴BC 2 =CD·AC,∴BC 2 =2CD·OE (3) ∵DE=2 BC=4 AB=4. tanC tanA= ![](https://iknow-pic.cdn.bcebos.com/6f061d950a7b0208c2506c5361d9f2d3562cc8e3?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , 设BD=AD ![](https://iknow-pic.cdn.bcebos.com/03087bf40ad162d94e7731ae12dfa9ec8b13cde3?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , 点评:本题考查直线与圆相切,相似三角形,三角函数,要求学生掌握直线与圆相切,会证明直线与圆相切,熟悉相似三角形的判定方法,会证明两个三角形相似 |
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