如图,点C是以AB为直径的⊙O上的一点,AD与过点C的切线互相垂直,垂足为点D. (1)求证:AC平分∠BAD;
如图,点C是以AB为直径的⊙O上的一点,AD与过点C的切线互相垂直,垂足为点D.(1)求证:AC平分∠BAD;(2)若CD=1,AC=,求⊙O的半径长....
如图,点C是以AB为直径的⊙O上的一点,AD与过点C的切线互相垂直,垂足为点D. (1)求证:AC平分∠BAD;(2)若CD=1,AC= ,求⊙O的半径长.
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傷靓哒9530
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(1)见解析;(2) ![](https://iknow-pic.cdn.bcebos.com/d0c8a786c9177f3e3118aabf73cf3bc79f3d5657?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . |
试题分析:(1)连接OC,由OA=OC得∠ACO=∠CAO,由切线的性质得出OC⊥CD,根据垂直于同一直线的两直线平行得到AD∥CO,由平行线的性质得∠DAC=∠ACO,等量代换后可得∠DAC=∠CAO,即AC平分∠BAD. 过点O作OE⊥AC于E.先在Rt△ADC中,由勾股定理求出AD=3,由垂径定理求出AE= ![](https://iknow-pic.cdn.bcebos.com/4bed2e738bd4b31c2b04c20984d6277f9e2ff84c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,再根据两角对应相等的两三角形相似证明△AEO∽△ADC,由相似三角形对应边成比例得到 ![](https://iknow-pic.cdn.bcebos.com/8601a18b87d6277f2e2a648a2b381f30e924fc4c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,求出AO= ![](https://iknow-pic.cdn.bcebos.com/d0c8a786c9177f3e3118aabf73cf3bc79f3d5657?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,即⊙O的半径为 ![](https://iknow-pic.cdn.bcebos.com/d0c8a786c9177f3e3118aabf73cf3bc79f3d5657?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . 试题解析:(1)证明:如图,连接OC, ∵OA=OC,∴∠ACO=∠CAO. ∵CD切⊙O于C,∴OC⊥CD. 又∵AD⊥CD,∴AD∥CO. ∴∠DAC=∠ACO. ∴∠DAC=∠CAO,即AC平分∠BAD. (2)如图,过点O作OE⊥AC于E. 在Rt△ADC中, ![](https://iknow-pic.cdn.bcebos.com/cdbf6c81800a19d8fa2fd2b730fa828ba61e4657?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∵OE⊥AC,∴AE= ![](https://iknow-pic.cdn.bcebos.com/dbb44aed2e738bd41819e483a28b87d6277ff94c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) AC= ![](https://iknow-pic.cdn.bcebos.com/4bed2e738bd4b31c2b04c20984d6277f9e2ff84c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . ∵∠CAO=∠DAC,∠AEO=∠ADC=90°, ∴△AEO∽△ADC. ∴ ![](https://iknow-pic.cdn.bcebos.com/8601a18b87d6277f2e2a648a2b381f30e924fc4c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,即 ![](https://iknow-pic.cdn.bcebos.com/a71ea8d3fd1f4134d91b53cc261f95cad1c85e57?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∴AO= ![](https://iknow-pic.cdn.bcebos.com/d0c8a786c9177f3e3118aabf73cf3bc79f3d5657?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,即⊙O的半径为 ![](https://iknow-pic.cdn.bcebos.com/d0c8a786c9177f3e3118aabf73cf3bc79f3d5657?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . |
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