如图,以△ABC的边AB为直径作⊙O,交BC于点D,且∠DAC=∠B.(1)求证:AC是⊙O的切线;(2)若点E是 的
如图,以△ABC的边AB为直径作⊙O,交BC于点D,且∠DAC=∠B.(1)求证:AC是⊙O的切线;(2)若点E是的中点,连接AE交BC于点F,当BD=5,CD=4时,求...
如图,以△ABC的边AB为直径作⊙O,交BC于点D,且∠DAC=∠B.(1)求证:AC是⊙O的切线;(2)若点E是 的中点,连接AE交BC于点F,当BD=5,CD=4时,求AF的值.
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(1)证明见解析;(2) ![](https://iknow-pic.cdn.bcebos.com/bf096b63f6246b6071474fc6e8f81a4c510fa233?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . |
试题分析:(1)证明△ADC∽△BAC,可得∠BAC=∠ADC=90°,继而可判断AC是⊙O的切线. (2)根据(1)所得△ADC∽△BAC,可得出CA的长度,继而判断∠CFA=∠CAF,利用等腰三角形的性质得出AF的长度,继而得出DF的长,在Rt△AFD中利用勾股定理可得出AF的长. (1)∵AB是⊙O的直径, ∴∠ADB=∠ADC=90°, ∵∠B=∠CAD,∠C=∠C, ∴△ADC∽△BAC, ∴∠BAC=∠ADC=90°, ∴BA⊥AC, ∴AC是⊙O的切线. (2)∵BD=5,CD=4, ∴BC=9, ∵△ADC∽△BAC(已证), ∴ ![](https://iknow-pic.cdn.bcebos.com/1c950a7b02087bf45280aa5bf1d3572c11dfcf10?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,即AC 2 =BC×CD=36, 解得:AC=6, 在Rt△ACD中,AD= ![](https://iknow-pic.cdn.bcebos.com/0bd162d9f2d3572c4d8e616e8913632762d0c310?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∵∠CAF=∠CAD+∠DAE=∠ABF+∠BAE=∠AFD, ∴CA=CF=6, ∴DF=CA-CD=2, 在Rt△AFD中,AF= ![](https://iknow-pic.cdn.bcebos.com/622762d0f703918fc3bdc078523d269759eec410?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . |
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