如图,AB是⊙O的直径,C是⊙O上的一点,过点A作AD⊥CD于点D,交⊙O于点E,且 = .(1)求证:CD是⊙O的
如图,AB是⊙O的直径,C是⊙O上的一点,过点A作AD⊥CD于点D,交⊙O于点E,且=.(1)求证:CD是⊙O的切线;(2)若tan∠CAB=,BC=3,求DE的长....
如图,AB是⊙O的直径,C是⊙O上的一点,过点A作AD⊥CD于点D,交⊙O于点E,且 = .(1)求证:CD是⊙O的切线;(2)若tan∠CAB= ,BC=3,求DE的长.
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(1)证明见解析;(2) ![](https://iknow-pic.cdn.bcebos.com/a08b87d6277f9e2f97b5acba1c30e924b999f384?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . |
试题分析:(1)连结OC,由 ![](https://iknow-pic.cdn.bcebos.com/09fa513d269759ee91d0d179b1fb43166c22df84?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,根据圆周角定理得∠1=∠2,而∠1=∠OCA,则∠2=∠OCA,则可判断OC∥AD,由于AD⊥CD,所以OC⊥CD,然后根据切线的判定定理得到CD是⊙O的切线; (2)连结BE交OC于F,由AB是⊙O的直径得∠ACB=90°,在Rt△ACB中,根据正切的定义得AC=4,再利用勾股定理计算出AB=5,然后证明Rt△ABC∽Rt△ACD,利用相似比先计算出AD= ![](https://iknow-pic.cdn.bcebos.com/86d6277f9e2f0708b9859bb2ea24b899a801f284?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,再计算出CD= ![](https://iknow-pic.cdn.bcebos.com/43a7d933c895d143c925559170f082025aaf0734?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ;根据垂径定理的推论由 ![](https://iknow-pic.cdn.bcebos.com/09fa513d269759ee91d0d179b1fb43166c22df84?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 得OC⊥BE,BF=EF,于是可判断四边形DEFC为矩形,所以EF=CD= ![](https://iknow-pic.cdn.bcebos.com/43a7d933c895d143c925559170f082025aaf0734?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,则BE=2EF= ![](https://iknow-pic.cdn.bcebos.com/a8014c086e061d9593c6868a78f40ad163d9ca84?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,然后在Rt△ABE中,利用勾股定理计算出AE=,再利用DE=AD﹣AE求解. 试题解析:(1)证明:连结OC,如图, ∵ ![](https://iknow-pic.cdn.bcebos.com/09fa513d269759ee91d0d179b1fb43166c22df84?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∴∠1=∠2, ∵OC=OA, ∴∠1=∠OCA, ∴∠2=∠OCA, ∴OC∥AD, ∵AD⊥CD, ∴OC⊥CD, ∴CD是⊙O的切线; (2)解:连结BE交OC于F,如图, ∵AB是⊙O的直径, ∴∠ACB=90°, 在Rt△ACB中,tan∠CAB= ![](https://iknow-pic.cdn.bcebos.com/7dd98d1001e93901d4e83c8c78ec54e737d1969e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , 而BC=3, ∴AC=4, ∴AB= ![](https://iknow-pic.cdn.bcebos.com/00e93901213fb80e8c3bd06535d12f2eb838949e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∵∠1=∠2, ∴Rt△ABC∽Rt△ACD, ∴ ![](https://iknow-pic.cdn.bcebos.com/203fb80e7bec54e7cf06abacba389b504ec26a9e?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/7aec54e736d12f2e42ef1fd24cc2d5628435689e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,解得 ![](https://iknow-pic.cdn.bcebos.com/55e736d12f2eb9386787cb40d6628535e4dd6f9e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∵ ![](https://iknow-pic.cdn.bcebos.com/37d12f2eb9389b50b21551e08635e5dde6116e9e?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/a8ec8a13632762d052be150da3ec08fa503dc684?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,解得 ![](https://iknow-pic.cdn.bcebos.com/908fa0ec08fa513d8c2add6c3e6d55fbb3fbd984?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∵ ![](https://iknow-pic.cdn.bcebos.com/09fa513d269759ee91d0d179b1fb43166c22df84?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∴OC⊥BE,BF=EF, ∴四边形DEFC为矩形, ∴ ![](https://iknow-pic.cdn.bcebos.com/503d269759ee3d6df846367940166d224e4ade84?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∴ ![](https://iknow-pic.cdn.bcebos.com/3c6d55fbb2fb4316dd9fcbc823a4462308f7d384?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∵AB为直径, ∴∠BEA=90°, 在Rt△ABE中, ![](https://iknow-pic.cdn.bcebos.com/b3fb43166d224f4a9219c2a10af790529922d184?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ∴ ![](https://iknow-pic.cdn.bcebos.com/fcfaaf51f3deb48f878c106cf31f3a292cf5789e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . 【考点】切线的判定. |
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