如图,在△ABC中,AB=AC,以AB为直径的⊙O与边BC、AC分别交于D、E两点, DF AC于F.(1)求证:DF为⊙O
如图,在△ABC中,AB=AC,以AB为直径的⊙O与边BC、AC分别交于D、E两点,DFAC于F.(1)求证:DF为⊙O的切线;(2)若,CF=9,求AE的长....
如图,在△ABC中,AB=AC,以AB为直径的⊙O与边BC、AC分别交于D、E两点, DF AC于F.(1)求证:DF为⊙O的切线;(2)若 ,CF=9,求AE的长.
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试题分析:(1)连接OD,AD,求出OD∥AC,推出OD⊥DF,根据切线的判定推出即可. (2)求出CD、DF,推出四边形DMEF和四边形OMEN是矩形,推出OM=EN,EM=DF=12,求出OM,即可求出答案. 试题解析:(1)连接OD,AD, ∵AB是⊙的直径,∴∠ADB=90°. 又∵AB=AC,∴BD=CD. 又∵OB=OA,∴OD∥AC. ∵DF⊥AC,∴OD⊥DF. 又∵OD为⊙的半径,∴DF为⊙O的切线. (2)连接BE交OD于M,过O作ON⊥AE于N,则AE=2NE, ∵ ![](https://iknow-pic.cdn.bcebos.com/43a7d933c895d1432aedb09170f082025baf07ec?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,CF=9,∴DC=15.∴ ![](https://iknow-pic.cdn.bcebos.com/d439b6003af33a87108c54a8c55c10385243b59b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . ∵AB是直径,∴∠AEB=∠CEB=90°. ∵DF⊥AC,OD⊥DF,∴∠DFE=∠FEM=∠MDF=90°.∴四边形DMEF是矩形. ∴EM=DF=12,∠DME=90°,DM=EF.即OD⊥BE. 同理四边形OMEN是矩形,∴OM=EN. ∵OD为半径,∴BE=2EM=24. ∵∠BEA=∠DFC=90°,∠C=∠C,∴△CFD∽△CEB. ∴ ![](https://iknow-pic.cdn.bcebos.com/5fdf8db1cb13495435febcd6554e9258d0094a9b?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/0e2442a7d933c895579eccc9d21373f08302009b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . ∴EF=9=DM. 设⊙O的半径为R, 则在Rt△EMO中,由勾股定理得: ![](https://iknow-pic.cdn.bcebos.com/8ad4b31c8701a18b377146fd9d2f07082938feee?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/f7246b600c3387440d5f7bce520fd9f9d62aa0ee?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . 则EN=OM= ![](https://iknow-pic.cdn.bcebos.com/c2fdfc039245d688bee1e616a7c27d1ed31b24a4?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . ∴AE=2EN=7. |
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