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(1)证明见解析;(2)7.
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://img.yulucn.com/upload/a/aa/aaa319163e17925407c660cfbce9e391_thumb.jpg)
,CF=9,∴DC=15.∴
![](https://img.yulucn.com/upload/6/fd/6fdc3c58c135a6499a4a586d91543b01_thumb.jpg)
.
∵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://img.yulucn.com/upload/7/1c/71c51ec5e460b4c58cd143ad94ae9487_thumb.jpg)
,即
![](https://img.yulucn.com/upload/d/5e/d5e8c2fb0fbbde50c71a5722823befe1_thumb.jpg)
.
∴EF=9=DM.
设⊙O的半径为R,
则在Rt△EMO中,由勾股定理得:
![](https://img.yulucn.com/upload/6/54/654a74ac4b002621e9c8f463e982a169_thumb.jpg)
,解得:
![](https://img.yulucn.com/upload/1/3a/13ab0c4798e89de28d15f1629fd88820_thumb.jpg)
.
则EN=OM=
![](https://img.yulucn.com/upload/3/eb/3ebf8cd1fd7998b2328bdcc027bfd3de_thumb.jpg)
.
∴AE=2EN=7.
1年前
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