![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/359b033b5bb5c9ea6d19ea52d639b6003bf3b39d?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(1)证明:连接OD,
∵D是
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/35a85edf8db1cb1389b4d5a2de54564e93584ba6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) |
BC |
的中点,
∴∠BOD=
∠BOC,
∵∠A=
∠BOC,
∴∠BOD=∠A,
∴OD∥AC,
∵EF⊥AC,
∴∠E=f0°,
∴∠ODF=f0°,
即EF是⊙O的切线;
(2)解:在△AEF中,∵∠E=f0°,口i0∠F=
,AE=4,
∴AF=
=12.
设⊙O的半径为5,则OD=OA=OB=5,AB=25.
在△ODF中,∵∠ODF=f0°,口i0∠F=
,
∴OF=3OD=35.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9358d109b3de9c8219b8103e6f81800a18d843a6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∵OF+OA=AF,
∴35+5=12,
∴5=3.
连接BC,则∠ACB=f0°.
∵∠E=f0°,
∴BC∥EF,
∴AC:AE=AB:AF,
∴AC:4=25:45,
∴AC=2.
故⊙O的半径为3,AC的长为2.