![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/8ad4b31c8701a18bb1e3fcfe9d2f07082838fe73?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(1)证明:如图1,连接OE.
∵点E是弧CB的中点,
∴∠CAE=∠EAO=∠OEA,
∴OE∥AC.
又∵EF⊥AC于F,
∴OE⊥EF.
又∵OE是⊙O的半径,
∴EF是圆点O的切线;
(2)如图2,连接BE.则BE=CE=6,∠AEB=90°,
又∵AE=8,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/1e30e924b899a901731fb5871e950a7b0208f55d?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴AB=10.
方法一:∵△FAE∽△EAB,
∴AE
2=AF?AB,
∴AF=6.4;
作OM⊥AF于M,则四边形MOEF是正方形,
∴AM=AF-OE=1.4,
∴AC=2AM=2.8,
∴
=
=
=
.
方法二:如图1,连接BC交OE于H,则BC∥EF,
OE⊥BC,则5
2-OH
2=6
2-(5-OH)
2=BH
2,
∴OH=1.4,
∴
=
=
=
.