(1)证明:∵C是弧BC的中点,∠AOB=l20°
∴∠AOC=∠BOC=60°,
又∵OA=OC=OB,
∴△OAC和△OBC都是等边三角形,
∴AC=OA=OB=BC,
∴四边形OACB是菱形.
∵∠AOC=60°,
∴
=
,
∴AM=
AO,
∴AB=
AO,
∵CO=AO,
∴
=
;
(2)解:连接CO并延长交⊙O于点M,连接BM,过点B作BF⊥CM于点F,
∵∠CDB=∠CMB,
sin∠CDB=,
∴sin∠CMB=
=
,
设BC=x,则MC=3x,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/503d269759ee3d6d1a97507a40166d224f4ade2a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
故BM=2
x,
∴BF?MC=BC?BM,
∴BF=
=