$证明:(1)连接AC，交OD于点H$
$∵AB是⊙O的直径，∴ AC⊥BC$
$∵ OD 平分∠AOC，∴ ∠AOD=∠COD\ $
$∴ {\widehat{AD}}={\widehat{CD}},∴OD⊥AC,∴OD//BC$
$(2)∵OE//BC,∴△OEF∽△BCF,∴\frac{OE}{BC}=\frac{OF}{BF}=\frac{5}{6}$
$∴设OE=5x，则BC=6x$
$∵AO=OB,OH//BC,∴AH=CH,∴OH=\frac{1}{2}BC=3x$
$∵PB是⊙O的切线，∴∠OBP=90°,∴∠PBO=∠AHO$
$∵∠BOP=∠HOA,∴ △POB∽△AOH,∴\frac{PO}{AO}=\frac{OB}{OH},解得x=\frac{3}{10}$
$经检验，x=\frac{3}{10}是原分式方程的解且符合题意$
$∴OE=\frac{3}{2},∴⊙O的半径为\frac{3}{2}$
