$证明:连接AD,延长DO交⊙O于点E,则 ∠AOD=∠BOE$
$所以{\widehat{AD}}={\widehat{BE}}$
$因为OD⊥AC, 所以{\widehat{AD}}={\widehat{CD}},所以{\widehat{CD}}={\widehat{BE}}$
$所以{\widehat{CD}}+{\widehat{BC}}= {\widehat{BE}}+{\widehat{BC}},所以{\widehat{BD}}={\widehat{CE}}$
$所以∠BAD=∠ODC$
$因 为∠BOD=2∠BAD,所以∠BOD=2∠ODC.$
