證明?$：(1)∵\(yùn)widehat{AB}=\widehat{AB}$?
?$∴∠ACB=\frac {1}{2}∠AOB$?
?$∵\(yùn)widehat{BC}=\widehat{BC}$?
?$∴∠BAC=\frac {1}{2}∠BOC$?
?$∵∠ACB=2∠BAC$?
?$∴∠AOB=2∠BOC$?
?$(2)$?如圖，過點(diǎn)?$O$?作?$OD⊥AB $?于點(diǎn)?$E，$?交?$⊙O $?于點(diǎn)?$D，$?連接?$BD，$?
則?$∠DOB=\frac {1}{2}∠AOB，AE=BE. $?
?$∵ ∠AOB=2∠BOC，$?
?$∴∠DOB=∠BOC，$?
?$∴ BD=BC.$?
?$∵AB=4，BC=\sqrt{5}$?
?$∴BE=2，DB=\sqrt{5}$?
∵ 在?$Rt△BDE $?中?$，∠DEB=90°，$?
?$∴DE= \sqrt{BD2-BE2}=1. $?
∵ 在?$Rt△BOE$?中?$，∠OEB=90°，$?
?$∴OB2=OE2+BE2，$?
即?$OB2=(OB-1)2+22，$?
?$∴ OB=\frac {5}{2}，$?
即?$⊙O$?的半徑是?$\frac {5}{2}$
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