證明?$：(1)$?如圖，連接?$AO$?并延長，交?$BC$?于點?$F，$?連接?$OC，$?
則?$OA=OB=OC，$?
?$∵ AB=AC，$??$OB=OC， $?
∴ 點?$A、$??$O$?在?$BC$?的垂直平分線上，
?$∴ AF⊥BC， $?
?$∴ ∠AFB=90°. $?
?$∵ AE//BC，$?
?$∴ ∠OAE=∠AFB=90°， $?
?$∴ OA⊥AE. $?
?$∵ OA$?是?$⊙O$?的半徑，
?$∴ AE$?是?$⊙O$?的切線
?$(2)∵AB=AC，$?
?$∴∠ACB=∠ABC=75°，$?
∴ 在?$△ABC，$?中，?$∠BAC=180°-∠ACB-∠ABC=30°.$?
?$∵ \widehat{BC}=\widehat{BC}，$?
?$∴ ∠BOC=2∠BAC=60°，$?
?$∴ ∠COD=180°-∠BOC=120°.$?
?$∵OB=OC，∠BOC=60°，$?
?$∴△BOC$?是等邊三角形，
?$∴OC=BC=2，$?
?$∴\widehat{CD}$?的長?$=\frac {120π×2}{180}=\frac {4π}{3}.$
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