證明 ：?$(1)$?連接?$OE. $?
?$∵ AB$?是?$⊙O$?的直徑，
?$∴ ∠ACB=90°.$?
?$∵ CE$?平分?$∠ACB，$?
?$∴ ∠ACE= \frac {1}{2} ∠ACB=45°. $?
?$∵ \widehat{AE}=\widehat{AE}，$?
?$∴ ∠AOE=2∠ACE=90°$?
?$∵ EF//AB，$?
?$∴ ∠AOE+∠FEO=180°，$?
?$∴ ∠FEO=90°，$?
?$∴ OE⊥FE.$?
又?$ ∵ OE$?是?$⊙O$?的半徑，
?$∴ EF $?與?$⊙O$?相切
?$(2)$?連接?$OG、$??$OC. $?
?$∵ ∠CAB=30°，$??$∠ACB=90°，$?
?$∴ ∠B=60°. $?
?$∵ OB=OC，$?
?$∴ △OBC$?為等邊三角形，
?$∴ ∠COB=60°，$?
?$∴ ∠AOC=180°-∠COB=120°. $?
?$∵ EG⊥AC，$??$∠ACE=45°，$?
?$∴ ∠MEC=45°. $?
?$∵\widehat{CG}=\widehat{CG}，$?
?$ ∴ ∠GOC=2∠MEC=90°，$?
?$∴ ∠AOG=∠AOC-∠GOC=30°. $?
?$∵ AB=8，$??$AB$?是?$⊙O$?的直徑，
?$∴ OA=OG=4，$?
?$∴ \widehat{AG}$?的長?$= \frac {30π×4}{180} =\frac {2π}{3}.$?
