證明：?$(1)$?連接?$OC. $?
?$∵\widehat{BC}=\widehat{BD}，$?
?$∴ ∠BOD=∠BOC.$?
?$∵ OA=OC，$?
?$∴ ∠OCA=∠A，$?
?$∴ ∠BOC=∠OCA+∠A=2∠A. $?
?$∵ ∠BOD=2∠F，$?
?$∴ ∠A=∠F. $?
?$∵ △AEG、$??$△FBG$?的內角和均為?$180°，$??$∠AGE=∠BGF，$?
?$∴∠AEG=∠GBF.$?
?$∵DE⊥AC，$?
?$∴ ∠AEG=90°，$?
?$∴ ∠GBF=90°，$?
?$∴ OB⊥BF.$?
?$∵ OB$?為?$⊙O$?的半徑，
?$∴ BF$?是?$⊙O$?的切線.
?$(2)△DGB$?為等腰三角形,理由:
?$∵ AB$?是?$⊙O$?的直徑，?$\widehat{BC}=\widehat{BD}，$?
?$∴\widehat{AC}=\widehat{AD}，$?
?$∴ ∠OBD=∠OBC. $?
?$∵ AB$?是?$⊙O$?的直徑，
?$∴ ∠ACB=90°$?
由?$(1)，$?得?$∠AEG=90°，$?
?$∴ ∠ACB=∠AEG，$?
?$∴ EF//BC，$?
?$∴ ∠DGB=∠OBC，$?
?$∴ ∠DGB=∠OBD，$?
?$∴ DB=DG，$?
?$∴ △DBG$?為等腰三角形
?$(3)$?由?$(1)，$?得?$∠GBF=90°，$?
由?$(2)，$?得?$∠DGB =∠OBD，$?
?$∴∠DGB +∠F = 90°，$??$ ∠OBD+∠FBD=90°，$?
?$∴∠F=∠FBD，$?
?$∴ BD=DF.$?
由?$(2)，$?得?$BD=DG.$?
?$∵BD=2，$?
?$∴ FG=DF+DG=2BD=4$?
