解：?$ (1)DE⊥BC$?，理由：
連接?$OD$?，?$BD$?
∵?$AB $?是?$⊙O$?的直徑
∴?$∠ADB=90°$?
∴?$BD⊥AC$?
∵?$AB= BC$?
∴?$AD=CD$?
∵?$OA=OB$?
∴?$OD//BC$?
∵?$DE$?是?$⊙O$?的切線(xiàn)
∴?$OD⊥DE$?
∴?$DE⊥BC$?
?$(2)$?在?$Rt△BCD$?中，?$∠DCB=30°$?
∴?$BC=2BD$?，?$BC2- BD2=CD2$?
∴?$(2BD)2-BD2=(2\sqrt 3)2$?
∴?$BD=2$?
∴?$AB=BC=2BD=4$?
∴?$OD=2$?
在?$Rt△DCE$?中，?$∠DCE= 30°$?
∴?$DE=\frac {1}{2}CD= \sqrt {3}$?
在?$Rt△DOE$?中，?$OE=\sqrt {OD2+DE2}=\sqrt {7}$?
∴?$b=-(AB+OE)= -(4+\sqrt {7})=-4-\sqrt {7}$?
?$(3)$?過(guò)?$D$?作?$DF⊥OB$?于?$F$?
∵?$AB=BC$?
∴?$∠A= ∠ACB=30°$?
∴?$∠DOB=2∠A=60°$?
∵?$OB=OD$?
∴?$△OBD$?為等邊三角形
∵?$AB=4$?
∴?$BD=OD=OB=\frac {1}{2}AB=2$?
∵?$DF⊥OB$?
∴?$OF=\frac {1}{2}OB=1$?
∴在?$Rt△OFD$?中，?$DF= \sqrt {OD^2-OF^2}=\sqrt {3}$?
∵?$S_{△BOD}=\frac {1}{2}OB · DF=\frac {1}{2}×2×\sqrt {3}=\sqrt {3}$?
?$S_{△BCD}=\frac {1}{2}BD · CD=\frac {1}{2}×2×2\sqrt 3=2\sqrt 3$?
?$S_{扇形BOD}=\frac {60π×2^2}{360}=\frac {2}{3}π$?
∴?$S_{△BOD}+S_{△BCD}=S_{扇形BOD}=3\sqrt 3-\frac 23π$?
∴線(xiàn)段?$CD$?、?$BC$?和?$\widehat {BD}$?所圍成的圖形的面積為?$3\sqrt {3}-\frac 23π$?
