解:?$(1)$?如圖,過(guò)?$A$?點(diǎn)作?$AM⊥BC$?于?$M,$?交?$EF$?于?$N$?
∵?$AB=AC=6,$??$BC=4$?
∴?$BM=MC=\frac {1}{2}BC=2$?
∴?$AM=\sqrt {AB^2-BM^2}=\sqrt {6^2-2^2}=4\sqrt {2}$?
∴?$sin B=\frac {AM}{AB}=\frac {4\sqrt {2}}6=\frac {2\sqrt 2}3$?
?$(2)$?∵點(diǎn)?$E、$??$F{分別} $?是?$AB、$??$AC$?的中點(diǎn)
∴?$AE=\frac {1}{2}AB=\frac {1}{2}AC=AF=3,$??$EF//BC,$??$EF=\frac {1}{2}BC=2$?
∵?$AM⊥BC$?
∴?$AM⊥EF,$?即?$AN⊥EF$?
∴?$EN=NF=\frac {1}{2}EF=1$?
∴?$AN^2=AE^2-EN^2=3^2-1^2=8$?
∵?$CD//AB,$??$EF//BC$?
∴四邊形?$BCDE$?是平行四邊形
∴?$DE=BC=4$?
∴?$DN=DE-EN=4-1=3$?
∴?$AD=\sqrt {AN^2+DN^2}=\sqrt {8+3^2}=\sqrt {17}$?
故線段?$AD$?的長(zhǎng)為?$\sqrt {17}$?