?$(1)$?如圖，令?$x=0，$?由?$y= -\frac {\sqrt 3}3x+1$?得?$y= 1$?
令?$y=0，$?由?$y=-\frac {\sqrt 3}3x+1$?得?$x=\sqrt 3$?
∴?$B$?點(diǎn)的坐標(biāo)為?$(\sqrt 3，$??$0)，$??$A$?點(diǎn)的坐標(biāo)為?$(0，$??$1)$?
?$(2) $?由?$(1)$?知?$OB=\sqrt 3，$??$OA= 1$?
∴?$tan∠OBA =\frac {OA}{OB}=\frac {\sqrt 3}3$?
∴?$∠OBA = 30°$?
∵?$△A BC$?和?$△ABO$?關(guān)于?$AB$?成軸對(duì)稱
∴?$BC = BO=\sqrt 3，$??$∠CBA=∠OBA = 30°$?
∴?$∠CBO = 60°$?
過點(diǎn)?$C$?作?$CM⊥x$?軸于?$M，$?則在?$Rt△BCM$?中
?$CM=BC×sin∠CBO=\sqrt 3×sin 60°=\frac 32$?
?$BM=BC×cos∠CBO=\sqrt 3×cos 60°=\frac {\sqrt 3}2$?
∴?$OM=OB-BM=\sqrt 3-\frac {\sqrt 3}2=\frac {\sqrt 3}2$?
∴?$C$?點(diǎn)坐標(biāo)為?$(\frac {\sqrt 3}2，$??$\frac 32)$?
連接?$OC$?
∵?$OB= CB，$??$∠CBO = 60°$?
∵?$△BOC$?為等邊三角形
過點(diǎn)?$C$?作?$CE//x$?軸，并截取?$CE= BC，$?則?$∠BCE = 60°$?
連接?$BE，$?則?$△BCE$?為等邊三角形
作?$EF⊥x$?軸于?$F，$?則?$EF= CM =\frac 32$?
?$BF= BM =\frac {\sqrt 3}2$?
?$OF=OB+ BF=\sqrt 3+\frac {\sqrt 3}2=\frac {3\sqrt 3}2$?
∴點(diǎn)?$E$?坐標(biāo)為?$(\frac {3\sqrt 3}2，$??$\frac 32)$?
∴?$D$?點(diǎn)的坐標(biāo)為?$(0，$??$0)$?或?$(\frac {3\sqrt 3}2，$??$\frac 32)$?
