?$ (1)$?解:∵?$\triangle ABC$?是等邊三角形
∴?$∠ACB = 60°$?
∵?$CE = CD,$?∴?$∠E=∠CDE$?
又∵?$∠ACB=∠E + ∠CDE$?
∴?$∠E=\frac 12∠ACB=\frac 12×60°=30°$?
?$ (2)$?證明:連接?$BD$?
∵?$\triangle ABC$?是等邊三角形,?$D$?是?$AC$?的中點
∴?$BD$?平分?$∠ABC,$??$∠ABC = 60°$?
∴?$∠DBC = 30°$?
?$ $?由?$(1)$?知?$∠E = 30°,$?∴?$∠DBC=∠E$?
∴?$DB = DE。$?
∵?$DM\perp BC,$?∴?$M$?是?$BE$?的中點