这道数学题怎样解答?
2*** 2012-02-02 11:48:40 举报
BE=2CD
作BC中点F,连接DF,则在直角三角形BDE中, DF=BF=EF=1/2BE, ∠DBC=∠BDF, ∠DFE=∠DBC+∠BDF, ∠DBF=1/2∠ABC, ∠DFE=∠ABC, ∠ABC=∠C, ∠DFE=∠C, DF=DC, CD=1/2BE