∠DCE = 90°-∠BCE = 15° ,
DE = CD·tan∠DCE = AB·tan15° = (2-√3)AB ,
AE = AD-DE = BC-DE = 2AB-(2-√3)AB = (√3)AB ,
tan∠ABE = AE/AB = √3 ,
可得:∠ABE = 60° ,
∠CBE = 90°-∠ABE = 30° ,
∠BEC = 180°-∠CBE-∠BCE = 75° = ∠BCE ,
所以,BE = BC .
∠DCE = 90°-∠BCE = 15° ,
DE = CD·tan∠DCE = AB·tan15° = (2-√3)AB ,
AE = AD-DE = BC-DE = 2AB-(2-√3)AB = (√3)AB ,
tan∠ABE = AE/AB = √3 ,
可得:∠ABE = 60° ,
∠CBE = 90°-∠ABE = 30° ,
∠BEC = 180°-∠CBE-∠BCE = 75° = ∠BCE ,
所以,BE = BC .