过点E作EF⊥BC于F,
则有：EF‖AD .
因为,EF = BE·sin∠CBE = (1/2)BE ,
所以,BE = 2EF = AD .又因为,EF‖AD ,AE = EC ,
所以,EF是△ACD的中位线,
可得：AD = 2EF = BE .