证明：延长BE交CD的延长线与点F∵BE平分∠ABC∴∠ABE=∠CBE∵AB平行于CD∴∠ABE=∠CFE∴∠CFEE=∠CBE∴CB=CF=CD+DF又∵CE分别平分角BCD∴BE=FE（三线合一）∵∠AEB=∠FED∴△ABE=△DFE∴AB=EF∴BC=AB+CD
