延长BE交CD于H∵AB∥CD∴∠ABE=∠H∵BE平分∠ABC∴∠ABE=∠CBE∴∠CBE=∠H∴BC=CH∵CE平分∠BCD∴BE=EH(等腰三角形三线合一）∵∠ABE=∠H,∠AEB=∠DEH∴△ABE≌△DHE(ASA)∴AB=DH∵CH=CD+DH=CD+ABBC=CH∴BC=AB+CD
