如图,已知AD//BC,AE平分∠DAB,EB平分∠ABC,E在CD上,求证:AB=AD+BC

证明：延长AE交BC的延长线于点F∵AE平分∠DAB∴∠DAE＝∠BAE∵AD//BC∴∠F＝∠DAE,∠FCE＝∠D∴∠F＝∠BAE∴AB＝BF∵BE平分∠ABC∴∠ABE＝∠FBE∴△ABE≌△FBE （ASA）∴AE＝EF∴△ADE≌△FCE （AAS）∴CF＝AD∵BF＝CF+BF∴BF＝AD+BC∴AB＝AD+BC