已知梯形ABCD中AD//BC AE平分角BAD BE平分角ABC 且AE BE交DC于点E 求证：AB＝AD＋BC

∵AE,BE为角平分线∴∠DAE=∠EAB,∠ABE=∠EBC又∵AD‖BC∴∠DAE=∠AFB=∠FAB∴AB=BF又∵∠ABE=∠EBC∴AE=EF在△ADE和△FCE中∠DAE=∠AFB∠AED=∠FECAE=EF∴△ADE≌△FCE∴AD=CF∴BC+CF=BF=AB=AD+BC...