Question

如图,AM//BN,AE平分角MAB,BE平分角ABC,求证：AB=AD+BC.

Answer 1

证明：在AB上截取AF=AD,连接EF∵AE平分∠MAB∴∠DAE=∠FAE又∵AD=AF,AE=AE∴∠AFE=∠ADE∵AM//BN∴∠ADE+∠BCE=180°∵∠AFE+∠BFE=180°∴∠BCE=∠BFE∵BE平分∠ABC∴∠CBE=∠FBE又∵BE=BE∴△BCE≌△BFE（AAS）∴BC=BF∴AB=AF+BF=AD+BC