Question

已知：AE是∠BAC的角平分线,AE平行于DM,CM=BM 求证：BF=CD

Answer 1

延长DM到N,使MN=DM,连接BN,∵BM=CM,∠BMN=∠CMD,

∴ΔBMN≌ΔCMD,∴CD=BN,∠D=∠N.

∵AE平分∠BAC,∴∠EAB=∠EAC,

∵AE∥DM,∴∠D=∠EAC,∠EAB=∠BFM,

∴∠D=∠BFM,∴∠N=∠BFM,

∴BN=BF

∴BF=CD