Question

三角形ABC中角B等于60度,角A的角平分线交BC于D,角C的角平分线交AB于E,求证AC=AE+DC

Answer 1

AD与CE交于点O,在AC上截取AF=AE,联接OF

∵∠EAO=∠FAO AO=AO AE=AF

∴AOE ≌△AOF

∴∠AOE=∠AOF

∵∠OAC=½∠BAC ∠OCA=½∠ACB

∴∠OAC+∠OCA=½(∠BAC+∠ACB)=½(180°-∠B)=60°

∴∠AOE=∠OAC+∠OCA=60°

∴∠AOF=60° ∠COD=∠AOE=60°

∴∠COF=180°-∠AOE+∠AOF=60°

∴∠COF=∠COD

∵∠OCF=∠OCD OC=OC

∴COF ≌△COD

∴CF=CD

∵AC=AF+CF

∴AC=AE+CD