在△ABC中,∠B=60°,△ABC的角平分线AD、CE相交于点O,求证:AE+CD=AC

2025-06-22 02:31:12
推荐回答(2个)
回答1:

证明:在AC上截取CF=CD,△ODC≌△COF,因为∠B=60°,∠BAC+∠BCA=120°
AD、CE是角平分线,∠CAD+∠ECA=60°=∠COD=∠COF,∠AOF=60°=∠AOE,
再证△AOE≌△AOF,AF=AE,AE+CD=AC

回答2:

在AC上截取AF = AE,
∵AD平分∠BAC,
∴∠EAO =∠FAO =∠BAC/2
∵AO =AO
∴△EAO≌△FAO
∴∠EOA=∠FOA
∵CE平分∠ACB
∴∠ACO = ∠BCO=∠ACB/2
∵∠BAC+∠ACB+∠B=180°
∵∠B=60°
∴∠BAC+∠ACB=120°
∴∠EOA = ∠FOA = ∠FAO + ∠ACO = 1/2(∠BAC+∠ACB) = 60°
∴∠COF = 60°= ∠EOA = ∠DOC
∵∠ACO = ∠BCO,OC=OC
∴△COF≌△COD
∴CD=CF
∴AC = AF + CF = AE+CD