首页 > 留学知识库

问题: 请高手帮忙,明天交,谢谢

在三角形ABC中,角B=60,AD,CE分别平分角BAC,角ACB。求证:AC=AE+DC

解答:

AD,CE分别平分∠BAC,∠ACB交于O点,过O点作∠AOC的平分线交AC于H点
∠BAC+∠ACB=120°AD,CE分别平分∠BAC,∠ACB
=>∠DAC+∠ACE=60°
=>∠AOC=120°
=>∠AOE=60°
OH平分∠AOC
=>∠AOH=∠COH=∠AOE=∠COD=60°
=>△AOE全等于△AOH,△COH全等于△COD
=>AE=AH,HC=DC
=>AC=AE+DC