一道关于三角形内心性质的问题

连接BI，CI
由正弦定理
AB/sin∠AIB = AI/sin∠ABI AB/AI = sin∠AIB/sin∠ABI
BE/sin∠BIE = BI/sin∠EBI BE/BI = sin∠BIE/sin∠EBI
I为内心，BI为∠ABC平分线
∠ABI = ∠EBI sin∠ABI = sin∠EBI
又∠AIB和∠BIE互为补角 sin∠AIB = sin∠BIE
所以 AB/AI = BE/BI
即 AI/IE = AB/BE

同理可证 AI/IE = AC/EC
所以 AI/IE = AB/BE = AC/EC