求证:AO:ON=AB:BN=AC:CN=(AB+AC):BC.
证明:延长BA到D,使AD=AC,连接CD.则:∠D=∠ACD=(1/2)∠BAC;
点O为内心,则:∠BAN=(1/2)∠BAC=∠D;可知AN∥DC.
∴BN:NC=AB:AD=AB:AC;…………………………[这是几何中常用到的"角平分线性质定理]
(再罗嗦一遍,即:若AN平分角BAC,则AB:AC=BN:NC)
同理:BO平分∠ABN,故AO:ON=AB:BN;(1)
CO平分∠ACN,故AO:ON=AC:CN;(2)
又AN∥DC,则⊿BAN∽⊿BDC,得BD:BC=AB:BN.
即:(AB+AD):BC=AB:BN, (AB+AC):BC=AB:BN.(3)
由(1),(2),(3)可知:AO:ON=AB:BN=AC:CN=(AB+AC):BC.
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