点O为三角形ABC的内心,连接AO交BC于M,证明AB/BM=AO/OM=AC/CM.
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点O为三角形ABC的内心,连接AO交BC于M,证明AB/BM=AO/OM=AC/CM.
点O为三角形ABC的内心,连接AO交BC于M,证明AB/BM=AO/OM=AC/CM.
点O为三角形ABC的内心,连接AO交BC于M,证明AB/BM=AO/OM=AC/CM.
先证明AB/BM=AC/CM 在△ABM 和△ACM中 ,由正弦定理 AB/BM=sinAMB/sinMAB=sinAMC/sinMAC=AC/MC
在△OBM和△OBA中,同理可得AB/BM=AO/OM 所以AB/BM=AO/OM=AC/CM.