如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN 求证:△ABM≌△CAN ∠AMB=∠CMD如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN求证:(1)△ABM≌△CAN(2)∠AMB=∠CMD

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如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN 求证:△ABM≌△CAN ∠AMB=∠CMD如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN求证:(1)△ABM≌△CAN(2)∠AMB=∠CMD

如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN 求证:△ABM≌△CAN ∠AMB=∠CMD如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN求证:(1)△ABM≌△CAN(2)∠AMB=∠CMD
如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN 求证:△ABM≌△CAN ∠AMB=∠CMD
如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN
求证:(1)△ABM≌△CAN
(2)∠AMB=∠CMD

如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN 求证:△ABM≌△CAN ∠AMB=∠CMD如图,AB⊥AC,NC⊥AC,AB=AC,M为AC中点,AN⊥BM交BC于D,BC平分∠ACN求证:(1)△ABM≌△CAN(2)∠AMB=∠CMD
因为AB//CN所以角NAB=角ANC,
角NAB+角ABM=90度,角ANC+角CAN=90度
所以角ABM=角CAN
又因为AB=AC 角ACN=角CAB=90度
根据ASA判别法得到两个三角形全等
因为全等所以角AMB=角CNA
角CNA=角NAB