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:△ABM≌△CAN∠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:△ABM≌△CAN∠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:△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:△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:△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:△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:△ABM≌△CAN∠AMB=∠CMD
证明:
∵AB⊥AC,NC⊥AB,
∴∠BAM=∠ACN=90º
∵∠MBA+∠BMA=90º
∠NAC+∠BMA=90º【AN⊥BM】
∴∠MBA=∠NAC
又∵AB=AC
∴⊿ABM≌⊿CAN(ASA)
∴AM=CN,∠AMB=∠N
∵AM=CM
∴CM=CN
∵BC平分∠ACN
∴∠MCD=∠NCD
又∵CD=CD
∴⊿MCD≌⊿NCD(SAS)
∴∠CMD=∠N
∴∠AMB=∠CMD