Question

如图,△ABC中,∠BAC=90°,AB=AC,点D是BC的一点.

Answer 1

（1）证明：做AO⊥BC于O,做EF⊥BC于F,设∠OAD = α ；AD = DE=m

则∠EDF=αOA = DF=m*cosαOD=FE = m*sinα

∵OC=OA=DF∴OC-DC = DF-DC即：OD = CF

又∵OD = EF∴ CF=EF∴RT△CFE为等腰直角三角形. 即∠ECF = 45°

又∵ ∠ACB = 45°∴ ∠ACE = 90°即：CE⊥AC

（2）做MH⊥BC于H,做AO⊥BC于O,即H为DC中点

∵N为BD中点,则NH = 1/2BC = OC = OA∴NH-OH = OC-OH即：ON=HC

∵HC=MH∴MH=ON∠AON=∠NHM=90°∴ △AON≌△NHM

∴AN=NM∠NAO=∠MNH∵在RT△AON中,∠NAO + ∠ANO = 90°

∴∠MNH+∠ANO =90°即：AN⊥MN

总之,MN与AN之间的关系是即相等也垂直!