如图,在RT△ABC中,∠A=90°,AB=AC,D为AC中点,AE⊥BD,垂足为E,延长AE交边BC于F,求证:∠ADB=∠CDF

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如图,在RT△ABC中,∠A=90°,AB=AC,D为AC中点,AE⊥BD,垂足为E,延长AE交边BC于F,求证:∠ADB=∠CDF
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如图,在RT△ABC中,∠A=90°,AB=AC,D为AC中点,AE⊥BD,垂足为E,延长AE交边BC于F,求证:∠ADB=∠CDF
如图,在RT△ABC中,∠A=90°,AB=AC,D为AC中点,AE⊥BD,垂足为E,延长AE交边BC于F,求证:∠ADB=∠CDF

如图,在RT△ABC中,∠A=90°,AB=AC,D为AC中点,AE⊥BD,垂足为E,延长AE交边BC于F,求证:∠ADB=∠CDF
证明:过A、D分别做BC的垂线,垂足分别为G、H.
设AG=1,那么CG=1,DH= ,BH= ,
tan∠DBH= ,
又∠GAF=∠DBH,∴GF= AG= ,
FH=GH-GF= - = ,
tan∠FDH= =
∴∠DBH=∠FDH
∵∠ADB=∠DBH+∠C,
∠CDF═∠FDH+∠CDH,
∴∠ADB=∠CDF.