已知:在△ABC中,∠A=90°,AB=AC,D为AC中点,AE⊥BD于E,延长AE交BC于F,求证:∠ADB=∠CDF.

问题描述:

已知:在△ABC中,∠A=90°,AB=AC,D为AC中点,AE⊥BD于E,延长AE交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=

1
2
,BH=
3
2

tan∠DBH=
1
3

又∠GAF=∠DBH,
∴GF=
1
3
AG=
1
3

FH=GH-GF=
1
2
-
1
3
=
1
6

tan∠FDH=
FH
DH
=
1
3

∴∠DBH=∠FDH
∵∠ADB=∠DBH+∠C,
∠CDF=∠FDH+∠CDH,
∴∠ADB=∠CDF.