Question

已知,如图,在△ABC中,∠BAC=90°,AB=AC,点D在AC上,点E在BA的延长线上,BD的延长线交CE与F

Answer 1

∵∠CAE=∠BAD=90°

AB=AC,BD=CE

∴RT△ABD≌RT△ACE(HL)

∴∠ABD=∠ECA=∠FCD

∵∠ADB=∠CDF

∴∠CFD=180°-(∠FCD+∠CDF)

=180°-(∠ABD+∠ADB)

=180°-(180°-∠BAD)

=90°

∴DF⊥CE

即BF⊥CE