Question

已知:如图△ABC中,∠ABC=90°,在AC上截取CD=CB,AE=AB,DF⊥BE于F.求证:DF=FB

Answer 1

∵BC=DC,∴∠CBD=1/2(180°-∠C)=90°-1/2∠C,

∵AB=AE,∴∠ABE=1/2∠180°-∠A)=90°-1/2∠A,

∴∠DBF=∠CBD+∠ABE-90°

=180°-1/2(∠C+∠A)-90°

=45° (∵∠ABC=90°,∴∠A+∠C=90°),

又DF⊥BE,

∴ΔBDF是等腰直角三角形,

∴DF=FB.