如图在RT△ABC中AB=AC∠A=90°BF平分∠ABC,CD⊥BF交BF的延长线于点D求证:BF=2CD

Question

如图在RT△ABC中AB=AC∠A=90°BF平分∠ABC,CD⊥BF交BF的延长线于点D求证:BF=2CD

Answer 1

证明：延长CD BA 交于E
∵BD⊥CE
∴∠E+∠EBD=90° ∠E+∠ECA=90°
∴∠EBD=∠ECA
∵∠BAC=∠CAE=90°
又∵AB=AC
∴BAF≌CAE
∴BF=CE
∵BD⊥CE 且 BF平分∠ABC
∴CD=1/2CE=1/2BF
∴BF=2CD