$證明:過點(diǎn)E作EG⊥BF,交BF的延長(zhǎng)線于點(diǎn)G$
$∵EG⊥BF,∴∠EGF=90°, ∴∠GBE+∠BEG=90°\ $
$∵BE⊥AB,∴∠ABE=90°=∠ABC+∠EBC\ $
$∴∠ABC=∠BEG$
$在△ABC和△BEG中$
$\begin{cases}{ ∠ACB=∠BGE }\ \\ { ∠ABC=∠BEG } \\{AB=BE } \end{cases}$
$∴△ABC≌△BEG(AAS),∴EG=BC\ $
$∵CD=BC,∴EG=CD$
$在△EGF和△DCF中$
$\begin{cases}{ ∠EGF=∠DCF }\ \\ { ∠EFG=∠DFC } \\{EG=DC } \end{cases}$
$∴△EGF≌△DCF(AAS),∴EF=DF $
