证明:
$\because △ ACD,$$△ BCE$分别是以$AC,$$BC$为底边的等腰三角形,
$\therefore AD=CD,$$CE=EB,$
$\therefore ∠ A = ∠ DCA。$
$\because ∠ A = ∠ CBE,$
$\therefore ∠ CBE = ∠ DCA,$
$\therefore CD// BE,$
$\therefore ∠ DCE = ∠ FEB。$
$\because EF=AD,$
$\therefore CD=EF。$
在$△ DCE$和$△ FEB$中,
$\begin{cases} CD=EF, \\ ∠ DCE=∠ FEB, \\ CE=EB, \end{cases}$
$\therefore △ DCE ≌ △ FEB \ (\mathrm{SAS}),$
$\therefore DE=FB。$