证明：$\because \angle ACB = 90^{\circ}，$$CD\perp AB，$
$\therefore \angle CBF+\angle CFB=\angle DBE+\angle DEB = 90^{\circ}.$
$\because BF$平分$\angle ABC，$
$\therefore \angle CBF = \angle DBE.$
$\therefore \angle CFB = \angle DEB.$
又$\because \angle FEC = \angle DEB，$
$\therefore \angle CFB = \angle FEC.$
$\therefore CE = CF$
