解：(1)证明：$\because AD$平分$\angle BAC，$$DE\perp AB，$$\angle C = 90^{\circ}，$
$\therefore DC = DE。$
在$Rt\triangle DCF$和$Rt\triangle DEB$中，
$\begin{cases}DF = DB \\DC = DE\end{cases}$
$\therefore Rt\triangle DCF\cong Rt\triangle DEB(HL)。$
$\therefore CF = EB。$
(2)$AF + BE = AE。$
理由：$\because\angle C=\angle DEA = 90^{\circ}，$$DC = DE，$$AD = AD，$
$\therefore Rt\triangle DCA\cong Rt\triangle DEA(HL)。$
$\therefore AC = AE。$
$\because AF + FC = AC，$$CF = EB，$
$\therefore AF + BE = AE。$
