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$128^{\circ}$

 解：$\because BF\perp AC，$$CE\perp AB，$$\therefore \angle BED = \angle CFD = 90^{\circ}.$

 在$\triangle BDE$和$\triangle CDF$中，

$\begin{cases}\angle BED=\angle CFD,\\\angle BDE=\angle CDF,\\BD = CD,\end{cases}$

 $\therefore \triangle BDE\cong\triangle CDF(\text{AAS})，$$\therefore DE = DF.$

 又$\because BF\perp AC，$$CE\perp AB，$
$\therefore$点$D$在$\angle BAC$的平分线上.

 解：(1) 证明：$\because PR\perp AB，$$PS\perp AC，$$PR = PS，$
$\therefore AP$平分$\angle BAC，$$\therefore \angle BAP = \angle CAP.$

 又$\because \angle CAP = \angle APQ，$$\therefore \angle BAP = \angle APQ，$
$\therefore QP// AR.$

 (2) 相等. 理由：

$\because PR\perp AB，$$PS\perp AC，$$\therefore \angle ARP = \angle ASP = 90^{\circ}.$

 在$\text{Rt}\triangle APR$和$\text{Rt}\triangle APS$中，

$\begin{cases}AP = AP,\\PR = PS,\end{cases}$

 $\therefore \text{Rt}\triangle APR\cong\text{Rt}\triangle APS(\text{HL})，$
$\therefore AR = AS.$