證明: (1)連接$AC，$在$\triangle ACE$和$\triangle ACF$中，
$\begin{cases}AE = AF \\CE = CF \\AC = AC\end{cases}$
$\therefore \triangle ACE\cong\triangle ACF(SSS)，$$\therefore \angle FAC=\angle EAC。$
$\because CB\perp AB，$$CD\perp AD，$$\therefore \angle B=\angle D = 90^{\circ}。$
$\because AC = AC，$$\therefore \triangle ACB\cong\triangle ACD(AAS)，$$\therefore CB = CD。$
(2)由
(1)得$\triangle ACE\cong\triangle ACF，$$CB = CD。$
$\because AE = 8，$$CD = 6，$
$\therefore S_{\triangle ACF}=S_{\triangle ACE}=\frac{1}{2}AE\cdot CB=\frac{1}{2}\times8\times6 = 24，$
$\therefore S_{四邊形AECF}=S_{\triangle ACF}+S_{\triangle ACE}=24 + 24 = 48。$
(3)$\angle DAB+\angle ECF = 2\angle DFC。$
證明：$\because \triangle ACE\cong\triangle ACF，$$\therefore \angle EAC=\angle FAC，$$\angle ACE=\angle ACF。$
$\because \angle DAB=\angle FAC+\angle EAC，$$\angle ECF=\angle ACF+\angle ACE，$
$\therefore \angle DAB+\angle ECF=\angle FAC+\angle EAC+\angle ACF+\angle ACE = 2\angle FAC+2\angle ACF = 2(\angle FAC+\angle ACF)。$
$\because \angle DFC=\angle FAC+\angle ACF，$
$\therefore \angle DAB+\angle ECF = 2\angle DFC。$
