4/
c/
$
b/ \Rightarrow \left\{\begin{matrix}
\widehat{ABE} = \widehat{DBE}\\
\widehat{AEB} = \widehat{DEB} (1)
\end{matrix}\right. \\ \widehat{AEF} = \widehat{DEC} (2) \\ (1) + (2) \Rightarrow \widehat{BEF} = \widehat{BEC} \\\triangle BEF = \triangle BEC (g-c-g) \Rightarrow EF = EC $
d/
$ BA = BD (gt) ; EA = ED (cmt) \Rightarrow BE $ là đường trung trực $ AD $