$ (1) $证明:∵$D$为边$BC$的中点,∴$BD = CD$
$ $又$BE//AC,$∴$∠EBD = ∠C$
$ $在$∆BDE$和$∆CDA$中
$ \begin {cases}∠EBD=∠C \\BD = CD \\∠BDE=∠CDA\end {cases}$
∴$∆BDE≌∆CDA(AS A)$
$ (2) $证明:由$(1),$得$∆BDE≌∆CDA,$∴$BE = CA$
$ $又$AD⊥BC,$$D$为$BC$的中点
∴$AD$垂直平分$BC,$即$BA = CA$
∴$BA = BE$