$(2)證明：∵?ABCD$
$∴AD=BC，∠BAD=∠DCB，AB=CD$
$∵?BEDF$
$∴ED=BF$
$∴AD-ED=BC-BF$
$即AE=FC$
$又∵AD//BC$
$∴?AEFC$
$∴AF//EC，∠DAF =∠BCE$
$∴?MENF$
$∴∠EMF=∠ENF$
$∴∠AMB=∠CND$
$∵∠DAF =∠BCE$
$∠BAD=∠DCB $
$∴∠BAF=∠DCE$
$在△ABM和△CDN中$
${{\begin{cases} {∠AMB=∠CND } \\ {∠BAF=∠DCE } \\ {AB=CD} \end{cases}}}$
$∴△ABM≌△CDN(AAS)$
