$(1)$证明：∵$AD = BE$
∴$AD + BD = BE + BD，$即$AB = DE$
在$\triangle ABC$和$\triangle DEF $中
$\begin {cases}AB = DE\\AC = DF\\BC = EF\end {cases}$
∴$\triangle ABC≌\triangle DEF(\mathrm {SSS})$
$(2)$解：由$(1)$得$\triangle ABC≌\triangle DEF$
∴$∠A=∠EDF$
又$∠A = 55°，$∴$∠EDF = 55°$
又$∠EDF+∠E+∠F = 180°，$$∠E = 45°$
∴$∠F=180°-∠E-∠EDF = 80°$
