$(1)$证明：$\because$四边形$ABCD$是平行四边形，
$\therefore AB$∥$CD$，
$\therefore \angle ABE=\angle DFE$，
$\because AE=DE$，$\angle AEB=\angle DEF$，
$\therefore \triangle AEB$≌$\triangle DEF\left(AAS\right)$，
$\therefore AB=DF$，
$\because AB$∥$DF$，
$\therefore$四边形$ABDF$是平行四边形，
$\because \angle BDF=90^{\circ}$，
$\therefore$平行四边形$ABDF$是矩形.
$(2)$$\because$四边形$ABCD$是平行四边形，四边形$ABDF$是矩形，
$\therefore CD=AB=DF=3$，$BF=AD=5$，
$\therefore CF=CD+DF=6,AB$∥$CF$，
$\because \angle BDF=90^{\circ}$，
$\therefore BD=\sqrt{B{F}^{2}-D{F}^{2}}=\sqrt{{5}^{2}-{3}^{2}}=4$，$BD\bot CF$，
$\therefore S_{梯形ABCF}=\frac{1}{2}\left(AB+CF\right)\cdot BD=\frac{1}{2}\times \left(3+6\right)\times 4=18$，
即四边形$ABCF$的面积为$18$.

$(1)$证明：$\because$四边形$ABCD$是平行四边形，
$\therefore AB$∥$CD$，
$\therefore \angle ABE=\angle DFE$，
$\because AE=DE$，$\angle AEB=\angle DEF$，
$\therefore \triangle AEB$≌$\triangle DEF\left(AAS\right)$，
$\therefore AB=DF$，
$\because AB$∥$DF$，
$\therefore$四边形$ABDF$是平行四边形，
$\because \angle BDF=90^{\circ}$，
$\therefore$平行四边形$ABDF$是矩形.
$(2)$$\because$四边形$ABCD$是平行四边形，四边形$ABDF$是矩形，
$\therefore CD=AB=DF=3$，$BF=AD=5$，
$\therefore CF=CD+DF=6,AB$∥$CF$，
$\because \angle BDF=90^{\circ}$，
$\therefore BD=\sqrt{B{F}^{2}-D{F}^{2}}=\sqrt{{5}^{2}-{3}^{2}}=4$，$BD\bot CF$，
$\therefore S_{梯形ABCF}=\frac{1}{2}\left(AB+CF\right)\cdot BD=\frac{1}{2}\times \left(3+6\right)\times 4=18$，
即四边形$ABCF$的面积为$18$.