(N/A) To prove that $AD \parallel CF$ and $AD = CF$:
$1$. In quadrilateral $ABED$,we are given that $AB = DE$ and $AB \parallel DE$. Since one pair of opposite sides is equal and parallel,$ABED$ is a parallelogram.
Therefore,$AD \parallel BE$ and $AD = BE$ (Opposite sides of a parallelogram are equal and parallel) ... $(1)$
$2$. In quadrilateral $BEFC$,we are given that $BC = EF$ and $BC \parallel EF$. Since one pair of opposite sides is equal and parallel,$BEFC$ is a parallelogram.
Therefore,$BE \parallel CF$ and $BE = CF$ (Opposite sides of a parallelogram are equal and parallel) ... $(2)$
$3$. From equations $(1)$ and $(2)$,we have $AD \parallel BE$ and $BE \parallel CF$,which implies $AD \parallel CF$.
Also,$AD = BE$ and $BE = CF$,which implies $AD = CF$.