In the figure,$AB \parallel DE$,$AB = DE$,$AC \parallel DF$ and $AC = DF$. Prove that $BC \parallel EF$ and $BC = EF$.

(N/A) Given: $AB \parallel DE$,$AB = DE$,$AC \parallel DF$ and $AC = DF$.
To prove: $BC \parallel EF$ and $BC = EF$.
Proof:
$1$. In quadrilateral $ACFD$,we are given $AC \parallel DF$ and $AC = DF$. Since one pair of opposite sides is equal and parallel,$ACFD$ is a parallelogram.
Therefore,$AD \parallel CF$ and $AD = CF$ (Opposite sides of a parallelogram are equal and parallel).
$2$. In quadrilateral $ABED$,we are given $AB \parallel DE$ and $AB = DE$. 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).
$3$. From the above results,we have $CF \parallel AD$ and $BE \parallel AD$. This implies $CF \parallel BE$.
Also,$CF = AD$ and $BE = AD$. This implies $CF = BE$.
$4$. In quadrilateral $BCFE$,we have $CF \parallel BE$ and $CF = BE$. Since one pair of opposite sides is equal and parallel,$BCFE$ is a parallelogram.
Therefore,$BC \parallel EF$ and $BC = EF$ (Opposite sides of a parallelogram are equal and parallel).
Hence,proved.