$E$ and $F$ are points on diagonal $AC$ of a parallelogram $ABCD$ such that $AE = CF$. Show that $BFDE$ is a parallelogram.

(N/A) Given: $A$ parallelogram $ABCD$. $E$ and $F$ are points on the diagonal $AC$ such that $AE = CF$.
To prove: $BFDE$ is a parallelogram.
Proof: Let the diagonals $AC$ and $BD$ of the parallelogram $ABCD$ intersect at point $O$.
Since the diagonals of a parallelogram bisect each other,we have:
$OA = OC$ ... $(1)$
$OD = OB$ ... $(2)$
Given that $AE = CF$ ... $(3)$
Subtracting equation $(3)$ from equation $(1)$,we get:
$OA - AE = OC - CF$
$OE = OF$ ... $(4)$
Now,in quadrilateral $BFDE$,the diagonals $BD$ and $EF$ bisect each other at point $O$ (from equations $(2)$ and $(4)$).
$A$ quadrilateral whose diagonals bisect each other is a parallelogram.
Therefore,$BFDE$ is a parallelogram.