Question

In the figure given below, $E$ and $F$ are points on diagonal $AC$ of a parallelogram $ABCD$ such that $AE=CF$. Show that $BFDE$ is a parallelogram.

Answer 1

$ABCD$ is a parallelogram and $AE=CF$

Join $BD,$ meet $AC$ at point $O.$

Since, diagonals of a parallelogram bisect each other.

$\therefore$ $OA=OC$

and $OD=OB$

Now, $OA=OC$

and $AE=CF$ [ Given ]

$\Rightarrow$ $OA-AE=OC-CF$

$\Rightarrow$ $OE=OF$

Thus, $BFDE$ is a quadrilateral whose diagonals bisect each other.

Hence, $BFDE$ is a parallelogram.