Question

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

Answer 1

let $BD$ meet $AC$ at point $O$

$ABCD$ is a parallelogram.

Diagonal of parallelogram bisect each other

$\therefore OA=OC$

$OD=OB$

Now, $OA=OC$

$AE=CF$

$OA-AE=OC-CF$

$\Rightarrow OE=OF$

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

Hence $BFDE$ is a parallelogram