Question

In the given figure 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

Hint: Here, we will join the other pair of diagonal. Further we will use the property of a parallelogram to get the desired result.

Complete step by step solution:
Given that:
$ABCD$ is a parallelogram and $AE = CF$
To show: $BFDE$is a parallelogram i.e.$OE = OF$
Construction: Join $BD$diagonal, such that it cut $AC$at point$O$.
Proof: Since, $ABCD$is a parallelogram so, the diagonal of a parallelogram bisect each other
i.e $OA = OC\,and\,\,OB = OD$ ……(i)
Now, $AE = CF$ (given) …..(ii)
Then we subtract equation (ii) from the equation (i), we have
$OA - AE = OC = CF$
$OE = OF$
Now, in quadrilateral$BFDE$, $OB = OD$
And $OE = OF$
As, the diagonals of quadrilateral $BFDE$ bisect each other.
Thus, $BFDE$ is a parallelogram.


Note: Students must note that in a quadrilateral if the diagonal bisects each other then, it is a parallelogram.Also,if both the opposite sides of a quadrilateral are equal and parallel ,then too it can be said that it is a parallelogram