If the diagonals of a parallelogram are equal,then show that it is a rectangle.

(N/A) Let $ABCD$ be a parallelogram such that its diagonals are equal,i.e.,$AC = BD$.
In $\Delta ABC$ and $\Delta BAD$:
$AC = BD$ [Given]
$BC = AD$ [Opposite sides of a parallelogram are equal]
$AB = AB$ [Common side]
Therefore,by $SSS$ congruence criterion,$\Delta ABC \cong \Delta BAD$.
By $CPCT$,we have $\angle ABC = \angle BAD$ ............. $(1)$
Since $ABCD$ is a parallelogram,$AD \parallel BC$ and $AB$ is a transversal. Therefore,the sum of consecutive interior angles is $180^{\circ}$:
$\angle ABC + \angle BAD = 180^{\circ}$ [Consecutive interior angles are supplementary] ............. $(2)$
From $(1)$ and $(2)$,we get:
$\angle ABC + \angle ABC = 180^{\circ}$
$2 \angle ABC = 180^{\circ}$
$\angle ABC = 90^{\circ}$
Since $ABCD$ is a parallelogram with one angle equal to $90^{\circ}$,it is a rectangle.