Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles,then it is a square.

(N/A) Let $ABCD$ be a quadrilateral such that its diagonals $AC$ and $BD$ are equal $(AC = BD)$ and bisect each other at right angles at point $O$ ($AO = OC$,$BO = OD$,and $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^{\circ}$).
$1$. To prove $ABCD$ is a rhombus:
In $\Delta AOD$ and $\Delta AOB$:
$AO = AO$ (Common)
$OD = OB$ (Given)
$\angle AOD = \angle AOB = 90^{\circ}$ (Given)
By $SAS$ congruence criterion,$\Delta AOD \cong \Delta AOB$.
Therefore,$AD = AB$ $(CPCT)$.
Similarly,we can prove $AB = BC$,$BC = CD$,and $CD = DA$.
Thus,$AB = BC = CD = DA$. Since all sides are equal,$ABCD$ is a rhombus.
$2$. To prove $ABCD$ is a square:
In $\Delta ABC$ and $\Delta BAD$:
$AC = BD$ (Given)
$BC = AD$ (Sides of rhombus are equal)
$AB = BA$ (Common)
By $SSS$ congruence criterion,$\Delta ABC \cong \Delta BAD$.
Therefore,$\angle ABC = \angle BAD$ $(CPCT)$.
Since $AD \parallel BC$ and $AB$ is a transversal,the sum of consecutive interior angles is $180^{\circ}$:
$\angle ABC + \angle BAD = 180^{\circ}$.
Since $\angle ABC = \angle BAD$,we have $2 \angle ABC = 180^{\circ}$,which implies $\angle ABC = 90^{\circ}$.
$A$ rhombus with one angle equal to $90^{\circ}$ is a square. Hence,$ABCD$ is a square.