Show that if the diagonals of a quadrilateral bisect each other at right angles,then it is a rhombus.

(N/A) Let $ABCD$ be a quadrilateral such that its diagonals $AC$ and $BD$ bisect each other at right angles at point $O$.
In $\Delta AOB$ and $\Delta AOD$:
$AO = AO$ (Common side)
$OB = OD$ (Given that $O$ is the midpoint of $BD$)
$\angle AOB = \angle AOD = 90^{\circ}$ (Given)
By $SAS$ congruence criterion,$\Delta AOB \cong \Delta AOD$.
Therefore,$AB = AD$ (Corresponding parts of congruent triangles are equal) ... $(1)$
Similarly,by considering other pairs of triangles:
In $\Delta AOB$ and $\Delta COB$,we get $AB = CB$ ... $(2)$
In $\Delta COB$ and $\Delta COD$,we get $CB = CD$ ... $(3)$
In $\Delta COD$ and $\Delta AOD$,we get $CD = AD$ ... $(4)$
From $(1), (2), (3),$ and $(4)$,we have $AB = BC = CD = DA$.
Since all sides of the quadrilateral $ABCD$ are equal,it is a rhombus.