Question

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

Answer 1

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle. i.e. OA=OC, OB=OD, and $\angle AOB=\angle BOC=\angle COD=\angle AOD={90}^{\circ }$. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal.
In $\Delta AODand\Delta COD$,
OA = OC (Diagonals bisect each other)
$\angle$ AOD =$\angle$ COD (Given)
OD=OD (Common)
$\therefore \Delta AOD\cong \Delta COD$ (By SAS congruence rule)
$\therefore$ AD=CD ……………(1)
Similarly, it can be proved that
AD=AB and CD = BC ………..(2)
From equations (1) and (2)
AB=BC=CD=AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that AB CD is a rhombus.