Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Let \(ABCD\) be a quadrilateral, where \(AC\) and \(BD\) are diagonals.
Diagonals bisect each other at right angles.
Diagonals bisect each other at right angles.
\(\Rightarrow OA = OC\) and \(OB = \) \(OD\) - - - - (I)
Also, \(\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ\)
We know by the theorem, "If the diagonals of a quadrilateral bisect each other, then it is a parallelogram".
So, \(ABCD\) is a parallelogram.
We know by the theorem, "If the diagonals of a quadrilateral bisect each other, then it is a parallelogram".
So, \(ABCD\) is a parallelogram.
\(\Rightarrow AB = CD\) and \(BC =\) - - - - (II)
In \(\Delta AOB\) and \(\Delta BOC\):
\(OA = OC\) [Using (I)]
\(OA = OC\) [Using (I)]
\(\Rightarrow \angle AOB = \angle\) \(BOC\) [Each \(90^\circ\)]
\(BO = BO\) [Common side]
Therefore, by congruence rule, \(\Delta AOB \cong \Delta BOC\)
Corresponding parts of congruence triangles are congruent.
\(\Rightarrow AB =\) - - - - (III)
From (II) and (III), we get that:
\(AB = BC = CD = AD\)
Here, all the sides of a parallelogram are equal.
Therefore, \(ABCD\) is a rhombus.
\(AB = BC = CD = AD\)
Here, all the sides of a parallelogram are equal.
Therefore, \(ABCD\) is a rhombus.