show that the diagonals of a square are equal and bisect each other at right angle then it's a rhombus
Answers
Answer:
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Answer:

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e.,
OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. To prove
ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are
equal.
In ΔAOD and ΔCOD,
OA = OC (Diagonals bisect each other)
∠AOD = ∠COD (Given)
OD = OD (Common)
So, ΔAOD ≅ ΔCOD (By SAS congruence rule)
Hence, AD = CD …………..1
Similarly, it can be proved that
AD = AB and CD = BC ………..2
From equation 1 and 2, we get
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 ABCD is a
rhombus.