Question

show that the diagonals of a square are equal and bisect each other at right angle then it's a rhombus​

Answer 1

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.