Question

SHOW THAT IF THE DIAGONAL OF A QUADRILATERAL BISECT EACH OTHER AT RIGHT ANGLE, THEN IT IS A RHOMBUS ​

Answer 1

Step-by-step explanation:

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.

Hope it helps!

Answer 2

Answer:

refer to the attachment