Question

If the diagonals of a quadrilateral bisect each other at right angles, show that it is a rhombus.

Answer 1

hey!

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)
∴ ΔAOD ≅ ΔCOD (By SAS congruence rule)
∴ 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 ABCD is a rhombus.

hope it helps uh

Answer 2

Consider a quadrilateral ABCD 
we know diagonal bisect at 90 degree so AO=OC and BO=OD

now in triagle AOB and COD
Angle O = angle O
AO=CO
BO=DO
triangles congruent (SAS criteria) so AB // CD 
therefore ABCD is a parallelogram

Now in Triangle AOB and BOC
Angle O = ANgle O
OB=OB
AO=OC
Triangle congruent (SAS criteria)
so AB=BC 
since AB = BC and ABCD is parallelogram
 AB=BC=CD=DA
 therefore ABCD is a rhombus

Plz manage as I couldnt upload a diagram coz I dont have a mobile phone, but I guess I did clean work so that you can understand easily.
Thank You