Question

show that if the diagonal of a quadrilateral bisect each other at right angle then it is a rhombus

Answer 1

We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.          

∴ In ΔAOB and ΔAOD, we have                

AO = AO[Common]                

OB = OD[Given that O in the mid-point of BD]                

∠AOB = ∠AOD[Each = 90°]                

ΔAOB ≌ ΔAOD[SAS criteria]          

Their corresponding parts are equal.

AB = AD...(1)

Similarly,AB = BC...(2)

BC = CD...(3)

CD = AD...(4)          

∴ From (1), (2), (3) and (4), we have AB = BC CD = DA          

Thus, the quadrilateral ABCD is a rhombus.

Answer 2

Step-by-step explanation:

Take quadrilateral ABCD , AC and BD are diagonals which intersect at O.

In △AOB and △AOD

DO=OB ∣ O is the midpoint

AO=AO ∣ Common side

∠AOB=∠AOD ∣ Right angle

So, △AOB≅△AOD

So, AB=AD

Similarly, AB=BC=CD=AD can be proved which means that ABCD is a rhombus.