Question

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

Answer 1

Answer:

Sol:   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

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.