Question

If in a quadrilateral ABCD, AC perpendicular to BD bisect each
other then prove 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.

Hope it helps.

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