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

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.

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quadrilaterals 4

Answer: In the given quadrilateral ABCD diagonals AC and BD bisect each other at right angle. We have to prove that AB=BC=CD=AD

quadrilaterals 5

So, AB=AD

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

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

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THANKS FOR HELP.....
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By

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use congruency dont mark my comment as unhelpful as i know the answer well but i dont have time to answer it
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can u explain how OD=DO common ?
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Bhai sun parallelogram ke diagonals jab intersect kar te hai tab vo equal hote hai
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Here is the right answer

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Ma
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I hoped it was helpful to u🤗🤗🤗🤗

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This may help

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Continuation

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Show that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus
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