Question

prove that if diagonal of parallelogram are bisecting each other at 90 degree then it is square.

Answer 1

We know that in a parallelogram, opposite sides are equal and the diagonals bisect each other. 
suppose ABCD is the parallelogram named in order, then consider the triangles COD and AOD
CO=OA [diagonals bisect each other]
∠COD=∠AOD= 90°. [given]
OD=OD [common]
∴Triangle COD congruent to Triangle AOD.
⇒ AD=CD [CPCT]
⇒CD=AB
⇒AD=BC
⇒AD=CD=BC=AB
∴All sides are equal.
∴ABCD is a SQUARE.

Thank You......!!!!!!
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Yours, Jahnavi.

Answer 2

Given :- ABCD is a square.

To proof :- AC = BD and AC ⊥ BD

Proof :- In △ ADB and △ BCA

AD = BC [ Sides of a square are equal ]

∠BAD = ∠ABC [ 90° each ]

AB = BA [ Common side ]

△ADB ≅ △BCA [ SAS congruency rule ]

⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]

In △AOB and △AOD

OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]

AB = AD [ Sides of a square are equal ]

AO = AO [ Common side ]

△AOB ≅ △ AOD [ SSS congruency rule ]

⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]

∠AOB + ∠AOD = 180° [ Linear pair ]

∠ AOB = ∠AOD = 90°

⇒ AO ⊥ BD

⇒ AC ⊥ BD

Hence proved, AC = BD and AC ⊥BD