ong>Answer:

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We have a QUADRILATERAL ABCD such that angleO is the mid-point of AC and BD. Also AC ⊥ BD.

Now, in ΔAOD and ΔAOB, we have

AO = AO

[Common]

OD = OB

[∵ O is the mid-point of BD]

∠AOD = ∠AOB

[Each = 90°]

∴ ΔAOD ≌ ∠AOB

[SAS criteria]

∴Their corresponding parts are equal.

⇒ AD = AB

...(1)

Similarly, we have

AB = BC

...(2)

BC = CD

...(3)

CD = DA

...(4)

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

∴Quadrilateral ABCD is having all sides equal.

In ΔAOD and ΔCOB, we have

AO = CO

[Given]

OD = OB

[Given]

∠AOD = ∠COB

[Vertically opposite angles]

∴

ΔAOD ≌ ΔCOB

⇒Their corresponding pacts are equal.

⇒ ∠1 = ∠2

But, they form a pair of interior alternate angles.

∴AD || BC

Similarly, AB || DC

∴ ABCD is' a parallelogram.

∵ Parallelogram having all of its sides equal is a rhombus.

∴ ABCD is a rhombus.

Now, in ΔABC and ΔBAD, we have

AC = BD

[Given]

BC = AD

[Proved]

AB = BA

[Common]

ΔABC ≌ ΔBAD

[SSS criteria]

Their corresponding angles are equal.

∠ABC = ∠BAD

Since, AD || BC and AB is a transversal.

∴∠ABC + ∠BAD = 180°

[Interior opposite angles are supplementary]

i.e. The rhombus ABCD is having one angle equal to 90°.

Thus, ABCD is a SQUARE.