Question

Answer this!!!..............​

Answer 1

Answer:

GIVEN ::

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

OA = OC, OB = OD

and

∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°

TO PROVE ::

ABCD is rhombus and AB = BC = CD = AD

PROOF ::

In ΔAOB and ΔCOB

OA = OC (Given)

∠AOB = ∠COB (Opposite sides of a parallelogram are equal)

OB = OB (Common)

Therefore, ΔAOB ≅ ΔCOB ( by SAS congruence)

condition.

Thus, AB = BC

Similarly we can prove

AB = BC = CD = AD

So Opposites sides of a quadrilateral are equal Hence ABCD is a parallelogram.

Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

Answer 2

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GIVEN ::

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

OA = OC, OB = OD

and

∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°

TO PROVE ::

ABCD is rhombus and AB = BC = CD = AD

PROOF ::

In ΔAOB and ΔCOB

OA = OC (Given)

∠AOB = ∠COB (Opposite sides of a parallelogram are equal)

OB = OB (Common)

Therefore, ΔAOB ≅ ΔCOB ( by SAS congruence)

condition.

Thus, AB = BC

Similarly we can prove

AB = BC = CD = AD

So Opposites sides of a quadrilateral are equal Hence ABCD is a parallelogram.

Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.