Question

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Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.​

Answer 1

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Let ABCD be a quadrilateral circumscribing a circle with centre O.

Now join AO, BO, CO, DO.

From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]

Let ∠DAO=∠BAO=1

Also ∠ABO=∠CBO [Since, BA and BC are tangents]

Let ∠ABO=∠CBO=2

Similarly we take the same way for vertices C and D.

Sum of the angles at the centre is 360°

Recall that sum of the angles in quadrilateral, ABCD = 360°  

=2(1+2+3+4)=360°

 =1+2+3+4=180°  

In ΔAOB,∠BOA=180−(1+2)

In ΔCOD,∠COD=180−(3+4)

∠BOA+∠COD=360−(1+2+3+4)

=360°–180°

=180°

Since AB and CD subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Hope helps uh buddy..!!

Answer 2

Answer:

Let ABCD be a quadrilateral circumscribing a circle with centre O.

Now join AO, BO, CO, DO.

From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]

Let ∠DAO=∠BAO=1

Also ∠ABO=∠CBO [Since, BA and BC are tangents]

Let ∠ABO=∠CBO=2

Similarly we take the same way for vertices C and D.

Sum of the angles at the centre is 360°

Recall that sum of the angles in quadrilateral, ABCD = 360°

=2(1+2+3+4)=360°

=1+2+3+4=180°

In ΔAOB,∠BOA=180−(1+2)

In ΔCOD,∠COD=180−(3+4)

∠BOA+∠COD=360−(1+2+3+4)

=360°–180°

=180°

Since AB and CD subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.