# `prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.`

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

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

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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

o

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

o =2(1+2+3+4)=360 o =1+2+3+4=180 o

In ΔAOB,∠BOA=180−(1+2)

In ΔCOD,∠COD=180−(3+4)

∠BOA+∠COD=360−(1+2+3+4)

=360 o –180 o =180 o

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.

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## Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. from Mathematics Circles Class 10 CBSE

Given : A circle with centre O touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.To prove : ∠AOB + ∠COD = 180°∠AOD + ∠BOC = 180° Const. : Join OP, OQ, OR and OS.Proof : Since, the two tangents drawn from an external point to a circle subtend equal angles at the centre.∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8Since sum of all the angles subtended at a point is 360°.∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8= 360°⇒2 ∠2 + 2 ∠3 + 2 ∠6 + 2 ∠7) = 360°⇒ 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°⇒ ∠2 + ∠3 + ∠6 + ∠7) = 180°⇒ (∠6 + ∠7) + (∠2 + ∠3) = 180°⇒ ∠AOB + ∠COD = 180°Similarly, we can prove ∠AOD + ∠BOC = 180°

## Circles

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

Given : A circle with centre O touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.

To prove : ∠AOB + ∠COD = 180°

∠AOD + ∠BOC = 180°

Const. : Join OP, OQ, OR and OS.

Proof : Since, the two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8

Since sum of all the angles subtended at a point is 360°.

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8

= 360°

⇒2 ∠2 + 2 ∠3 + 2 ∠6 + 2 ∠7) = 360°

⇒ 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°

⇒ ∠2 + ∠3 + ∠6 + ∠7) = 180°

⇒ (∠6 + ∠7) + (∠2 + ∠3) = 180°

⇒ ∠AOB + ∠COD = 180°

Similarly, we can prove ∠AOD + ∠BOC = 180°

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## Ex 10.2, 13

Ex 10.2,13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Given : Let A..

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## Ex 10.2, 13 - Chapter 10 Class 10 Circles (Term 2)

Last updated at Feb. 25, 2017 by Teachoo

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### Transcript

Ex 10.2,13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Given : Let ABCD be the quadrilateral circumscribing the circle with centre O. ABCD touches the circle at points P,Q,R and S To prove: Opposite sides subtend supplementary angles at centre i.e. ∠ AOB + ∠ COD = 180° & ∠ AOD + ∠ BOC = 180° Construction: Join OP, OQ, OR & OS Proof: Let us rename the angles In Δ AOP and Δ AOS AP = AS AO = AO OP = OS ∴ Δ AOP ≅∆ AOS ∠ AOP = ∠ AOS i.e. ∠ 1 = ∠ 8 Similarly, we can prove ∠2 = ∠3 ∠5 = ∠4 ∠6 = ∠7 Now ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360° ∠ 1 + ∠ 2 + ∠ 2 + ∠ 5 + ∠ 5 + ∠ 6 + ∠ 6 + ∠ 1 = 360° 2 (∠ 1 + ∠ 2 + ∠ 5 + ∠ 6) = 360° ∠ 1 + ∠ 2 + ∠ 5 + ∠ 6 = (360°)/2 (∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180° ∠ AOB + ∠ COD =180° Hence both angle are supplementary Similarly, we can prove ∠ BOC + ∠ AOD =180° Hence proved

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### Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.

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