# prove that opposite sides of a quadrilateral c 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.

Byju's Answer Standard X Mathematics

Summary: Tangents and Its Properties

Prove that op... Question

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**Step 1: Draw appropriate diagram and use the concept of congruence:**

As we know that by theorem the tangents from the external point are equal

In △OAP,△OAS So, AP = AS ⇒OA = OA

(It is the common side)

⇒OP = OS

(They are the radii of the circle)

So, by

SSS congruency △OAP ≅ △OAS

Thus, ∠POA = ∠AOS This shows that ∠7 = ∠8

In this same manner, the angles which are equal are

⇒∠4 = ∠3 ⇒∠2 = ∠1 ⇒∠6 = ∠5

**Step 2: Use the concept of complete angle**

On adding these angles we get,

∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360° [Complete angle]

On rearranging,

(∠7+∠8)+(∠2+∠1)+(∠4+∠3)+(∠6+∠5) = 360°

⇒

2∠1+2∠8+2∠5+2∠4 = 360°

On taking out 2 as common we get, ⇒

(∠1+∠8)+(∠5+∠4) = 180°

Thus, ∠BOC+∠DOA = 180° Similarly, ∠AOB+∠DOC = 180°

**Hence, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.**

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SIMILAR QUESTIONS

**Q.**2. How to prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Q.**Question 13

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

**Q.**Prove that opposite sides of a quadrilatral circumscribing a ciecle subtend supplementary angle at the centre of the circle.

**Q.**

**Question 13**

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

**Q.**STATEMENT - 1 : The opposite side of a quadrilateral circumscribing a circle subtend supplementary angle at the centre of the circle.

STATEMENT - 2 : The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

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Summary: Tangents and Its Properties

Standard X Mathematics

## 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..

**Check sibling questions**

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

Last updated at March 16, 2023 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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