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# 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.

Hard Open in App Solution Verified by Toppr

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.

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

Summary: Tangents and Its Properties

Prove that op... Question

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

Open in App Solution

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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## 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)

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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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