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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Question

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.

Byju's Answer Standard X Mathematics

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.

Suggest Corrections 63

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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Theorems MATHEMATICS Watch in App EXPLORE MORE

Summary: Tangents and Its Properties

Standard X Mathematics

स्रोत : byjus.com

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 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 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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