NCERT Solutions for Class 10 Maths Chapter 10

Important NCERT Questions

Circles

NCERT Books for Session 2022-2023

CBSE Board and UP Board Others state Board

EXERCISE 10.2

Page No:214

Questions No:13

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

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Quadrilateral ABCD is circumscribing a circle with centre O, touching at point P, Q, R and S.

join the points P, Q, R and S from the centre O.

In ΔOAP and ΔOAS,

OP = OS [radill of same circle]

AP = AS [Tangents drawn from point A]

AO = AO [Common]

ΔOPA ≅ ΔOCA [SSS Congruency rule]

Hence, ∠POA = ∠SOA or ∠1 = ∠8

Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7

Sum of all angles at point O is 360°. Therefore

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

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

⇒ 2∠1 + 2∠2 + 2∠5 + 2∠6 = 360

⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360

⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180°

⇒ ∠AOB + ∠COD = 180°

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

Hence, the opposite sides subtend supplementary angles at the centre.