Theorem 10.8 : The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle Class 9

Given: a circle with centre O. Arc PQ of this circle subtended angles POQ at centre O and ∠PAQ at a point A remaining part of a circle.

To prove :  ∠POQ = 2∠PAQ

Construction: join AO and extend it to point B.

Proof: there are two general cases in this question 

CASE 1:

⇒In △AOP

OA = OP (Radius Of The Circle)

∠APO = ∠PAO (Angle opposite to equal sides are equal)     

⇒By Exterior angle property

⇒(Exterior angle is the sum of interior opposite angle)

⇒∠POB = ∠APO + ∠PAO

⇒∠POB = ∠PAO + ∠PAO

⇒∠POB = 2∠PAO                 ….. (1)

⇒In △AOQ

OA = OQ (Radius Of The Circle)

∠AQO = ∠OAQ (Angle opposite to equal sides are equal)     

⇒By Exterior angle property

⇒(Exterior angle is the sum of interior opposite angle)

⇒∠QOB = ∠AQO + ∠OAQ

⇒∠QOB = ∠OAQ + ∠OAQ

⇒∠QOB = 2∠OAQ                 ….. (2)

Adding equations (1) and (2)

⇒∠POB + ∠QOB = 2∠PAO + 2∠OAQ 

⇒∠POQ = 2(∠PAO + ∠OAQ )

⇒∠POQ = 2∠PAQ

Hence proved 

CASE 2:

⇒In △AOP

⇒ OA = OP (Radius Of The Circle)

⇒∠APO = ∠PAO (Angle opposite to equal sides are equal)     

⇒By Exterior angle property

⇒(Exterior angle is the sum of interior opposite angle)

⇒∠POB = ∠APO + ∠PAO

⇒∠POB = ∠PAO + ∠PAO

⇒∠POB = 2∠PAO                 ….. (1)

⇒In △AOQ

⇒ OA = OQ (Radius Of The Circle)

⇒∠AQO = ∠OAQ (Angle opposite to equal sides are equal)     

By Exterior angle property

(Exterior angle is the sum of interior opposite angle)

⇒∠QOB = ∠AQO + ∠OAQ

⇒∠QOB = ∠OAQ + ∠OAQ

⇒∠QOB = 2∠OAQ                 ….. (2)

Adding equations (1) and (2)

⇒ ∠POB + ∠QOB = 2∠PAO + 2∠OAQ 

Reflex angle ∠POQ = 2(∠PAO + ∠OAQ )

Reflex angle ∠POQ = 2∠PAQ

⇒ 360°- ∠POQ = 2∠PAQ

Hence proved

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