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