**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