Theorem 9

Transcript

Theorem 9 The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc. Given: Arc PQ of this circle subtends angles POQ at centre O & ∠ PAQ at a point A remaining part of circle. To Prove: ∠POQ = 2∠PAQ Proof: There are two general cases – minor arc and major arc Proving Case I Join AO and extend it to point B In ∆APO OP = OA ∴ ∠OPA = ∠OAP Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOP = ∠OPA + ∠OAP ∠BOP = ∠OAP + ∠OAP ∠BOP = 2∠OAP In ∆AQO OQ = OA ∴ ∠OQA = ∠OAQ Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOQ = ∠ OQA + ∠ OAQ ∠ BOQ = ∠OAQ + ∠OAQ ∠BOQ = 2∠OAQ Adding (3) and (4) ∠BOP + ∠BOQ = 2∠OAP + 2∠OAQ ∠POQ = 2(∠OAP + ∠OAQ) ∠ POQ = 2∠PAQ Hence Proved Proving Case II Join AO and extend it to point B In ∆APO OP = OA ∴ ∠OPA = ∠OAP Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOP = ∠OPA + ∠OAP ∠BOP = ∠OAP + ∠OAP ∠BOP = 2∠OAP In ∆AQO OQ = OA ∴ ∠OQA = ∠OAQ Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOQ = ∠ OQA + ∠ OAQ ∠ BOQ = ∠OAQ + ∠OAQ ∠BOQ = 2∠OAQ Adding (3) and (4) ∠BOP + ∠BOQ = 2∠OAP + 2∠OAQ Reflex angle ∠POQ = 2 (∠OAP + ∠OAQ) Reflex angle ∠POQ = 2 ∠PAQ 360° − ∠POQ = 2∠PAQ Hence Proved