Question

Theorem 10.8 class 9th circle

Answer 1

Given : A circle with center at O. 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 Construction : Join AO and extend it to point B Proof : There are two general cases Solving Case I Adding (3) and (4) ∠BOP + ∠BOQ = 2∠OAP + 2∠OAQ ⇒ ∠POQ = 2(∠OAP + ∠OAQ) ∠ POQ = 2∠PAQ Hence Proved. CASE II Copy from case I till line with pencil ⇒ reflex angle ∠POQ = 2 (∠OAP + ∠OAQ reflex angle ∠POQ = 2 ∠PAQ or 360° − ∠POQ = 2∠PAQ Hence, Proved Solving Case I 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 Theorem : Angle subtended by a diameter/semicircle on any point of circle is 90° Given : A circle with centre at 0. PQ is the diameter of circle subtending ∠PAQ at point A on circle. To Prove : ∠PAQ = 90° Proof : Now, POQ is a straight line passing through center O. ∴ Angle subtended by arc PQ at O is ∠POQ = 180° Also, By 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. Thus, ∠ POQ = 2∠PAQ ∠ POQ﷮2﷯ = ∠PAQ 180° ﷮2﷯ = ∠PAQ 90° = ∠PAQ ∠PAQ = 90° Hence, Proved.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