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1 angle subtended by diameter semicircle is 90.JPG 2 angle paq = 90.JPG

  1. Chapter 10 Class 9 Circles
  2. Serial order wise
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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. 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.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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