# Theorem 10.8

Last updated at Nov. 23, 2017 by Teachoo

Last updated at Nov. 23, 2017 by Teachoo

Transcript

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 ∠ POQ2 = ∠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.