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

Theorem : Angle subtended by a diameter/semicircle on any point of circle is 90° right angle 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° = ∠PAQ 90° = ∠PAQ ∠PAQ = 90° Hence, Proved.

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