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


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.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.