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Angle subtended by Diameter is 90°
Last updated at May 26, 2026 by Teachoo
Class 9
Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja
- Definitions
- Symmetries of Circle
- Number of Circles
- Exercise Set 5.1
- Think, Draw and Infer (Page 98)
- Angle Subtended by Chords
- Exercise Set 5.2
- Mid-points and Perpendicular Bisectors of Chords
- Exercise Set 5.3
- Distance of Chords from the Centre
- Exercise Set 5.4
- Distance of unequal chords from center
- Exercise Set 5.5
- Angle subtended by an arc
- Exercise Set 5.6
- Concyclicity of Points and Cyclic Quadrilateral
-
Theorems (and their Proofs)
- End-of-Chapter Exercises
Transcript
Theorem Angle subtended by a diameter/semicircle on any point of circle is 90° Given: A circle with centre at O. 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° By Theorem 9: 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
