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Ex 5.3, 1
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
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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
Ex 5.3, 1 Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that ∠CMA=∠CMB=90^∘. You need to show that AM=BM.) Let’s do the proof of the converse Given: AB is a chord And OM ⊥ AB To Prove: OM bisects chord AB i.e. AM = BM Proof: We prove ∆OAM & ∆OBM congruent In ∆OAM & ∆OBM ∠OMA = ∠OMB OA = OB OM = OM ∴ ∆OAM ≅ ∆OBM Thus, by Corresponding parts of Congruent Triangles (CPCT) AM = BM Hence proved
