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Theorem 4
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 4 The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord Given: AB is chord of circle & M is mid-point of AB i.e. AM = BM To Prove: OM ⊥ AB i.e. ∠ OMB = ∠ OMA = 90° Proof: We prove ∆AOM & ∆BOM congruent In ∆AOM & ∆BOM OA = OB OM = OM AM = BM ∴ ∆AOM ≅ ∆BOM Thus, by Corresponding parts of Congruent Triangles (CPCT) ∠ AMO = BMO Now, in line AB, ∠AMO and ∠BMO form Linear Pair So, we can write ∠AMO + ∠BMO = 180° Putting ∠AMO = ∠BMO from (1) ∠AMO + ∠AMO = 180° 2 ∠AMO = 180° ∠AMO = (180°)/2 ∠AMO = 90° Therefore, ∠AMO = ∠BMO = 90° Thus, OM ⊥ AB Hence proved
