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Theorem 6
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 6 Chords of a circle having the same length are all at the same distance from the centre of the circle Given: A circle with center at O; AB and CD are two equal chords of circle i.e. AB = CD; OX & OY are perpendicular distance of AB & CD from center O respectively To Prove: OX = OY Proof : We make ∆ AOX & ∆ COY congruent, and by CPCT we can prove OX = OY We know that Perpendicular from the center to the chord, bisects the chord Now, given that AB = CD Dividing both sides by 2 𝐴𝐵/2 = 𝐶𝐷/2 AX = CY Since OX ⊥ AB We can write AX = BX = (𝑨𝑩 )/𝟐 Since OY ⊥ CD We can write CY = DY = (𝑪𝑫 )/𝟐 In ∆ AOX and ∆COY ∠OXA = ∠OYC OA = OC AX = CY ∴ ∆AOX ≅ ∆COY Thus, by Corresponding parts of Congruent Triangles (CPCT) OX = OY Hence proved (Both 90°, given) (Radius) (From (1)) (by R.H.S rule)
