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  1. Chapter 10 Class 9 Circles
  2. Serial order wise
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Theorem 10.6 Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres). Given : A circle with center at O. AB and CD are two equal chords of circle i.e. AB = CD & OX and OY are perpendiculars to AB & CD respectively. To Prove : OX = OY Proof : Since OX ⊥ AB Perpendicular from the center to the chord, bisects the chord AX = BX = (𝐴𝐵 )/2 Since OY ⊥ CD Perpendicular from the center to the chord, bisects the chord CY = DY = (𝐶𝐷 )/2 Now, given that AB = CD 𝐴𝐵/2 = 𝐶𝐷/2 AX = CY In ∆ AOX and ∆COY ∠OXA = ∠OYC OA = OC AX = CY ∴ ∆AOX ≅ ∆COY OX = OY Hence, Proved.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.