Theorem 10.6 - Class 9 - Equal chords are equidistant from the centre.jpg

2 Theorem 10.6 - Class 9 - In AOX and DOY OXA = OYD (Both given).jpg

  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 : Now, given that AB = CD 𝐴𝐵/2 = 𝐶𝐷/2 AX = DY In ∆ AOX and ∆DOY ∠OXA = ∠DOY OA = DO AX = DY ∴ ∆AOX ≅ ∆DOY OX = DY Hence, Proved.

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