Theorem 10.7 - Chords equidistant from centre of circle are equal.jpg

2 Theorem 10.7 - Perpendicular from center the chord bisects the chord.jpg
3 Theorem 10.7 - AB = CD  (from 3 and 4) Hence proved.jpg

  1. Chapter 10 Class 9 Circles
  2. Serial order wise

Transcript

Theorem 10.7 Chords equidistant from the centre of a circle are equal in length. Given : C is 𝑎 circle with center at 0. AB and CD are two Chords of the circle where OX is distance of chord AB from center i.e. OX ⊥ AB & OY is distance of chord AB from center i.e. OY ⊥ CD & OX = OY To Prove : AB = CD Proof : In ∆AOX and ∆CDY ∠OXA = ∠OYC OA = OC OX = OY ∴ ∆AOX ≅ ∆COY AX = CY For Chord AB OX ⊥ AB Perpendiculars from center to the Chord Bisects the Chord ∴ X bisects AB. ∴ AB = 2AX For Chord CD OY ⊥ CD Perpendiculars from center to the chord bisects the Chord ∴ Y bisects CD. ∴ CD = 2CY From (1) AX = CY 2AX = 2CY AB = CD 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.