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

Theorem 10.4 The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Given : A circle with center at O. AB is chord of circle & OX bisects AB i.e. AX = BX To Prove : OX ⊥ AB Proof : In ∆AOX & ∆BOX OA = OB OX = OX AX = BX ∴ ∆AOX ≅ ∆BOX ∠ AXO = BXO In line AB, Hence, ∠AXO and ∠BXO form linear Pair ∠AXO + ∠BXO = 180° ∠AXO + ∠AXO = 180° 2 ∠AXO = 180° ∠AXO = (180°)/2 ∠AXO = 90° ∴ ∠AXO = ∠BXO = 90° ⇒ OX ⊥ AB 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.