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Theorem 9.7 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo