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Theorems

Theorem 9.1

Theorem 9.2 Important

Theorem 9.3 Important

Theorem 9.4

Theorem 9.5 Important

Theorem 9.6 You are here

Theorem 9.7 Important

Theorem 9.8

Theorem 9.9 Important

Theorem 9.10

Theorem 9.11 Important

Angle in a semicircle is a right angle Important

Only 1 circle passing through 3 non-collinear points Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

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.