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Theorem 9.5 - Chapter 9 Class 9 Circles
Last updated at May 29, 2023 by Teachoo
Theorems
Theorem 9.1
Theorem 9.2 Important
Theorem 9.3 Important
Theorem 9.4
Theorem 9.5 Important You are here
Theorem 9.6
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
Transcript
Theorem 9.6 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 : Since OX ⊥ AB Perpendicular from the center to the chord, bisects the chord AX = BX = (𝐴𝐵 )/2 Since OY ⊥ CD Perpendicular from the center to the chord, bisects the chord CY = DY = (𝐶𝐷 )/2 Now, given that AB = CD 𝐴𝐵/2 = 𝐶𝐷/2 AX = CY In ∆ AOX and ∆COY ∠OXA = ∠OYC OA = OC AX = CY ∴ ∆AOX ≅ ∆COY OX = OY Hence, Proved.
