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Theorems

Theorem 10.1

Theorem 10.2 Important

Theorem 10.3 Important Deleted for CBSE Board 2023 Exams

Theorem 10.4

Theorem 10.5

Theorem 10.6 Important You are here

Theorem 10.7

Theorem 10.8 Important

Theorem 10.9

Theorem 10.10 Important

Theorem 10.11

Theorem 10.12 Important

Angle in a semicircle is a right angle Important

Chapter 10 Class 9 Circles

Serial order wise

Last updated at Aug. 25, 2021 by Teachoo

Theorem 10.6 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.