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Theorem 10.6 - Chapter 10 Class 9 Circles (Term 2)
Last updated at March 23, 2023 by Teachoo
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
Transcript
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.
