Last updated at Aug. 13, 2018 by Teachoo

Transcript

Theorem 10.7 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.

Theorems

Theorem 10.1

Theorem 10.2 Important

Theorem 10.3 Important

Theorem 10.4

Theorem 10.5 Deleted for CBSE Board 2022 Exams

Theorem 10.6 Important

Theorem 10.7 You are here

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 (Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.