Theorem 10.7 - Chords equidistant from centre of circle are equal.jpg

2 Theorem 10.7 - Perpendicular from center the chord bisects the chord.jpg
3 Theorem 10.7 - AB = CD  (from 3 and 4) Hence proved.jpg

  1. Chapter 10 Class 9 Circles
  2. Serial order wise
Ask Download

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 from center i.e. OX ⊥ AB & OY is distance 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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail