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Theorem 10.3 - Chapter 10 Class 9 Circles (Term 2)
Last updated at Aug. 25, 2021 by Teachoo
Transcript
Theorem 10.3 The perpendicular from the center of a circle to a chord bisects the chord. Given : C is a circle with center at O. AB is a chord such that OX ⊥ AB To Prove : OX bisect chord AB i.e. AX = BX Proof : In ∆OAX & ∆OBX ∠OXA = ∠OXB OA = OB OX = OX ∴ ∆OAX ≅ ∆OBX AX = BX Hence, Proved.
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Theorems
Theorem 10.1
Theorem 10.2 Important
Theorem 10.3 Important You are here
Theorem 10.4
Theorem 10.5 Deleted for CBSE Board 2022 Exams
Theorem 10.6 Important
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 (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.