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Theorem 11
Last updated at May 26, 2026 by Teachoo
Class 9
Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja
- Definitions
- Symmetries of Circle
- Number of Circles
- Exercise Set 5.1
- Think, Draw and Infer (Page 98)
- Angle Subtended by Chords
- Exercise Set 5.2
- Mid-points and Perpendicular Bisectors of Chords
- Exercise Set 5.3
- Distance of Chords from the Centre
- Exercise Set 5.4
- Distance of unequal chords from center
- Exercise Set 5.5
- Angle subtended by an arc
- Exercise Set 5.6
- Concyclicity of Points and Cyclic Quadrilateral
-
Theorems (and their Proofs)
- End-of-Chapter Exercises
Transcript
Theorem 11 The sum of two opposite angles of a cyclic quadrilateral is 180°. Given: A circle with center Where ABCD is a cyclic quadrilateral To Prove: ∠ BAD + ∠ BCD = 180° Note: We need to prove sum of 1 pair equal to180°, the other pair can be proved via angle sum property Proof: Join OD & OB We use the theorem Angle subtended by chord at center is double the angle subtended at any other point For Arc DAB Thus, ∠ DOB = 2 × ∠ BAD For Arc DCB Thus, Reflex angle ∠ DOB = 2 × ∠ BCD We know that Angle around the center = 360° Thus, we can write ∠ DOB + Reflex angle ∠ DOB = 360° 2 × ∠ BAD + 2 × ∠ BCD = 360° 2 × (∠ BAD + ∠ BCD) = 360° ∠ BAD + ∠ BCD = (360° )/2 ∠ BAD + ∠ BCD = 180° Hence proved
