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Ex 5.3, 2
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
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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
Ex 5.3, 2 An isosceles triangle ABC is inscribed in a circle, with AB=AC. Show that the altitude from A to BC passes through the centre of the circle. Given: ∆ ABC inscribed in circle With AB = AC To Prove: Altitude from A to BC passes through the centre of the circle i.e. AM passes through point O Proof: In any isosceles triangle, The altitude is also the median Thus, point M is the mid-point of chord BC Now, we know from Theorem 4 that The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord Thus, OM ⊥ BC But AM ⊥ BC Now we have two lines, OM and AM, that are both perpendicular to the line BC at the exact same point (M) Since a line can have only one perpendicular drawn to it through a specific point, the line segments OM and AM must lie on the exact same straight line. Therefore, points A, O, and M are collinear, Which means the Altitude AM passes directly through the centre O Hence proved
