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Ex 5.4, 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
- Exercise Set 5.3
- Distance of Chords from the Centre
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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 5.4, 2 Consider Fig. 5.15. If CE is perpendicular to AB,CH is perpendicular to GH, and CE =CH, show that AB=GF. Given: : 𝐶𝐸⊥𝐴𝐵 and 𝐶𝐻⊥𝐺𝐹. And, distances are equal i.e. CE = CH To Prove: AB = GF Proof: We make ∆ CAE & ∆ CGH congruent, and by CPCT we can prove AE = GH, which are half of AB & GF In ∆CAE and ∆CGH ∠CEA = ∠CHG CA = CG CE = CH ∴ ∆CAE ≅ ∆CGH Thus, by Corresponding parts of Congruent Triangles (CPCT) AE = GH We know that Perpendicular from the center to the chord, bisects the chord For Chord AB Since CE ⊥ AB, ∴ E bisects AB. So, we can write AB = 2AE For Chord GF Since CH ⊥ GF ∴ H bisects GF. So, we can write GF = 2GH From (1) AE = GH Multiplying by 2 both sides 2AE = 2GH Putting 2AE = AB & 2GH = GH from (2) &(3) AB = GF Hence proved
