# Ex 10.6, 10 (Optional) - Chapter 10 Class 9 Circles (Term 2)

Last updated at Feb. 27, 2019 by Teachoo

Last updated at Feb. 27, 2019 by Teachoo

Transcript

Ex 10.6, 10 (Optional) In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC Circumcircle is a circle where all 3 vertices of triangle are on the circle Given: In ∆ABC, AD is angle bisector of ∠A and OD is perpendicular bisector of BC, intersecting each other at point D. To Prove: D lies on the circle Construction: Join OB and OC Proof: Since BC is a chord of the circle Hence, its perpendicular bisector will pass through centre O of the circumcircle. ∴ OE ⊥ BC & E is mid-point of BC Also, Chord BC subtends twice the angle at center, as compared to any other point. BC subtends ∠BAC on the circle & BC subtends ∠BOC on the center ∴ ∠BAC = 1/2 ∠ BOC In ∆ BOE and ∆COE BE = CE ∠BEO = ∠CEO OE = OE ∴ ∆BOE ≅ ∆COE ∴ ∠BOE = ∠COE Now, ∠BOC = ∠BOE + ∠COE ∠BOC = ∠BOE + ∠BOE ∠BOC = 2 ∠BOE (OD bisects BC) (Both 90°, as OD ⊥ BC) (Common) (SAS Congruency rule) (CPCT) …(2) Also, given that AD is angle bisector of ∠A ∴ ∠BAC = 2∠BAD From (1) ∠BAC = 1/2 ∠BOC Putting ∠ BAC = 2 ∠BAD & ∠ BOC = 2 ∠BOE 2 ∠BAD = 1/2 (2∠BOE) 2 ∠BAD = ∠BOE ∠BAD = 1/2 ∠BOE Now, BD subtends ∠BOE at center and half of its angle at Point A So, BD must be a chord ∴ D lies on circle Hence Proved

Ex 10.6 (Optional)

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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.