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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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