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

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.