In any triangle ABC, if the angle bisector of angle A and perpendicular bisector of BC intersect, prove
that they intersect on the circumcircle of the triangle ABC
Answers
Answer:
∠BAE=∠CAE ....Angles inscribed in the same arc
∴arc BE=arc CE
∴ chord BE≅ chord CE ...(1)
In △BDE and △CDE,
BE=CE ....from (1)
BD=CD ...ED is perpendicular bisector of BC
DE=DE ...common side
∴△BDE≅△CDE ...SSS test of congruence
⇒∠BDE=∠CDE .....c.a.c.t.
Also, ∠BDE+∠CDE=180
∘
...angles in linear pair
∴∠BDE=∠CDE=90
∘
Therefore, DE is the right bisector of BC.
Answer:
Given: In ∆ABC, AD is the angle bisector of ∠A and OD is the perpendicular bisector of BC, intersecting each other at point D.
To Prove: D lies on the circle
Construction: Join OB and OC
Proof:
BC is a chord of the circle.
The perpendicular bisector will pass through centre O of the circumcircle.
∴ OE ⊥ BC & E is the midpoint of BC
Chord BC subtends twice the angle at the centre, as compared to any other point.
BC subtends ∠BAC on the circle & BC subtends ∠BOC on the centre
∴ ∠BAC = 1/2 ∠ BOC
In ∆ BOE and ∆COE,
BE = CE (OD bisects BC)
∠BEO = ∠CEO (Both 90°, as OD ⊥ BC)
OE = OE (Common)
∴ ∆BOE ≅ ∆COE (SAS Congruence rule)
∴ ∠BOE = ∠COE (CPCT)
Now,
∠BOC = ∠BOE + ∠COE
∠BOC = ∠BOE + ∠BOE
∠BOC = 2 ∠BOE …(2)
AD is angle bisector of ∠A
∴ ∠BAC = 2∠BAD
From (1)
∠BAC = 1/2 ∠BOC
2 ∠BAD = 1/2 (2∠BOE)
2 ∠BAD = ∠BOE
∠BAD = 1/2 ∠BOE
BD subtends ∠BOE at centre and half of its angle at Point A.
Hence, BD must be a chord.
∴ D lies on the circle.
