Asked by anusarika mohanty
| 22nd Dec, 2013,
04:58: PM
Expert Answer:
Let perpendicular bisector of side BC and angle bisector of
A meet at point D.
Let perpendicular bisector of side BC intersects it at E.
Perpendicular bisector of side BC will pass through circum centre O of circle. Now,
BOC and
BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively.
We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
BOC = 2
BAC = 2
A ... (1)
In
BOE and
COE
OE = OE (common)
OB = OC (radii of same circle)
OEB =
OEC (Each 90o as OD
BC)
BOE
COE (RHS congruence rule)
BOE =
COE (by CPCT) ... (2)
But
BOE +
COE =
BOC
BOE +
BOE = 2
A [Using equations (1) and (2)]
BOE = 2
A
BOE =
A

BOE =
COE =
A
The perpendicular bisector of side BC and angle bisector of
A meet at point D.
BOD =
BOE =
A ... (3)
Since AD is the bisector of angle
A
BAD = 
2
BAD =
A ... (4)
From equations (3) and (4), we have
BOD = 2
BAD
It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.
Therefore, the perpendicular bisector of side BC and angle bisector of
A meet on the circum circle of triangle ABC.

Let perpendicular bisector of side BC intersects it at E.
Perpendicular bisector of side BC will pass through circum centre O of circle. Now,


We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.



In


OB = OC (radii of same circle)







But























From equations (3) and (4), we have


It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.
Therefore, the perpendicular bisector of side BC and angle bisector of

Answered by
| 22nd Dec, 2013,
09:55: PM
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