In any triangle ABC,if the angle bisector of

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 = 2A                 ... (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 = 2A
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 =  
 2BAD = 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.
    

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

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.