Q. 103.7( 39 Votes )

In any triangle ABC, if the angle bisector of A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

Answer :

Let perpendicular bisector of side BC and angle bisector of A meet at point D. Let the perpendicular bisector of side BC intersect it at E


Perpendicular bisector of side BC will pass through circum centre O of the circle


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


BOC = 2 BAC = 2 A (i)


In ΔBOE and ΔCOE,


OE = OE (Common)


OB = OC (Radii of same circle)


OEB = OEC (Each 90° as OD perpendicular to BC)


ΔBOE COE (RHS congruence rule)


BOE = COE (By CPCT) (ii)


However,


BOE + COE = BOC


BOE + BOE = 2 A [From (i) and (ii)]


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 (iii)


Since AD is the bisector of angle A


BAD =


2 BAD = A (IV)


From (iii) and (iv), we get


BOD = 2 BAD


Hence, proved


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