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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

triangle ABC.


See pic for reference.

Suppose angle bisector of angle A meets circumcircle with centre O at P, and OP is joined to meet BC at D.                                                            ...(i)

 Angle BAP = Angle CAP (Because AP is angle bisector)
Angle BOP = 2 Angle BAP
Angle COP = 2 Angle CAP
So, Angle BOP = Angle COP

Therefore in triangles BDO and CDO,

BO = CO                          (Radii of the same circle)
Angle BOP = Angle COP (Proven above)
DO = DO                          (Common side

So triangles BDO and CDO are congruent. (SAS criterion)
So, BD = CD and angle BDO = angle CDO (CPCTE)
But angles BDO and CDO form linear pair.
So, angle BDO + angle CDO = 180 degree
So, angle BDO = half of 180 = 90 degree

With BD = CD and angle BDO = 90 degree,
OD is perpendicular bisector of BC
OP is perpendicular bisector of CD                                                  ...(ii)

By (i) and (ii),
P lies on angle bisector of angle A and perpendicular bisector of BC and is a point on the circle.

Hence proven.
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