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.
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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
OR
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.
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
OR
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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What is meant by direct election?
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Direct election is one in which the voters directly choose their representatives to the Legislature.
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