Question

How do I prove that the angle bisector of any angle of a ∆ and the perpendicular bisector of the opposite side, will intersect on the circumcircle of the triangle?

Answer 1

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GIVEN:

A triangle ABC, AP is the bisector of angle A. AP intersects triangle's circumcircle with centre O at P.

TO PROVE THAT:

OP is the perpendicular bisector of BC.

PROOF:

If we prove that OP is the perpendicular bisector of BC, then we can say that the common point ‘P' of AP & OP lies on the circumcircle.

Since angle BOP = 2*angleBAP……(1) ( as angle subtended by a chord at the centre is twice the angle subtended by the same chord on the remaining arc of the circle.

Similarly, angle COP = 2*angleCAP…….(2)

But, angle BAP = angle CAP ( given)

So, angle BOP = angle COP ( by 1 & 2)

Now, in triangle COQ & triangle BOQ

CO = BO ( radii of the same circle)

OQ = OQ ( common side)

angle COQ = angle BOQ ( proved above)

So, triangle COQ is congruent to triangle BOQ ( by SAS congruence criterion)

So, CQ = BQ ( corresponding parts of congruent triangles)

And also, angle CQO= BQO ( cpct)

But angle CQO + angle BQO =180° ( straight angle)

So, each angle = 90°

Therefore, OQ ie OP is perpendicular bisector of BC.

This way, we can say that P lies on the circumcircle.

[ PROVED]