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Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.

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According to the question,

Triangle ABC and l is perpendicular bisector of BC.

To prove:

Angles bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ΔABC.

Proof:

Let the angle bisector of ∠A intersect circumcircle of ΔABC at D.

Construction: Join BP and CP.

Since, angles in the same segment are equal

We have, ∠BAP = ∠BCP

We know that,

AP is bisector of ∠A.

Then,

∠BAP = ∠BCP = ½ ∠A …(1)

Similarly,

We have,

∠PAC = ∠PBC = ½ ∠A …(2)

From equations (1) and (2),

We have

∠BCP = ∠PBC

We know that,

If the angles subtended by two Chords of a circle at the centre are equal, the chords are equal.

So,

BP = CP

Here, P is on perpendicular bisector of BC.

Hence, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ΔABC.

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