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.