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.
Proof:
Given: \(△ABC\) is inscribed in a circle.
The bisector of \(∠A\) and perpendicular bisector of \(BC\) intersect at point \(O\).
To prove: The angle bisector of \(∠A\) and perpendicular bisector of \(BC\) intersect on the circumcircle of \(△ ABC\).
Proof:
Join \(BO\) and \(OC\).
Let the angle bisector of \(∠A\) intersect the circumcircle of \(△ ABC\) at point \(D\).
Now, join \(DC\) and \(DB\).
Since, angles in the same segment are equal.
Therefore, \(∠BCD = ∠\)
That is, \(∠BCD = ∠\) \(=\frac{1}{2} ∠\) .....(1)
Similarly, \(∠DBC = ∠\) \(= \frac{1}{2} ∠\) .....(2)
From (1) and (2),
\(∠DBC = ∠\)
\(BD = \) [Sides opposite to equal angles are equal]
Thus, \(D\) lies on the perpendicular bisector of \(BC\).
Therefore, the angle bisector of \(∠A\) and perpendicular bisector of \(BC\) intersect on the circumcircle of \(△ ABC\).
Hence, proved.