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

Explanation

Given `DeltaABC` is inscribed in a circle. Bissecter of `angleA` and perpendicular bisector of BC intersect at point Q.
To prove A, B, Q and C are non-cyclic. Constrution Join BQ and QC.
Proof We have assumed that, Q lies outside the circle.
In `DeltaBMQ and DeltaCMQ`,
BM=CM [QM is the perpendicular bisector of BC]
`angleBMQ=angleCMQ ["each" 90^(@)]`
MQ=MQ [common side]
`:. DeltaBMQ cong DeltaCMQ` [by SAS congruence rule]
`:. BQ=CQ` [by CPCT]..(i)
Also, `angleBAQ=angleCAQ` [given]..(ii)
From Eqs. (i) and (ii), we cen say that Q lies on the circle.
[equal chords of a circle subtend equal angles at the circumference.]
Hence, A, B, Q and C are non-cyclic.