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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.
Given $\triangle A B C$ is inscribed in a circle. Bisecter of $\angle A$ and perpendicular bisector of $B C$ intersect at point $Q$.
To prove $A, B, Q$ and $C$ are con-cyclic.
Construction Join $B Q$ and $Q C$.
Proof We have assumed that, $Q$ lies outside the circle.
In $\triangle B M Q$ and $\triangle C M Q$,
$B M=C M$ $[Q M$ is the perpendicular bisector of $B C]$
$\angle B M Q=\angle C M Q$ [each 90°]
$M Q=M Q$ [common side]
$\therefore$ $\triangle B M Q \equiv \triangle C M Q$ [by SAS congruence rule]
$\therefore \quad B Q=C Q$ [by CPCT]...(i)
Also, $\angle B A Q=\angle C A Q$ [given]...(ii)
From Eqs. (i) and (ii), we can 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 con-cyclic.