Prove that angle bisector of any angle of

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. 

Hence proved.