$A, B$ and $C$ are three points on a circle. Prove that the perpendicular bisectors of $AB, BC$ and $CA$ are concurrent.

(N/A) Given: Three non-collinear points $A, B$ and $C$ lie on a circle.
To prove: The perpendicular bisectors of $AB, BC$ and $CA$ are concurrent.
Construction: Join $AB, BC$ and $CA$. Draw the perpendicular bisectors $ST$ of $AB$,$PM$ of $BC$,and $QR$ of $CA$. Since points $A, B$ and $C$ are not collinear,the lines $ST, PM$ and $QR$ are not parallel and will intersect at some point $O$.
Proof:
$1$. Since point $O$ lies on $ST$,which is the perpendicular bisector of $AB$,we have $OA = OB$ (Any point on the perpendicular bisector of a line segment is equidistant from its endpoints) ... $(1)$
$2$. Similarly,since $O$ lies on $PM$,the perpendicular bisector of $BC$,we have $OB = OC$ ... $(2)$
$3$. Also,since $O$ lies on $QR$,the perpendicular bisector of $CA$,we have $OC = OA$ ... $(3)$
From $(1)$,$(2)$,and $(3)$,we get $OA = OB = OC = r$ (where $r$ is the radius).
This implies that $O$ is the center of the circle passing through $A, B$ and $C$. Since the perpendicular bisectors of the sides of a triangle intersect at a unique point (the circumcenter),the perpendicular bisectors of $AB, BC$ and $CA$ must be concurrent at point $O$.