Question

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

Answer 1

Given:

Three non-collinear points $A,B$ and $C$ are on a circle.

To prove:

Perpendicular bisector of $AB,BC$ and $CA$ are concurrent.

Construction:

Join $AB,BC$ and $CA$

Draw perpendicular bisector $ST$ of $AB,PM$ of $BC$ of $CA$ are respectively,

As point $A,B$ and $C$ are not collinear, so $ST.PM$ and $QR$ are not parallel and will intersect.

Proof:

Since, $O$ lies on $ST$. the perpendicular bisector of $AB$.

$\Rightarrow OA=OB$...(1)

Similarly. $O$ lies on $PM$ the perpendicular bisector of $BC$

$\Rightarrow OB=OC$...(2)

And, $O$ lies on $QR$ the perpendicular bisector of $CA$

$\Rightarrow OC=0A$...(3)

From (1). (2) and (3),

$OA=OB=OC=r$ (say)

With $O$ as a centre and $r$ as the radius, draw circle $C(0,r)$ which will pass through $A.B$ and $C$.

This prove that there is a circle passing through the points$A,B$ and $C$.

Since $ST,PM$ or $QR$ can cut each other at one and only one point $O$.

$\Rightarrow O$ is the only point equidistant from $A,B$ and $C$.

Hence, the perpendicular bisectors of $AB,BC$ and $CA$ are concurrent.