Question

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

Answer 1

Given. Three non-collinear points $A,B\mathrm{and}C$ are on a circle.

To prove. The perpendicular bisector of $AB,BC\mathrm{and}CA$ are concurrent.

Construction. Join $AB,BC\mathrm{and}CA$

Draw $ST$ which is the perpendicular bisector of $AB$, $PM$ which is the perpendicular bisector of $BC$ and, $QR$ which is the perpendicular bisector of $CA$.

Proof:

Since $O$ lies on $ST$. The perpendicular bisector of $AB$.

$\therefore OA=OB...\left(1\right)$

Since $O$ lies on $PM$. The perpendicular bisector of $BC$.

$\therefore OB=OC...\left(2\right)$

Since $O$ lies on $QR$. The perpendicular bisector of $CA$.

$\therefore OC=OA...\left(3\right)$

From equation $\left(1\right),\left(2\right)\mathrm{and}\left(3\right)$.

$OA=OB=OC=r\left(\mathrm{say}\right)$

With $O$ as a center and $r$ as the radius, draw circle which will pass through $A.B\mathrm{and}C$.

It proves that there is a circle passing through the points A. B and C.

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

Therefore, $O$ is the only point equidistant from $A.B\mathrm{and}C$.

Hence, the perpendicular bisectors of $AB,BC\mathrm{and}CA$ are concurrent.