Question

Question 3
A, B and C are three points on a circle, prove that perpendicular bisectors of AB, BC and CA are concurrent.

Answer 1

Given a circle passing through three points A, B and C.
Construction : Draw the perpendicular bisectors of AB and AC and let they meet at a point O. Join OA, OB and OC.
Meet at a point O. Join OA, OB and OC
To prove that perpendicular bisector of BC also passes through O i.e., LO, ON and OM are concurrent.

Proof
In $\Delta$ OEA and $\Delta$ OEB
AE = BE [OL is the perpendicular bisector of AB]
$\angle AEO=\angle BEO\left[eachangleis{90}^{\circ }\right]$
And OE = OE [common side]
$\therefore \Delta OEA\cong OEB$ [by SAS congruence rule]
OA = OB [by CPCT]

Similarly, $\Delta OFA\cong \Delta OFC$ [by SAS congruence rule]
$\therefore$ OA = OC
$\therefore$ OA = OB = OC = r [by CPCT]

Now, we draw a perpendicular from O to the BC and join them.
$In\Delta OMBand\Delta OMC$
OB = OC [proved above]
OM = OM [Common side]
And $\angle OMB=\angle OMC\left[eachangleis{90}^{\circ }\right]$
$\therefore$ $\Delta OMB\cong \Delta OMC$ [by RHS congruence rule]
$\Rightarrow$ BM = MC [by CPCT]
Hence, OM is the perpendicular bisector of BC.
Hence, OL, ON and OM are concurrent.