The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?                                                                                                                                            [Page No. 94]

The locus of all points equidistant from two given points A and B is the perpendicular bisector of the segment AB, the straight line through the midpoint of AB that is perpendicular to it. To be sure this line is exactly the locus, two things must be checked.

(i) Every point on the perpendicular bisector is equidistant from A and B.
Let M be the midpoint of AB and P any point on the bisector.
In $\Delta PMA$ and $\Delta PMB$ we have $PM = PM$ (common), $MA = MB$ (M is the midpoint) and
 $\angle PMA = \angle PMB = 90^\circ$.
By the SAS congruence, $\Delta PMA \cong \Delta PMB$, so $PA = PB$.

(ii) Every point equidistant from A and B lies on the perpendicular bisector. Suppose $PA = PB$, and let M be the midpoint of AB. In $\Delta PMA$ and $\Delta PMB$, $PA = PB$ (given),  $MA = MB$ and $PM$ is common, so by the SSS congruence $\Delta PMA \cong \Delta PMB$ and $\angle PMA = \angle PMB$.
These two angles lie along the line AB and add up to $180^\circ$, so each is $90^\circ$; hence $PM \perp AB$ and $P$ lies on the perpendicular bisector.
Since both directions hold, the perpendicular bisector is precisely the required locus.