A version of this answer, with diagrams and notations, is on my blog, at http://mathdoesnthurt.blogspot.com/2014/10/a-question-was-posed-about-following.html.
First, we must understand the problem.
What are we asked to find? (Alternately, what is the unknown?)
We are asked to find a place, with reference to three towns.
What are the data?
We are told only that there are three towns. Notably, we know nothing of the towns’ relative locations.
What is the condition?
Our solution place must be equally distant from each of the three towns.
Is it possible to satisfy the condition?
We should note that the towns cannot lie on a straight line. If they did, any point equally distant from the two farthest towns would be closer to the town between them.
As a final step in understanding the problem, we should introduce suitable notation and draw a diagram.
We will call our towns A, B, and C, and we will represent them as noncollinear points in a plane. We will call the equidistant point that we are looking for P.
Having understood the problem, we must next devise a strategy.
Do we know a related problem?
We may not know how to find a point equidistant from three other points, but we do know how to find a point equidistant from two other points: it is the midpoint of the line segment between them. Let's focus on towns A and B. We will draw a line segment AB¯¯¯¯¯ and construct its midpoint, which we'll call M.
Point M isn't the only point that's equidistant from A and B, though. Any point that lies on the perpendicular bisector of AB¯¯¯¯¯ (let's call such a point N) is also equidistant from A and B.
Point N is equidistant from A and B, wherever N lies on the perpendicular bisector of AB¯¯¯¯¯, because ?AMN is congruent to ?BMN by the side-angle-side postulate. BM¯¯¯¯¯¯¯=AM¯¯¯¯¯¯¯, ∠BMN≅∠AMN, and MN¯¯¯¯¯¯¯=MN¯¯¯¯¯¯¯.
Now we have an infinite number of points to work with, which are all equidistant from A and B. If one of these points is also equidistant from C, then we will have found our solution. Our strategy will be to use the same process to find such a point.
So let's execute our strategy.
As we did to find a point equidistant from A and B, we'll draw the line segment BC¯¯¯¯¯ and construct its midpoint Q.
What are we asked to find? (Alternately, what is the unknown?)
We are asked to find a place, with reference to three towns.
What are the data?
We are told only that there are three towns. Notably, we know nothing of the towns’ relative locations.
What is the condition?
Our solution place must be equally distant from each of the three towns.
Is it possible to satisfy the condition?
We should note that the towns cannot lie on a straight line. If they did, any point equally distant from the two farthest towns would be closer to the town between them.
As a final step in understanding the problem, we should introduce suitable notation and draw a diagram.
We will call our towns A, B, and C, and we will represent them as noncollinear points in a plane. We will call the equidistant point that we are looking for P.
Having understood the problem, we must next devise a strategy.
Do we know a related problem?
We may not know how to find a point equidistant from three other points, but we do know how to find a point equidistant from two other points: it is the midpoint of the line segment between them. Let's focus on towns A and B. We will draw a line segment AB¯¯¯¯¯ and construct its midpoint, which we'll call M.
Point M isn't the only point that's equidistant from A and B, though. Any point that lies on the perpendicular bisector of AB¯¯¯¯¯ (let's call such a point N) is also equidistant from A and B.
Point N is equidistant from A and B, wherever N lies on the perpendicular bisector of AB¯¯¯¯¯, because ?AMN is congruent to ?BMN by the side-angle-side postulate. BM¯¯¯¯¯¯¯=AM¯¯¯¯¯¯¯, ∠BMN≅∠AMN, and MN¯¯¯¯¯¯¯=MN¯¯¯¯¯¯¯.
Now we have an infinite number of points to work with, which are all equidistant from A and B. If one of these points is also equidistant from C, then we will have found our solution. Our strategy will be to use the same process to find such a point.
So let's execute our strategy.
As we did to find a point equidistant from A and B, we'll draw the line segment BC¯¯¯¯¯ and construct its midpoint Q.
And, again as before, we'll draw the perpendicular bisector b of BC¯¯¯¯¯.
The point where a and b intersect lies on both lines. Since every point on line a is equidistant from A and B, and every point on line b is equidistant from B and C, that intersection must be point P, our solution.
But wait, do we need to consider CA¯¯¯¯¯? Nope. We know PA¯¯¯¯¯=PB¯¯¯¯¯, and PB¯¯¯¯¯=PC¯¯¯¯¯, so, by the transitive property, PA¯¯¯¯¯=PC¯¯¯¯¯.
If we want to be fancy, we can call P the circumcenter of ?ABC, which is the center of the circumcircle of ?ABC, which in turn is the circle P, R=PA¯¯¯¯¯ that circumscribes ?ABC.
