This question is heavily related to the one here Rigid Motions uniquely specified by triangles?. I am struggling with a different aspect of it than that poster. The book's proof is as follows:
Given three points A, B, C (distinct and non-collinear) and three distances a, b, c, there is at most one point distance a from A, b from B, and c from C. For the circles of radius a, centre A, and radius b, centre B meet in at most two points: these are not equidistant from C, so at most one of them is distance c from C. Since rigid motions do not change distances, it follows that once we know what happens to a triangle we know what happens to everything else.
I understand that a single point wouldn't suffice 'cause all would seem to be a translation. By counter example, a line segment (or two points) wouldn't suffice since a reflection about its own line is indistinguishable from the identity transform. But why do 3 points suffice?
In the proof I see that if you had 3 points A, B, and C and drew circles distances a, b, and c from each respective point then the three circles could intersect in at most 1 place. I also understand by definition that rigid motions preserve distance. But I'm lost at the connection of how those two facts imply 3 points are sufficient? Can someone explain a little more?
You have three (not aligned) reference points $A$, $B$, $C$ and their transformed $A'$, $B'$, $C'$ under an arbitrary isometry. Take now any other point $P$ in the plane and set: $AP=a$, $BP=b$, $CP=c$. To find $P'$, the transformed of $P$ under the isometry, notice that $A'P'=AP=a$, $B'P'=BP=b$ and $C'P'=CP=c$. Hence you can just construct the circles of centres $A'$, $B'$, $C'$ and radii $a$, $b$, $c$ respectively: you know that these circles will intersect at a single point, which must then be $P'$.
In other words: once you know the action of the isometry on $ABC$, you can construct the transformed of any other point in the plane. No other information is needed.