Show that isometries of $\mathbb{S^2}$ are determined by the image of three points $A,B,C$ not in a "line".

Question

Given information: If points $P,Q \in \mathbb{S^2}$ have the same distace from three points $A,B,C \in \mathbb{S^2}$ not in a "line", then $P=Q$.

Deduce from the given information that an isometry of $\mathbb{S^2}$ is determined by the images of the three points $A,B,C$ not in a "line".

My understanding of the given information is that the points that are equidistant from $A,B,C$ are given by the intersection of three great circles. These three great circles intersect at two points, one is between the triangle spherical triangle $A,B,C$ and the other is it's antipodal point on the other side of the sphere. So we know that any point that is equidistant from $A,B,C$ by a given distance $k$ is unique. But how does this relate to isometries of $\mathbb{S^2}$?

Answer 1

Let $d(X,Y)$ be the distance between points $X$ and $Y$.

Then the given information is saying that $d(P,A)=d(Q,A), d(P,B)=d(Q,B),$ and $d(P,C)=d(Q,C)$. It's not saying that $P$ is equidistant from $A,B,C$ (and same for $Q$).

You are being asked to show that a point is uniquely determined by its distances from three points not in a line.

(Hint: assume that two distinct points $P$ and $Q$ have the same distances to three points not on a line, and see if you can get a contradiction).

(Hint 2: Draw the line that is equidistant from $P$ and $Q$. Where are $A,B,C$ relative to that line?)