Proof : all points are in the set are in the same line if three of them are

Question

The set S contains three points. Each round we can choose three points from this set( for example ABC) and add the symmetry of A from the perpendicular bisector of the line BC to the set. If at some point we realise there are three points in one line, prove that all points in S are in one line. (Sorry English is my second language and i didn't know the proper mathematical expressions i should use here. I'mjust trying to convey the meaning. ) My guess was that we can proceed from each three points that are not on same lime form a triangle and we know that a triangle has three perpendicular bisectors which meet inside the the triangle and can never be parallel while if three dots are in a line then we have two parallel perpendicular bisectors. But i don't know how to proceed from there.

Answer 1

Every three dots in the set form a triangle. We can put the triangle in a circle in a way that all the vertices are on the perimiter of the circle. it can be proved that the symmetry of each vertex of a triangle from the other two vertices perpendicular bisector will end up on the same circle. And the the new added line can again form a circle with the former dots and the same logic would apply. So there never will be three dots on one line. Therefore if three dots end up on one line, all dots should be on one line from the start. Sorry again. English is my second language and i Don't know the correct mathematical expressions. )