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.
Proof : all points are in the set are in the same line if three of them are
discrete mathematicsgeometryinvariance
Related Solutions
First off, the other two answers here are wrong because you can prove the statement using the same flaw if P was outside the triangle. The proof using the same flaw but done in the outside of the triangle is here: [https://www.youtube.com/watch?v=Yajonhixy4g][1]
The video clearly does this proof with P on the outside. If P were on the segments of the triangle, then the proof will still hold because triangle AEP will be congruent to triangle AFP by SAA (shared sides, bisected angles, 90 degree angles). You know that P and D are the same points because the definition of P is where the perpendicular bisector meets the bisector of the opposite angle, and provided the case where P's on BC, it can only possibly be a triangle if P and D are on the same place. D is the bisector of BC, so so is P. EP = FP because we proved that AEP = AFP. BP = CP because P bisects and is between BC. Since EBP and FCP are right triangles and the Hypotenuse and one of the legs are congruent, we have triangle EBP is congruent to triangle FCP. From here, basic elementary algebra can be use to prove AB = AC.
So, what is wrong with this proof? The answer is betweenness. Every single one of these proofs rely on a certain order of points your drawing above should be:
We still have AE=AF, PE=PF, and PB=PC, and it's still true that BE=FC, but AB=AC is not true. This is because F is still between A and C, but E is not between A and B. Since E is not between A and B, SAA does not follow, and neither does SAS for this particular case.
Let $R$ be the another point (different of $P$) where the $CP$ meets the circumcircle of $APQ$, let $F$ be the point where $BQ$ and $CP$ meets. Since $A$, $P$, $R$ and $Q$ lies in the same circumference it follows $$\measuredangle CRQ =180^{\circ}-\measuredangle PRQ=\measuredangle PAQ=\measuredangle BAC$$ Also, $\measuredangle BAC=\measuredangle PCB=\measuredangle RCB$ for construction, so $\measuredangle CRQ=\measuredangle RCB$ and $QR$ is parallel to $BC$. Then $\angle QRF=\angle FCB=\angle BAC=\angle QBC =\angle FQR$, i.e. $\triangle FBC$ and $\triangle FQR$ are isosceles. It follows that the bisector of $BC$ is perpendicular to $BC$ and passes by $F$, the same for the bisector of $QR$, so the bisector of $BC$ is the bisector of $QR$ too, and it is the line by the circumcircles of $\triangle ABC$ and $\triangle BPQ$
Best Answer
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. )