Prove that the perpendicular bisectors of all 3 sides of a triangle intersect in one point

Question

I don't know where to start. Ceva's theorem?

Answer 1

You don't need Ceva's theorem. Just build two perpendicular bisector of the triangle $ABC$ (bisector of $AB$ and bisector of $AC$). They will meet at the point "$O$". By definition, $OA=OB$ (because $O$ is in the bisector of $AB$) and also $OA=OC$ (because $O$ is in the bisector of $AC$) and then $OB=OC$. It means that, by definition, $O$ is in the bisector of $BC$.

Answer 2

Alternative approach: the perpendicular bisectors are the altitudes of the median triangle, by Thales' theorem. The orthocenter of the median triangle exists and is unique by Trig Ceva's theorem.

Answer 3

A point P is on the right bisector of AB iff the distances |AP| and |BP| are equal. So if P is on the right bisector of AB and also on the right bisector of BC then |AP|=|BP| and |BP|=|CP|, so |AP|=|CP|, implying that P also lies on the right bisector of AC.