I don't know where to start. Ceva's theorem?
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The perpendicular bisector is the locus of points such that $\ldots$. It follows that that the intersection between the perpendicular bisectors of $AB$ and $AC$ also lies on the perpendicular bisector of $BC$. – Jack D'Aurizio Mar 01 '17 at 02:03
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How did you plan to apply Ceva's theorem to three lines that are not cevians? – Jack D'Aurizio Mar 01 '17 at 02:03
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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 $BC$.
Arnaldo
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@Gerard L. That come from Pythagoras theorem. $OA^2=OB^2=OM^2+MA^2=OM^2+MB^2$. $M$ is the midpoint of $AB$. Is it clear? – Arnaldo Mar 01 '17 at 02:42
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But then how does "O is in(on) the bisector of BC) mean that all three bisectors intersect at one point? – Gerard L. Mar 01 '17 at 02:53
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Because $O$ is built as the intersection of bisectors of $AB$ and $AC$ and I just proved that it is also in the bisector of $BC$. So all three bisectors goes through $O$. – Arnaldo Mar 01 '17 at 02:56
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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.
Jack D'Aurizio
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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.
DanielWainfleet
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