# What are the conditions required for the perpendicular bisectors of all sides of a quadrilateral to intersect?

What are the conditions required for the perpendicular bisectors of all sides of a quadrilateral to intersect?

Actually, this question came in my mind while I was thinking about how a circle can pass through 4 given points. I think the properties should be same as of the cyclic quadrilaterals, but I am not sure or they all can bisectors even if it's not a cyclic quadrilateral.

• If all four perpendicular bisectors meet at the same point, then the quadrilateral is cyclic (and the proof is trivial). Commented Jul 18, 2019 at 7:07

The perpendicular bisector of a segment $$AB$$ is exactly the set of points that have equal distance to $$A$$ and $$B$$. If the bisectors of $$AB$$ and $$BC$$ intersect in a point $$P$$, then the three distances are equal: $$\lvert P,A\rvert=\lvert P,B\rvert=\lvert P,C\rvert$$ (and $$P$$ will automatically lie on the bisector of $$AC$$, which is the reason the perpendicular bisectors of a triangle intersect in a single point). If the bisectors over $$CD$$ and $$DA$$ also pass through $$P$$, then the distances from $$P$$ to $$A,B,C,D$$ are all the same, hence they lie on a circle around $$P$$. Conversely, if they lie on a circle, then the center of the circle has equal distance to all four points, hence lies on all the perpendicular bisectors.