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.
Yes, the conditions are exactly the same as those for cyclic quadrilaterals.
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.
