The perpendicular bisectors of the three sides of any triangle must be concurrent, as is well-known. However, this is not true of quadrilaterals. For quadrilaterals, the four perpendicular bisectors may be concurrent, or not. So when must they be?

The answer: when a pair of opposite angles in the quadrilateral are right angles.

Why is this the case? Well, it’s a consequence of what happens in triangles to this point of concurrence, called the circumcenter. In right triangles — and only in right triangles — the circumcenter falls on a side of the triangle, and that side is always the hypotenuse, with the circumcenter located at its midpoint. If a quadrilateral has two right angles as a pair of opposite angles, as ABCD does in the diagram above, then it can be split into two right triangles with a common hypotenuse, as shown — and that hypotenuse’s midpoint will then be the point of concurrence of all four perpendicular bisectors of the sides of the quadrilateral.

[Later edit:  my friend Andrew make the following comment, when I posted a link to this post on Facebook. I appreciate it when my friends make such corrections.]

“Actually, the Perpendicular Bisectors are concurrent for any Cyclical Quadrilateral. (Opposite angles sum to 180 degrees). Even Non-convex Cyclical Quadrilaterals have this property (Note a Non-convex Cyclical Quadrilateral must be self-intersecting). All Cyclical Quadrilaterals can be circumscribed by a circle.”

For proof, consider the cyclical or non-cyclical quadrilateral (or higher polygon, as well), together with its circumscribed circle. All sides of this polygon are chords of this circle, and perpendicular bisectors of chords pass through the circle’s center — the point of concurrency.