# Perpendicular bisector construction of a quadrilateral

In geometry, the perpendicular bisector construction of a quadrilateral is a construction which produces a new quadrilateral from a given quadrilateral using the perpendicular bisectors to the sides of the former quadrilateral. This construction arises naturally in an attempt to find a replacement for the circumcenter of a quadrilateral in the case that is non-cyclic.

## Definition of the construction

Suppose that the vertices of the quadrilateral $Q$  are given by $Q_{1},Q_{2},Q_{3},Q_{4}$ . Let $b_{1},b_{2},b_{3},b_{4}$  be the perpendicular bisectors of sides $Q_{1}Q_{2},Q_{2}Q_{3},Q_{3}Q_{4},Q_{4}Q_{1}$  respectively. Then their intersections $Q_{i}^{(2)}=b_{i+2}b_{i+3}$ , with subscripts considered modulo 4, form the consequent quadrilateral $Q^{(2)}$ . The construction is then iterated on $Q^{(2)}$  to produce $Q^{(3)}$  and so on.

An equivalent construction can be obtained by letting the vertices of $Q^{(i+1)}$  be the circumcenters of the 4 triangles formed by selecting combinations of 3 vertices of $Q^{(i)}$ .

## Properties

1. If $Q^{(1)}$  is not cyclic, then $Q^{(2)}$  is not degenerate.

2. Quadrilateral $Q^{(2)}$  is never cyclic. Combining #1 and #2, $Q^{(3)}$  is always nondegenrate.

3. Quadrilaterals $Q^{(1)}$  and $Q^{(3)}$  are homothetic, and in particular, similar. Quadrilaterals $Q^{(2)}$  and $Q^{(4)}$  are also homothetic.

3. The perpendicular bisector construction can be reversed via isogonal conjugation. That is, given $Q^{(i+1)}$ , it is possible to construct $Q^{(i)}$ .

4. Let $\alpha ,\beta ,\gamma ,\delta$  be the angles of $Q^{(1)}$ . For every $i$ , the ratio of areas of $Q^{(i)}$  and $Q^{(i+1)}$  is given by

$(1/4)(\cot(\alpha )+\cot(\gamma ))(\cot(\beta )+\cot(\delta )).$

5. If $Q^{(1)}$  is convex then the sequence of quadrilaterals $Q^{(1)},Q^{(2)},\ldots$  converges to the isoptic point of $Q^{(1)}$ , which is also the isoptic point for every $Q^{(i)}$ . Similarly, if $Q^{(1)}$  is concave, then the sequence $Q^{(1)},Q^{(0)},Q^{(-1)},\ldots$  obtained by reversing the construction converges to the Isoptic Point of the $Q^{(i)}$ 's.