This is all correct. Actually, for a general quadrilateral ABCD, the locus of all points whose reflections in the sides of the quadrilateral are concyclic is a cubic, the so-called isoptic cubic of the complete quadrilateral formed by the lines AB, BC, CD, DA. But if the quadrilateral ABCD is a rectangle - but also, more generally, if the quadrilateral ABCD is an arbitrary parallelogram! -, then this cubic degenerates into the union of a rectangular hyperbola and the line at infinity. Since we will consider only finite points here, we ignore the line at infinity, and thus we will just show that the locus is a rectangular hyperbola.

In fact, I begin with some preliminaries:

We will use directed angles modulo 180° throughout the following.

If ABC is a triangle, and g is a line which doesn't pass through any of the triangle's vertices, then the isogonal conjugate of the line g with respect to triangle ABC (i. e., the locus of the isogonal conjugates of all points on g with respect to triangle ABC) is a conic passing through the vertices A, B, C of triangle ABC. If the line g passes through the circumcenter of triangle ABC, then this conic is a rectangular hyperbola. One example of a line g passing through the circumcenter of triangle ABC is, of course, the perpendicular bisector of the side CA. Thus, the isogonal conjugate of the perpendicular bisector of the side CA with respect to triangle ABC is a rectangular hyperbola passing through the vertices A, B, C of triangle ABC. This hyperbola will be called the B-medial hyperbola of triangle ABC. Its center is the midpoint of the side CA. Of course, we can similarly define a C-medial and an A-medial hyperbola of triangle ABC (by considering the perpendicular bisectors of the sides AB and BC rather than of the side CA).

Now, the B-medial hyperbola is an important hyperbola and has lots of equivalent definitions:

Theorem 1. Let M be a point in the plane of a triangle ABC. Consider the following possible properties of the point M:

(a) The isogonal conjugate of the point M with respect to triangle ABC lies on the perpendicular bisector of the segment CA.
(b) We have < MCB = < BAM.
(c) If Z and X are the orthogonal projections of the point M on the sides AB and BC of triangle ABC, respectively, then the midpoint of the segment CA lies on the perpendicular bisector of the segment ZX.
(d) If the point D is such that the quadrilateral ABCD is a parallelogram (so, the point D is the point where the parallel to the line AB through the point C meets the parallel to the line BC through the point A), then < BMC = < AMD.
(e) If Q is the point satisfying the vector equation (in other words, the point Q is the image of the point M under the translation along the translation vector ), then the points A, M, D, Q are concyclic.
(f) The orthogonal projections of the point M on the lines AB, BC, CD, DA are concyclic.
(g) The reflections of the point M in the lines AB, BC, CD, DA are concyclic.

Then, each of these properties characterizes the B-medial hyperbola of triangle ABC. In other words, each of these properties holds if and only if the point M lies on the B-medial hyperbola of triangle ABC.

I will give a proof of Theorem 1 without the property (c); actually, I had given property (c) just for the sake of completeness (we won't need it anywhere further; you can prove (c) as a nice exercise). Since property (a) is clearly equivalent to M lying on the B-medial hyperbola of triangle ABC (since the B-medial hyperbola of triangle ABC is the isogonal conjugate of the perpendicular bisector of the segment CA), it is enough to show that the properties (b), (d), (e), (f), (g) are all equivalent to property (a). We will show this by subsequently proving the following equivalences:

(a) (b);
(b) (e);
(e) (d);
(d) (f);
(f) (g).

The proof of (a) (b) is very easy: If M' is the isogonal conjugate of the point M with respect to triangle ABC, then the lines CM and CM' are symmetric to each other with respect to the angle bisector of the angle BCA. Thus, < MCB = < ACM'. Also, the lines AM and AM' are symmetric to each other with respect to the angle bisector of the angle CAB. Thus, < BAM = < M'AC. Hence, the condition < MCB = < BAM is equivalent to < ACM' = < M'AC. But we have < ACM' = < M'AC if and only if the triangle CM'A is isosceles, i. e. if and only if CM' = AM', what is clearly equivalent to the assertion that the point M' lies on the perpendicular bisector of the segment CA. Hence, we have < MCB = < BAM if and only if the isogonal conjugate M' of the point M lies on the perpendicular bisector of the segment CA. Hence, the equivalence (a) (b) is proven.

Now we are going to prove (b) (e). In fact, since , the lines MQ and BA are parallel, and thus < QMA = < (MQ; AM) = < (BA; AM) = < BAM. On the other hand, since the quadrilateral ABCD is a parallelogram, we have ; together with , this yields . Therefore, the quadrilateral CMQD is a parallelogram, so that DQ || CM. On the other hand, DA || CB since the quadrilateral ABCD is a parallelogram. Thus, < QDA = < (DQ; DA) = < (CM; CB) = < MCB. Altogether we have proved < QMA = < BAM and < QDA = < MCB. Thus, we have < MCB = < BAM if and only if < QDA = < QMA, what is clearly equivalent to the assertion that the points A, M, D, Q are concyclic. This proves the equivalence (b) (e).

Now we will establish (e) (d). In fact, using , we have shown DQ || CM, and similarly, from , we can derive AQ || BM. Thus, < AQD = < (AQ; DQ) = < (BM; CM) = < BMC. Hence, < AQD = < AMD holds if and only if < BMC = < AMD holds. But < AQD = < AMD clearly means that the points A, M, D, Q are concyclic. Hence, the points A, M, Q, D are concyclic if and only if we have < BMC = < AMD. Thus, the equivalence (e) (d) is proven.

Let's now settle (d) (f). In fact, let R, S, T, U be the orthogonal projections of the point M on the lines AB, BC, CD, DA. We must show that < BMC = < AMD if and only if the points R, S, T, U are concyclic. Well, the points R, S, T, U are concyclic if and only if < STU = < SRU. Thus, it remains to show that < BMC = < AMD if and only if < STU = < SRU.

Well, since < MSC = 90° and < MTC = 90°, the points S and T lie on the circle with diameter MC, and thus < STM = < SCM. In other words, < STM = < BCM. Similarly, the points T and U lie on the circle with diameter MD, and thus < MTU = < MDU, so that < MTU = < MDA. Hence, < STU = < STM + < MTU = < BCM + < MDA. Analogously, < SRU = < CBM + < MAD. Thus,

< STU - < SRU = (< BCM + < MDA) - (< CBM + < MAD)
= (< BCM - < CBM) - (< MAD - < MDA)
= (< (BC; CM) - < (BC; BM)) - (< (AM; DA) - < (DM; DA))
= < (BM; CM) - < (AM; DM) = < BMC - < AMD.

This immediately shows that < BMC = < AMD if and only if < STU = < SRU. And thus, the equivalence (d) (f) is proven.

Finally we have to verify (f) (g). But this is almost trivial: The reflections of the point M in the lines AB, BC, CD, DA are nothing but the orthogonal projections of the point M on the lines AB, BC, CD, DA, streched with center M and stretch factor 2. Hence, these reflections are concyclic if and only if the projections are concyclic. This proves the equivalence (f) (g).

Altogether, Theorem 1 (without part (c)) is established.

Now, of course, for an arbitrary parallelogram ABCD, Theorem 1 part (g) answers the question on the locus of all points M such that the reflections of the point M in the lines AB, BC, CD, DA are concyclic: This locus is the B-medial hyperbola of triangle ABC. Completely analogously, we can see that this locus is also the C-medial hyperbola of triangle BCD, the D-medial hyperbola of triangle CDA, and the A-medial hyperbola of triangle DAB. Thus, as a bonus, we have obtained the following fact: For any parallelogram ABCD, the B-medial hyperbola of triangle ABC, the C-medial hyperbola of triangle BCD, the D-medial hyperbola of triangle CDA, and the A-medial hyperbola of triangle DAB are actually one and the same hyperbola. I propose calling this hyperbola the medial hyperbola of the parallelogram ABCD. The center of this hyperbola is the center of the parallelogram ABCD (because of the symmetry!), and it is a rectangular hyperbola.

We can thus summarize what we have established: For every parallelogram ABCD, the locus of all points M such that the reflections of the point M in the lines AB, BC, CD, DA are concyclic is the medial hyperbola of the parallelogram ABCD.

Now, the original problem of Pestich asked only for the case when the parallelogram ABCD is a rectangle. Actually, not much changes in this special case (except for the very special case when the rectangle ABCD is a square - in this case, the locus actually degenerates to the union of the two diagonals of the square, as you can easily see). For a generic rectangle ABCD, the locus is indeed a rectangular hyperbola through the vertices of the rectangle, and Grobber's construction of the vertices of this hyperbola directly follows from the fact that the hyperbola is the locus of all points whose reflections in the lines AB, BC, CD, DA are concyclic. Thus, everything is proven.

Darij