Proving that four lines (which are perpendicular bisectors of chords) meet a point

Question

In the diagram above, each of the four lines is a perpendicular bisector of one of the circles' chord. There are two pairs of circles which touch each other, and of course, as shown in the diagram, those four circles mutually intersect in a point. There must be an interesting connection in this configuration, but I can't see what it is, so I'd appreciate a hint.

Let me emphasise that I did not pose this question before doing any work: I have tried everything under the Sun to prove this claim,

Let $ABCD$ be a convex quadrilaterals such that the diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. [$USAMO\ 1993/2$]

and have arrived at this stage. Those four circles are obtained in the process of constructing the reflection points.

Edit: Put a new, hopefully less ambiguous diagram.

Answer 1

Let $ABCD$ be the given quadrilateral with perpendicular diagonals meeting at $E$, whose vertices are the centers of the four circles passing through $E$ mentioned in the question. Let then $PQRS$ be the quadrilateral whose vertices are the reflections of $E$ across the sides of $ABCD$ and whose sides are the chords mentioned in the question (see picture below). Connecting $E$ with $ABCD$ and $PQRS$ we get eight angles of vertex $E$, named $\alpha$ through $\theta$ in the picture.

Consider now the angle $\angle SPQ$ as the sum of four angles: of these we know that $\angle APE=\beta$ and $\angle BPE=\gamma$. Notice then that $AP=AS=AE$, so that $A$ is the center of circle $PSE$ and $\angle PAS=2\alpha+2\beta$. It follows that $\angle APS=\pi/2-\alpha-\beta$ and by a similar argument we also get $\angle BPQ=\pi/2-\gamma-\delta$.

Summing all those angles we finally get $$ \angle SPQ = (\pi/2-\gamma-\delta)+\gamma+\beta+(\pi/2-\alpha-\beta) =\pi-\alpha-\delta. $$

An analogous reasoning would show that $\angle SRQ=\pi-\theta-\epsilon$, so that: $$ \angle SPQ+\angle SRQ=2\pi-(\alpha+\theta)-(\delta+\epsilon) =2\pi-\pi/2-\pi/2=\pi. $$ If follows that $PQRS$, having two opposite angles supplementary, is a cyclic quadrilateral.

Answer 2

Too long for a comment :

• For starters, notice that the entire quadrilateral is circumscribable, i.e., all of its four vertices $($blue$)$ are concyclic.

• Secondly, by reflecting the $($red$)$ intersection of its diagonals O with regard to its four sides, we get four $($orange$)$ points, which lie on the same circle as the four $($green$)$ points determined by reflecting O with regard to the $($green$)$ midpoints of its four sides.

• By reflecting O with regard to the four $($blue$)$ vertices, we get another group of four concyclic $($red$)$ dots.

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Regarding the second observation, I was thinking that since by reflecting O with regard to the midpoint of each side we get a parallelogram $($specifically, a rectangle$)$, it is easy to see how all four reflections form a bigger rectangle, and we know that rectangles are circumscribable, i.e, all their four vertices are concyclic. So if there would be a way to show that reflection with regard to a side is “the same” $($from this perspective$)$ as reflecting with regard to its midpoint, then we're done. I also suspect that there is a deep connection between the first observation, and the second.