It is well known that $Q_1, Q_2$ are cyclic. (Follows directly from easy angle chase) Let $O_1, O_2$ be the circumcenter of $Q_1, Q_2$, respectively. When $M$ is the midpoint of $O_1O_2$, we will prove that $M$ is the circumcenter of $ABCD$. Assume that the angle bisecters of $\angle DAB$, $\angle CDA$ meets at $X, X_1$, and the angle bisectors of $\angle ABC$, $\angle BCD$ meets at $Y, Y_1$. ($X_1, Y_1$ are the intersection of external bisectors) Defiine $Z,W$ similarly with $X,Y$ so that $Q_1$ is $XZYW$. When $X_2$ and $Y_2$ are the midpoints of $XX_1$ and $YY_1$, $X_2$ and $Y_2$ are each circumcenters of $AXDX_1$ and $BYCY_1$. Also, since $Q_1$ and $Q_2$ are similar quadrilaterals, it is easy to see that $O_1X \parallel O_2X_1$, and $O_1Y \parallel O_2Y_1$. This implies that $MX_2 \parallel O_1X$ and $MY_2 \parallel O_1Y$. It suffices to prove that $O_1X \perp AD$, as this gives that $M = O$ since $X_2$ lies on the perpendicular bisector of segment $AD$, and $MX_2 \perp AD$ means that $M$ lies on the perpendicular bisector of $AD$. Doing this all four sides give the desired result. Let $\angle O_1XZ = \alpha$, $\angle O_1ZY = \beta$, $\angle O_1YW = \gamma$, and $\angle O_1WX = \delta$. We can see that $\angle XZY = \alpha + \beta$, and it goes the same with other angles. Using the relations like $\angle XZY = \pi - \dfrac{1}{2} (\angle CDA + \angle BCD)$, we can derive to a conclusion that gives a linear equation about $\alpha, \beta, \gamma, \delta$ where the constants are only the four angles of $ABCD$. Solving this, we can easily get $\alpha = \frac{\pi}{2} - \angle CDZ$, which implies that $O_1X \perp AD$, as desired.

mb i am on the train on the phone so.i will sketch solution..i also dont have diagram on paper so if my memory is bad oops i will change latwr at home lemma: if a, b are lines that pass through two distinct fixed point at infinity and moving at a constant speed each then their intersection moves at a constant speed. proof: let lines be l1 and l2. consider a fixed line m perpndicular to l1 and n perpendicular to l2. note that m intersect l1 and n intersect l2 also move at fixed speed because l1 at time t is parallel to l1 at time t' and use addendo in parallel lines with transversal m and the line/vector that l1 is moving in, same for l2. now an affine transform such l1 l2 are perpendicular means we can set m,n as directions of x,y axis such m intersect n is origin. conclusion is now obvious (say bc tan^-1 speed l1/speedl2 is fixed) now we angle chase akin to one of my previous post on an rmm 3 or look at above. it is easy to see that the diagonals of Q1 and Q2 coincide and are perpendicular. now take with reference to Q2. well-known Q1 negatively similar to Q2. thus we define map from Q2 to.Q1 by reflecting across one.diagonal of Q2 then dilating at intersection of diagonals. now on for convenience let PQRS be Q2 and transformation involves reflect in PR wlog. then suppose P goes to P'. etc. now move P' constant speed i.e on PR easy to see that A,B.are feet of S' on PQ, QR and C,D feet of Q' onto RS,SP. now clearly CDSQ' cyclic. hence check the slope note direction of Q'D which is fixed. because rightangle to PS. now CDQ' = CSQ by bowtie whcih is fixed. hence CD is fixed slope. similarly AB fixed slope. now by our lemma, perp bisector of AB, CD are fixed slope hence intersection, which is centre of (ABCD), moves on a line. now by similarity, Q1 has P' moves just expands in size so centre moves on a line through PR intersect QS now centre of Q2 is just fixed. now this means the points are deg 1,1,0 hence dilating from centre of Q2 by 2 the centrw of ABCD we get our phantom point which is deg 1. this means we check (1+1)+1=3 cases by combi null lol. use P' as PR intersect QS, P and P reflect over PR intersect QS. ill update soon why rhis wokrs im home slay, now we deal with the special cases. P'=P: direct symmetry of Q1 and Q2 over PR, so the midpoint is true because (ABCD) has centre on PR by symmetry. P' = P reflect over PR intersect QS: same as P=P', symmetry over QS P=PR intersect QS. Let this point be X. because PXQ and RXS are right, they add to 180 so X is on the isoptixook cubic of PQRS hence skibdii isognonal conjugable. the conjugate is centre of (PQRS) because of altitude and cirucmcentre isogonals. yet as Q1 is now just 0 radius so centre is X, we use SIX POINT CIRCLE (xonk) lemma so midpoint of circumcentres of Q1 Q2 is centre of PEDAL CIRCLE xoink!!!! now we are done

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as a sidenote because this problem generalizes the six point circle configuration, #2's synth solve shows shows an elementary proof of midpoint of isogonal conjugates is centre of pedal circle