Here is an approach with complex numbers. Let $X$ and $Y$ be the centers of the circumcircles of $ABPQ$ and $CDPQ$, so $XAB$ and $YCD$ are congruent isosceles triangles. The problem wants us to show that there exists a point $Z$ on the radical axis of these two circles over all choices of $X,Y$. Set this in the complex plane with $z=0$ and $\frac{x-b}{x-a}=\frac{y-d}{y-c}=\omega$ for some $|\omega|=1$. The power from $Z$ to the circle centered at $X$ passing through $A$ is $XZ^2-AX^2=x\overline x-(x-a)(\overline x-\overline a)=x\overline a+a\overline x-a\overline a$. Substituting $x=\frac{b-a\omega}{1-\omega}$ gives that this is $\frac{\overline ab-a\overline a\omega}{1-\omega}+\frac{a\overline b\omega-a\overline a}{\omega-1}-a\overline a=\frac{\overline ab-a\overline b\omega}{1-\omega}$. We want this to be equal to $\frac{\overline cd-c\overline d\omega}{1-\omega}$ over all choices of $\omega$, so we need $\overline ab=\overline cd$. Conversely, as long as this condition is satisfied, then the radical axis will pass through $z=0$. Now given generic $a,b,c,d$ we want to show the existence of $z$ such that $(\overline a-\overline z)(b-z)=(\overline c-\overline z)(d-z)$. This rearranges to $\overline ab-\overline cd=(\overline a-\overline c)z+(b-d)\overline z$. It is not too hard to see that this has a solution (potentially at infinity), so we are done.

proposed by Sergey Berlov, Russia it was also RMM Shortlist 2017 G3

Any synthetic solution?

also true for $$\angle AQB=180-\angle DQC$$and$$\angle APB=180-\angle DPC$$

ABCDE wrote: so $XAB$ and $YCD$ are congruent isosceles triangles The triangles are similar, but not necessarily congruent.

Is there any solution without using complex number?

bump.... can someone please post a synthetic solution

Here's a proof using Casey's theorem: we will prove a somewhat generalized version, let $M,N$ be the midpoints of $AB,CD$ and $O_1,O_2$ be points on the perpendicular bisector of $AB,CD$ such that $\frac{AB}{MO_1}=\frac{CD}{NO_2}$ and are on different sides of their respective segments. then the radical axis of circles with center $M,N$ and radius $MO_1,NO_2$ pass through a fixed point.(note that if the circles intersect, they give us $P,Q$) proof: $AB\cap CD=E$ and let $l$ be a line that the ratio of distances of all points on it from $AB,CD$ be $\frac{CD}{AB}$ let the radical axis of $w_1,w_2$ intersect the line $l$ at $X$. we prove that $X$ is on the radical axis of all such configurations. let $S,T$ be the centers of circles $w'_1,w'_2$. we know that by similarity of triangles, $\frac{SO_1}{TO_2}=\frac{AB}{CD}$. now using Casey's theorem on $w_1,w'_1$ and $w_2,w'_2$ on point $X$ we conclude that: $$P_{w_1}^{X}-P_{w'_1}^{X}=2h\cdot SO_1, P_{w_2}^{X}-P_{w'_2}^{X}=2g\cdot TO_2$$where $h,g$ are the distances of $X$ from $AB,CD$. we know that $\frac{SO_1}{TO_2}=\frac{AB}{CD}=\frac{g}{h}$ and $P_{w_1}^{X}=P_{w_2}^{X}$ thus $P_{w'_1}^{X}=P_{w'_2}^{X}$ which gives us that $X$ lies on the given radical axis.

Solved with Max Lu, Eric Shen, and Derek Liu. Obviously $P, Q$ are concyclic with $(ABPQ)$ and $(CDPQ)$. Let the two centers of these circles be $O_1, O_2$, and let $AB = a, CD = b$, and the distance from $O_1$ to $AB$ and $O_2$ to $CD$ be $a\lambda$ and $b\lambda$ respectively. Now, for two different values of $\lambda$, let the radical axises of these two circles intersect at $R$. I claim $R$ has the same power over all $\lambda$. The power of $R$ with respect to $(ABPQ)$ minus the power of $R$ with respect to $(CDPQ)$ for any $\lambda$ is \[h_1^2 - h_2^2 + (x-a\lambda)^2 - (y-b\lambda)^2 - (\frac14 + \lambda^2)(a^2 - b^2)\]where $h_1, h_2$ are the distance from $R$ to the perp bisectors of $AB,CD$ respectively, and $x, y$ are the distance from the foot of $R$ to the perp bisectors to the midpoints of $AB, CD$, respectively. Now, this is a linear equation, and it is true for two distinct values of $\lambda$, so it is true for all $\lambda$. We conclude $R$ lies on the radical axis for all $\lambda$.

see the following link

Here's another solution using the properties of $\gamma$ in post #71 . Note that the line $PQ$ is in fact the radical axis of $(APB),(CPD)$.It's well-known that the locus of points $P$ st $\angle APB = \angle CPD$ is a cubic curve passing through $A,B,C,D,AB\cap CD$.Now consider two points $P,Q$ satisfying the condition and let that curve be $\Gamma$. Let $PQ \cap \Gamma = S$ and consider the curve $\gamma$ in post #71 for $S$ ; which is the locus of points $R$ st $S$ is on the radical axis of $(ARB),(CRD)$. Since it's a cubical curve(again by the property 2 of post #71 ) ,since there's at most one cubical curve passing through 7 points , $\Gamma = \gamma$ and the point $S$ is the desired point by the definition of $\gamma$ and we're done.

Let $AC \cap BD = E$ and let $\odot (APB)$ meet $BD, AC$ again at $R,S$, $\odot (CQD)$ meet $AC, BD$ again at $T,U$. It suffices to show that there is a fixed point $X$ on the radical axis of $\odot (APB)$ and $\odot (CQD)$. Let $f(X) = \text{Pow}(X,\odot(CQD))-\text{Pow}(X,\odot(APB))$, so by generalized linearity of PoP we only need to find constants $u,v,w$ (not all equal to zero) such that $uf(E)+vf(A)+wf(B)=0$. Let $AE=a$, $BE=b$, $CE=c$, $DE=d$, $ER=x$ (in $\overrightarrow{EB}$ direction). Plugging $ES=\frac{bx}{a}$, $ET=\frac{dx}{a}$, $EU=\frac{cx}{a}$ in the equation above, we get $$uf(E)+vf(A)+wf(B)=[u(\frac{cd}{a}-b)+v(d+\frac{cd}{a})+w(\frac{bc}{a}+\frac{cd}{a})]x+v(a^2+ac)+w(b^2+bd)=0.$$This is equivalent to the following system of equations (in terms of $u,v$ and $w$) \[\left\{\begin{aligned}&u(\frac{cd}{a}-b)+v(d+\frac{cd}{a})+w(\frac{bc}{a}+\frac{cd}{a})=0,\\&v(a^2+ac)+w(b^2+bd)=0.\end{aligned}\right.\]which clearly has a solution $(u,v,w)$ where $u,v,w$ are not all zero.$\blacksquare$ (Note: $u+v+w=0$ corresponds to the points at infinity, in which case the lines $PQ$ are all parallel.)

Cayley-Bacharach on \[\begin{bmatrix} P & Q & R \\ A & B & M \\ I & J & X \end{bmatrix}\]where $R$ is the fixed point, $X = AD \cap BC$, $M$ is the Miquel point of quadrilateral $ABCD$ and $I$, $J$ are the circle points.