CGMO 2020: Prove that $X, P, Q, Y$ are concyclic.
Here's the hint.
($1$) The colored line are of importance. Think what the color means.
($2$) Make use of parallel line ratio.
This is a full proof following the natural wish in the OP to use the power of the point $C$ w.r.t. the points that should be on the circle.
The picture first (and try to figure out a property of the line $P'Q'$ without further reading):
Here, many elements are needed only for having a faithful picture. The points needed in the proof are the red ones:
$\color{red}Z$ is the intersection of the lines $EPQF$ and $AC$,
$\color{red}{P'}$ is $AB\cap CX$, and $\color{red}{Q'}=AD\cap CY$.
We compute $CX\cdot CP$, trying to express it in a "symmetricaly way" w.r.t. the given symmetry of the figure. First, since there is a right angle in $B$ in $\Delta BCP'$ we have $$ CB^2= CX\cdot CP'\ . $$ So it is natural to try to deal with the proportion $CP:CP'$ or with some derivated form of it.
A further hint so far:
Using for instance for the equality marked $(!)$ below the sine theorem in $\Delta AEZ$ and $\Delta AFZ$ we get: $$ \tag{$1$} \frac {PE}{QF}= \frac {AE}{AF}\overset{(!)}{=\!=} \frac {ZE}{ZF}= \frac {ZP}{ZQ}\ . $$
Lemma: $$ \tag{$2$} \color{red}{P'Q'}\|EF\ . $$ Proof: Menelaos in $\Delta EAZ$ for the "secant" line $CPP'$, respectively in $\Delta FAZ$ for the "secant" line $CQQ'$ gives: $$ \begin{aligned} 1&= \frac{PZ}{PE}\cdot \color{blue}{\frac{P'E}{P'A}}\cdot \frac{CA}{CZ} \ , \\ 1&= \frac{QZ}{QE}\cdot \color{blue}{\frac{Q'E}{Q'A}}\cdot \frac{CA}{CZ} \ , \end{aligned} $$ and the middle blue proportions are equal, since the others are correspondingly. (Use $(1)$.) Thus the claimed parallelism.
$\square$
The finish is now: $$ \begin{aligned} \frac {CX\cdot CP}{CY\cdot CQ} &= \frac {CX\cdot CP'}{CY\cdot CQ'}\qquad\text{ since }PQ\|P'Q' \\ &= \frac {CB^2}{CD^2} =1\ . \end{aligned} $$ $\square$
Note: The green region suggests that we are trying to "move proportions" from the line $CPP'$ to the line $CZA$ by using conveniently triangles "based" on the one or the other line.
Let $EF$ cut $AC$ at $R$. Then
- $AR$ is angle bisector for $\angle EAF$ so ${AE\over AF} = {ER\over RF}$ and thus $${EP\over PR} ={FQ\over QR}\;\;\;(*)$$
- Reflect $E,P$ and $X$ across $AC$, we get $E',P'$ and $X'$. Because of $(*)$ we have $E'F||P'Q$ and $Y,X',C,D$ are concyclic.
- Let $\angle CDX'= \phi$, then $\angle CYX' = \phi $ and $\angle X'DA = 90-\phi$, so $\angle QYX' = 180-\phi $ and $\angle X'P'Q = \phi$ which means $X',Y , Q$ and $P'$ are concyclic.
- By PoP with respect to point $C$ we see that $P,X,Y,Q$ are conyclic.
