TITLE: CGMO 2020: Prove that $X, P, Q, Y$ are concyclic.
QUESTION [6 upvotes]: In the quadrilateral $ABCD$, $AB=AD$, $CB=CD$, $\angle ABC =90^\circ$. $E$, $F$ are on $AB$, >$AD$ and $P$, $Q$ are on $EF$($P$ is between $E, Q$), satisfy $\frac{AE}{EP}=\frac{AF}{FQ}$. $X, Y$ are on $CP, CQ$ that satisfy $BX \perp CP, DY \perp CQ$. Prove that $X, P, Q, Y$ are concyclic.

My Progress: Couldn't proceed much . I noted that $ABCD$ is cyclic quad with diameter $AC$ .  I feel to use POP on C , so it is enough to show that $CX\cdot CP= CY\cdot CQ$ . But I am not sure about how to use "$\frac{AE}{EP}=\frac{AF}{FQ}$" criteria .
Please post hints rather than solution. It really helps me a lot.
Thanks in advance.

REPLY [4 votes]: 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.