MEMO 2022 I3

Let $ABCD$ be a parallelogram with $\angle DAB < 90$ Let $E$ be the point on the line $BC$ such that $AE = AB$ and let $F$ be the point on the line $CD$ such that $AF = AD$ . The circumcircle of the triangle $CEF$ intersects the line $AE$ again in $P$ and the line $AF$ again in $Q$ . Let $X$ be the reflection of $P$ over the line $DE$ and $Y$ the reflection of $Q$ over the line $BF$ . Prove that $A, X, Y$ lie on the same line.