Question 79
Let $ABCD$ be a parallelogram with $\angle DAB \lt 90^{\circ}$. Let $E \neq B$ be the point on the line $BC$ such that $AE = AB$ and let $F \neq D$ 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 $A, X$ and $Y$ lie on the same line.