Let ABCDABCD be a parallelogram with DAB<90\angle DAB < 90^{\circ}. Let EBE \neq B be the point on the line BCBC such that AE=ABAE = AB and let FDF \neq D be the point on the line CDCD such that AF=ADAF = AD. The circumcircle of the triangle CEFCEF intersects the line AEAE again in PP and the line AFAF again in QQ. Let XX be the reflection of PP over the line DEDE and YY the reflection of QQ over the line BFBF. Prove that A,XA, X and YY lie on the same line.