Question 145
In a parallelogram $ABCD$, angle $ABC$ is obtuse. Let $\omega$ be the circumcircle of triangle $ABC$. Line $AD$ intersects $\omega$ at points $A$ and $E$, and line $CD$ intersects $\omega$ at points $C$ and $F$. Prove that the circumcenter of triangle $DEF$ lies on $\omega$. (T. Emelyanova)