Call $H$ orthocenter of $DEF$, then that point is $(DH) \cap \omega = T$ We can angle chase, note that let $DH \cap \omega$ at $K$ then $A, T, K$ are collinear

Let $D'$ be reflection of $D$ through $I$; $H$ be orthocenter of $\triangle DEF$; $Q$ $\equiv$ $HD'$ $\cap$ $(\omega)$ $(Q \not \equiv D')$ We have: $\angle{FQH} = \angle{FDD'} = 90^o - \angle{FD'D} = 90^o - \angle{DEF} = 90^o - \angle{BFD} = \angle{BXH}$ Then: $H$ $\in$ $(FXQ)$ So: $\angle{EQX} = \angle{EQH} + \angle{HQX} = \angle{EDD'} + \angle{HFB} = \dfrac{\angle{ACB}}{2} + \angle{AFH} = \dfrac{\angle{ACB}}{2} + \dfrac{\angle{ABC}}{2} + 90^o - \dfrac{\angle{BAC}}{2}$ $= 180^o - \angle{BAC}$ or $Q$ $\in$ $(AEX)$ Similarly: $Q$ $\in$ $(AEY)$