Let $\omega$ touch $CD,BA$ at $G,H.$ If $AB,CD,EF$ concur, $EGFH$ is harmonic $\Longrightarrow$ $AD,BC,GH$ concur at $K$ and by Newton's theorem $AC,BD,EF,GH$ concur at $P.$ If $EE_1$ cuts $AD$ at $U,$ we have $UA \cdot UD=UE \cdot UE_1=UF^2,$ but from the complete $ABCD,$ we have $(A,D,F,K)=-1$ $\Longrightarrow$ $U$ is midpoint of $KF$ and similarly $FF_1$ goes through the midpoint $V$ of $KE.$ Now, by the symmetry of the K-isosceles $\triangle KEF,$ we obtain $EF \parallel E_1F_1.$

My solution: Let $ I $ be the incenter of $ ABCD $ . Let $ X, Y $ be the tangent point of $ (I) $ with $ AB, CD $, respectively . Let $ P, Q, R, S $ be the midpoint of $ EX, EY, FX, FY $, respectively . Let $ T=AD \cap BC \cap XY, U=PQ \cap BC, V=RS \cap AD $ . Since $ EP \cdot PX=BP \cdot PI, EQ \cdot QY=CQ \cdot QI $ , so we get $ PQ $ is the radical axis of $ \{ (I), (BCI) \} $ , hence the midpoint $ U $ of $ TE $ is the radical center of $ \{ (I), (BCI), (BCF) \} $ , so $ U $ lie on the radical axis $ FF_1 $ of $ \{ (I), (BCF) \} $ . Similarly, we can prove the midpoint $ V $ of $ TF $ lie on $ EE_1 $ . Since $ \{E, F \}, \{U, V \} $ are symmetry WRT $ TI $ , so we get $ \{ E_1\equiv EV \cap (I) , F_1 \equiv FU \cap (I) \} $ are symmetry WRT $ TI $ . ie. $ EF \parallel E_1F_1 $ Q.E.D