Collinearity in a Mixtilinear Configuration

Consider a triangle $\triangle{ABC}$, its Incircle, $\psi$, and its $B$-mixtilinear incircle, $\omega$. Let $T$ be the point of tangency of $\psi$ and $\omega$. Let $K$ and $L$ be the points of tangency of $\omega$ with the sides $AB$ and $BC$, respectively. Call $D$ the point of tangency of $\psi$ with $AC$. Call $E$ the  second intersection (further to $B$) of $\psi$ with $BT$. Call $F$ the second intersection of the circumcircle of $\triangle{EDI}$ with $BT$. Prove that $K$, $F$ and $I$ are collinear. 

Proof. $\angle{ATI}=\angle{CTI}$ and $\angle{ATB}=\angle{CTD}$ (well-known). We have $\angle{BTD}=\angle{ATC}-\angle{ATB}-\angle{CTD}$. Notice that $\angle{ATB}=\angle{CTD}=\angle{ACB}$ and $\angle{ATC}=180^\circ-\angle{ABC}$, then, $\angle{BTD}=\angle{BAC}-\angle{ACB}$. Moreover, $TI$ is an angle bisector of $\angle{BTD}$, therefore, $D$ is the reflection of $E$ around $TI$, which means $TE=TD$. Said this, and having in mind that $EDIF$ is cyclic, we find $\angle{DIF}=\frac{\angle{ABC}}{2}+\angle{ACB}$.

On the other hand, $K$, $I$ and $L$ are collinear (well-known), consequently, $\angle{AKL}=\angle{AKI}=90^\circ+\frac{\angle{ABC}}{2}$. Finally, some angle chase in quadrilateral $AKID$ give us $\angle{DIK}= \frac{\angle{ABC}}{2}+\angle{ACB}$. As $\angle{DIF}=\angle{DIK}$, we are done.